In the 1840s, in Geneva, the Swiss physicist Daniel Colladon set up a demonstration that looked more like a magic trick than an experiment. He filled a tank with water and let it drain through a small hole in the side, so that a smooth, curving jet of water arched out into the darkened room. Then he sent a beam of light into the tank, aimed straight down the axis of the jet.
The audience expected the light to shoot out in a straight line and hit the far wall, because light, as everyone knew, travels in straight lines. Instead, the light stayed inside the water. It followed the jet as it bent and fell, and the glowing arc of water landed in a shining pool on the floor. The light had been carried along a curved path, without any mirror bending it at each point.
Nothing mysterious was happening. Light was simply striking the inner surface of the water jet at a large angle, again and again, and each time it was being thrown back completely into the water instead of escaping into the air. That single effect, understood properly, is the reason a modern optical fibre can carry your phone call across a continent with almost no loss.
Q1. Why does light stay trapped inside a curving jet of water or inside a thin glass fibre, instead of leaking out through its sides? Q2. Why does a pencil kept slanting in a glass of water look broken, and why does the bottom of a pond look shallower than it really is? Q3. Why are the mirrors used in a search light, a torch or a car headlight made parabolic instead of spherical? Q4. On a hot day, why does a shining “pool of water” appear far ahead on a dry road and disappear when we reach that spot? Q5. In a rainbow, why is red always the outer arch and violet the inner one, and why do we see rainbows only in the morning or the evening? Q6. Why can a single magnifying glass never make letters look more than a few times bigger, and how do microscopes and telescopes get around this limit?
Each of these questions is answered somewhere in this chapter. Keep them in mind as you read, and check your answers against the Curiosity Corner boxes that appear along the way.
9.1 Introduction
“See it to believe it” is a popular saying. But seeing anything at all requires one thing first: light. Without light entering our eyes, no object around us would be visible, no matter how close it is.
This raises two questions that run through the whole chapter. What exactly is light? And how does light let us see the objects around us? We begin answering both here, and the answers are completed in higher classes.
You already know that the science of sound is called acoustics. In the same way, the science of light is called optics. Both sound and light travel as waves, but they are not the same kind of wave — sound needs a material medium such as air or water to travel through, while light does not. You have studied the nature of sound waves earlier, and the wave nature of light will be taken up in a later chapter.
Before starting this chapter, recall from earlier classes: (1) What are the laws of reflection and refraction? (2) What is dispersion of light? (3) What is refractive index? (4) What is total internal reflection? (5) How does light refract at a curved surface? (6) How does a rainbow form?
9.2 Nature of Light
How our understanding of light has changed
For a long time, light was described simply by what it does to us: it is that form of radiant energy which makes objects visible by stimulating the retina of the eye. This is a useful description, but it does not tell us what light actually is.
The next step was to recognise light as a form of energy that can travel through a material medium and also through empty space where there is no medium at all. Something that carries energy from place to place in this way is called a wave.
At the beginning of the twentieth century it was established that these waves are electromagnetic (EM) waves — waves made of varying electric and magnetic fields, needing no material medium to travel through. This is why sunlight reaches us across the vacuum of space, while sound from the Sun never can.
Later, quantum theory established the particle nature of light. According to this picture, light is carried by particles called photons, which act as energy carriers. Experiments performed with a countable number of photons have made it an established fact that light possesses a dual nature.
In simple words, we can say that light consists of energy-carrying photons that are guided by the rules of EM waves. Neither picture alone is complete; which one we use depends on the phenomenon we are trying to explain.
In the 1860s James Clerk Maxwell showed mathematically that electric and magnetic fields can travel together as a wave, and that this wave should move at almost exactly the measured speed of light — strong evidence that light itself is an EM wave. About two decades later, Heinrich Hertz produced and detected such waves in his laboratory, confirming Maxwell’s prediction experimentally.
The speed of light
In vacuum, these waves (or photons) travel with a speed
\[c = 299792458 \ \text{m s}^{-1}\]
According to Einstein’s special theory of relativity, this is the maximum possible speed for any object. For practical purposes, in numerical problems we write it as
\[c = 3 \times 10^{8} \ \text{m s}^{-1}\]
No object or signal can travel faster than the speed of light in vacuum, \(c = 3 \times 10^{8} \ \text{m s}^{-1}\). This is a result of Einstein’s special theory of relativity.
In a material medium the speed of EM waves is smaller, and is given by
\[c = \sqrt{\frac{1}{\epsilon \mu}}\]
where the permittivity \(\epsilon\) and the permeability \(\mu\) are constants that depend on the electric and magnetic properties of that medium. In other words, the medium itself decides how fast light can move through it, and it does so through its electrical and magnetic character.
Once the speed of light in a medium is known, we compare it with the speed in vacuum. The ratio
\[n = \frac{c}{v}\]
is called the absolute refractive index of that medium. Here \(c\) is the speed of light in vacuum and \(v\) is the speed of light in the medium, both in m s\(^{-1}\). Since \(n\) is a ratio of two speeds, it has no unit.
The absolute refractive index \(n\) of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium. It is a property of the medium.
The speed of light in a certain transparent liquid is measured to be \(2.25 \times 10^{8} \ \text{m s}^{-1}\). Calculate the absolute refractive index of the liquid. State the unit of your answer.
Three ways of studying light
Commonly observed phenomena concerning light can be broadly split into three categories, depending on which picture of light is needed to explain them.
(I) Ray optics or geometrical optics: A particular direction of propagation of energy from a source of light is called a ray of light. We use ray optics for understanding phenomena like reflection, refraction, double refraction, total internal reflection, and so on. Almost all of this chapter belongs to this category.
A ray of light is a particular direction along which energy propagates from a source of light.
(II) Wave optics or physical optics: For explaining phenomena like interference, diffraction, polarization and the Doppler effect, we must treat light energy as being in the form of EM waves. Wave theory is discussed in detail in a later chapter.
(III) Particle nature of light: Phenomena like the photoelectric effect, emission of spectral lines and the Compton effect cannot be explained using classical wave theory. These involve the interaction of light with matter. For such phenomena we have to use the quantum nature of light, which is also taken up later.
9.3 Ray Optics or Geometrical Optics
In geometrical optics we mainly study image formation by mirrors, lenses and prisms. Instead of worrying about what light is made of, we simply follow the directions along which its energy travels, and ask where those directions finally meet.
The whole of this study rests on four fundamental laws or principles, all of which you have already met in earlier classes. Everything derived later in this chapter — mirror formula, lens formula, prism formula — is built on these four statements, so they are worth stating carefully.
The four fundamental laws
(i) Rectilinear propagation. Light travels in a straight line in a homogeneous and isotropic medium. Here homogeneous means that the properties of the medium are the same everywhere in the medium, and isotropic means that the properties are the same in all directions. If either condition fails — for example, if the density of air changes gradually with height — light will no longer travel in a straight line, and we shall use exactly this fact later while explaining a mirage.
In a homogeneous and isotropic medium, light travels along a straight line.
(ii) Independence of rays. Two or more rays can intersect at a point without affecting their paths beyond that point. Each ray continues as though the other had never been there. This is what allows us to draw two or three convenient rays from an object and take their meeting point as the image, without asking whether the rays disturb one another.
(iii) Laws of reflection:
The reflected ray lies in the plane formed by the incident ray and the normal drawn at the point of incidence, and the two rays lie on either side of the normal.
The angles of incidence and reflection are equal.
Laws of reflection: (a) the incident ray, the reflected ray and the normal at the point of incidence all lie in one plane, with the incident and reflected rays on opposite sides of the normal; (b) the angle of incidence is equal to the angle of reflection.
(iv) Laws of refraction: These apply at the boundary between two media.
The refracted ray lies in the plane formed by the incident ray and the normal drawn at the point of incidence, and the two rays lie on either side of the normal.
The angle of incidence \(\theta_1\) in a medium of refractive index \(n_1\) and the angle of refraction \(\theta_2\) in a medium of refractive index \(n_2\) are related by Snell’s law,
\[(n_1)\sin\theta_1 = (n_2)\sin\theta_2\]
Here \(\theta_1\) and \(\theta_2\) are measured from the normal, and \(n_1\), \(n_2\) are the (unitless) refractive indices of the first and the second medium. Snell’s law tells us not only that light bends at a boundary, but exactly by how much — and note that the product \(n\sin\theta\) has the same value on both sides of the boundary. That single fact will later give us total internal reflection, the prism formula and the rainbow.
Laws of refraction (at the boundary between two media): (a) the incident ray, the refracted ray and the normal at the point of incidence lie in one plane, with the incident and refracted rays on opposite sides of the normal; (b) \((n_1)\sin\theta_1 = (n_2)\sin\theta_2\).
Interestingly, all four laws stated above can be derived from a single principle called Fermat’s (pronounced “Ferma”) principle. It says that “while travelling from one point to another by one or more reflections or refractions, a ray of light always chooses the path of least time.” Ideally it is the path of extreme time, that is, the path of minimum or maximum time. You are strongly recommended to go through a suitable reference book that will give you the proof of \(i = r\) during reflection, and of Snell’s law during refraction, using Fermat’s principle.
Why the “least time” idea is powerful
It is worth pausing on what Fermat’s principle really claims. Light does not bend at a glass surface because it “wants” to; it bends because, given that it moves slower inside glass, the bent path is the one that gets it from the starting point to the finishing point in the shortest time. A straight path through the glass would spend too long in the slow medium.
This also explains why the two sets of laws — reflection and refraction — look so different yet come from one source. Reflection involves a single medium, so least time simply means least distance, and that forces \(i = r\). Refraction involves two different speeds, and least time then forces exactly the sine relation that Snell wrote down.
The glass of a spectacle lens is 2 mm thick and the refractive index of its glass is 1.5. Calculate the time taken by light to cross this thickness. Express your answer with the most convenient prefix attached to the unit ‘second’.
Solution: We first need the speed of light inside the glass, because light slows down there. From the definition of refractive index, \(n = c/v\), so \(v = c/n\).
Speed of light in vacuum, \(c = 3 \times 10^{8} \ \text{m/s}\)
\(n_{glass} = 1.5\)
\(\therefore\) Speed of light in glass
\[v_{glass} = \frac{c}{n_{glass}} = \frac{3 \times 10^{8}}{1.5} = 2 \times 10^{8} \ \text{m/s}\]
Distance to be travelled by light in glass,
\[s = 2 \ \text{mm} = 2 \times 10^{-3} \ \text{m}\]
Now we simply use time = distance / speed, with the speed inside the glass — not the speed in vacuum. Using the vacuum value here is the commonest mistake in such problems.
\(\therefore\) Time \(t\) required by light to travel this distance,
\[t = \frac{s}{v_{glass}} = \frac{2 \times 10^{-3}}{2 \times 10^{8}} = 10^{-11} \ \text{s}\]
The most convenient prefix to express this small time is pico (p) \(= 10^{-12}\)
\[\therefore t = 10 \times 10^{-12} = 10 \ \text{ps}\]
A ray of light travelling in air strikes the flat surface of a glass slab of refractive index 1.5 at an angle of incidence of \(30^\circ\). Using Snell’s law, calculate the angle of refraction inside the glass. Also state whether the ray has bent towards or away from the normal, and justify your answer using the speed of light in the two media.
9.3.1 Cartesian Sign Convention
While using geometrical optics it is necessary to use some sign convention. The reason is simple. If we work only with numerical values of \(u\), \(v\) and \(f\), the relation between them for a spherical mirror (or for a lens) would come out different for different positions of the object and for different types of mirror. We would then need to memorise a separate formula for every case.
Here \(u\) and \(v\) are the distances of the object and the image respectively from the optical centre, and \(f\) is the focal length. A properly used sign convention enables us to use the same formula for all the different particular cases. Thus, while deriving a formula and also while using that formula, it is necessary to use the same sign convention throughout.
The most convenient sign convention is the Cartesian sign convention, because it is exactly analogous to coordinate geometry, which you already know. As shown in Fig. 9.2, we simply place a pair of coordinate axes on the optical device and read off signs the way we read them on a graph.
The rules of the convention
According to this sign convention (Fig. 9.2):
i) All distances are measured from the optical centre or pole. For most of the optical objects such as spherical mirrors, thin lenses, etc., the optical centre coincides with their geometrical centre.
ii) Figures should be drawn in such a way that the incident rays travel from left to right.
A diverging beam of incident rays corresponds to a real point object, as shown in Fig. 9.3 (a). Such a beam is actually spreading out from the object, which is therefore a genuine source of light lying to the left of the pole.
A converging beam of incident rays corresponds to a virtual object, as shown in Fig. 9.3 (b). Here the incident rays are already heading towards a point on the far side of the device; they would have met there if the device were not in the way. That meeting point is treated as the object even though no real source sits there.
A parallel beam corresponds to an object at infinity. Thus, a real object should be shown to the left of the pole (Fig. 9.3 (a)), and a virtual object or image to the right of the pole (Fig. 9.3 (b)).
iii) The \(x\)-axis can be conveniently chosen as the principal axis, with the origin at the pole.
iv) Distances to the left of the pole are negative and those to the right of the pole are positive.
v) Distances above the principal axis (\(x\)-axis) are positive, while those below it are negative.
Unless specially mentioned, we shall always consider objects to be real for further discussion. So in almost every problem you solve, the object distance \(u\) will carry a negative sign.
Cartesian sign convention: all distances are measured from the pole (taken as the origin), with the incident light travelling from left to right; distances measured to the right of the pole and above the principal axis are positive, while distances to the left of the pole and below the principal axis are negative.
Why a virtual object is not a strange idea
Students often accept a virtual image easily but find a virtual object puzzling. The distinction is decided entirely by the incident rays reaching the device.
If the rays arriving at the mirror or lens are diverging, they must have come from a real point, and that point is a real object. If the rays arriving are already converging, they were bent that way by some earlier mirror or lens, and the point towards which they were heading acts as the object for the present surface. Since that point lies to the right of the pole, its distance is positive by rule (iv). This situation occurs constantly in instruments such as microscopes and telescopes, where the image formed by one lens becomes the object for the next.
A converging beam of light is incident on a lens and, in the absence of the lens, the rays would have met at a point 12 cm behind it. State the sign and value of the object distance \(u\) to be used for this lens, and state whether the object is real or virtual.
9.4 Reflection
When light travelling in one medium falls on a surface and returns into the same medium, we say the light has been reflected. The laws that govern this return were stated in the previous section: the incident ray, the reflected ray and the normal lie in one plane, and the angle of incidence equals the angle of reflection.
Those two laws hold at every point of every reflecting surface, whether the surface is flat or curved. What changes from one surface to another is only the direction of the normal at each point. On a plane mirror the normal points the same way everywhere, so a parallel beam stays parallel after reflection. On a curved mirror the normal swings around as we move along the surface, so different parts of the beam are turned through different angles — and that is precisely what allows a curved mirror to bring light to a focus.
For this reason we study reflection in two steps. In the next subsection we look at reflection from a plane surface, including the images formed by two inclined mirrors. After that we take up reflection from curved (spherical) mirrors, where the mirror formula, magnification and image defects are developed.
9.4.1 Reflection from a Plane Surface
a) If the object is in front of a plane reflecting surface, the image formed is virtual and laterally inverted. It is of the same size as that of the object, and at the same distance as that of the object, but on the other side of the reflecting surface.
The word virtual here means that the reflected rays only appear to come from the image position; they never actually pass through it. That is why such an image can be seen by the eye but cannot be caught on a screen kept behind the mirror. Notice in Fig. 9.3 (c) that the reflected rays are diverging, and it is their backward extensions that meet behind the mirror.
b) If we are standing on the bank of a still water body and look for our image formed by the water (or if we are standing on a plane mirror and look for our image formed by the mirror), the image is laterally reversed, of the same size, and on the other side.
c) If an object is kept between two plane mirrors inclined at an angle \(\theta\) (as in a kaleidoscope), a number of images are formed due to multiple reflections from both the mirrors. Each mirror forms an image of the object, then each of those images acts as an object for the other mirror, and so the images multiply.
The exact number of images depends upon the angle between the mirrors and also upon where exactly the object is kept between them, as shown in Fig. 9.3 (d). It can be obtained as follows (Table 9.1).
9.4.2 Reflection from Curved Mirrors
In order to focus a parallel or a divergent beam by reflection, we need curved mirrors. A plane mirror simply turns a beam around without changing how spread out it is, so it can never gather light to a point.
You might have noticed that the reflecting mirrors used behind a torch bulb or in the headlights of a vehicle, and the rear-view mirrors of vehicles, are not plane but concave or convex. Mirrors used for a search light are parabolic. In this book we shall restrict ourselves to spherical mirrors only, which can be studied using simple mathematics.
Such mirrors are parts of a sphere polished from outside (giving a convex mirror) or from inside (giving a concave mirror). So the reflecting surface of a concave mirror bulges away from the incoming light, while that of a convex mirror bulges towards it.
Radius of curvature and focal length
The radius of the sphere of which a mirror is a part is called the radius of curvature (\(R\)) of the mirror. Only for spherical mirrors, half of the radius of curvature is the focal length of the mirror,
\[f = \frac{R}{2}\]
where \(f\) and \(R\) are both measured in metres (or centimetres, if the data are in centimetres).
For a concave mirror, the focal length is the distance at which parallel incident rays converge after reflection. For a convex mirror, it is the distance from where parallel rays appear to be diverging after reflection — the rays themselves never actually pass through that point.
The radius of curvature \(R\) of a spherical mirror is the radius of the sphere of which the mirror is a part. Its focal length is \(f = R/2\).
Now we must attach signs. According to the sign convention, the incident rays travel from left to right, and they should face the polished surface of the mirror. With the mirror placed accordingly, a convex mirror sends the reflected rays apart so that they appear to come from a focus lying behind the mirror, that is, on the positive side of the pole. Thus the focal length of a convex mirror is positive, as seen in Fig. 9.4 (a).
For a concave mirror placed the same way, the reflected rays actually cross in front of the mirror, that is, on the negative side of the pole. Thus the focal length of a concave mirror is negative, as seen in Fig. 9.4 (b).
Relation between f, u and v
For a point object, or for a small finite object, the focal length of a small spherical (concave or convex) mirror is related to the object distance and the image distance as
\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \qquad \text{--- (9.1)}\]
Here \(u\) is the object distance and \(v\) the image distance, both measured from the pole with proper signs, and \(f\) is the focal length. All three are lengths, measured in metres in SI units.
By a small mirror we mean that its aperture (diameter) is much smaller — at least one tenth — than the values of \(u\), \(v\) and \(f\). If this condition is not satisfied, Eq. (9.1) begins to fail, for reasons that will become clear when we discuss spherical aberration later in this section.
Focal power
Converging or diverging ability of a lens or of a mirror is defined as its focal power. It is measured as
\[P = \frac{1}{f}\]
In SI units it is measured in dioptre, where \(f\) must be substituted in metre.
\[\therefore 1 \ dioptre \ (D) = 1 \ m^{-1}\]
Focal power of a mirror or lens is its converging or diverging ability, given by \(P = 1/f\), with \(f\) in metre. Its SI unit is the dioptre (D), where \(1 \ D = 1 \ m^{-1}\).
The physical sense of this definition is worth noting. A mirror with a short focal length bends the beam sharply and brings it to a focus quickly, so it is optically “powerful”. A mirror with a long focal length bends the beam gently. Since power grows as focal length shrinks, taking \(P = 1/f\) is the natural choice.
Lateral magnification
The ratio of the linear size of an image to that of the object, measured perpendicular to the principal axis, is defined as the lateral magnification,
\[m = \frac{v}{u}\]
Being a ratio of two lengths, \(m\) has no unit. Its sign tells us the orientation of the image (erect or inverted), and its magnitude tells us whether the image is enlarged or diminished.
Lateral magnification is the ratio of the linear size of the image to the linear size of the object, measured perpendicular to the principal axis: \(m = v/u\).
For any position of the object, a convex mirror always forms a virtual, erect and diminished image, with \(m < 1\). In the case of a concave mirror, the nature of the image depends upon the position of the object. The following Table 9.2 will help you refresh your knowledge.
Table 9.2
| Position of object |
Position of image |
Real (R) or Virtual (V) |
Lateral magnification |
| \(u = \infty\) |
\(v = f\) |
R |
\(m = 0\) |
| \(u > 2f\) |
\(2f > v > f\) |
R |
\(m < 1\) |
| \(u = 2f\) |
\(v = 2f\) |
R |
\(m = 1\) |
| \(2f > u > f\) |
\(v > 2f\) |
R |
\(m > 1\) |
| \(u = f\) |
\(v = \infty\) |
R |
\(m = \infty\) |
| \(u < f\) |
\(\lvert v \rvert > \lvert u \rvert\) |
V |
\(m > 1\) |
The rear-view mirror of a car or a scooter is convex for exactly the reason listed above: whatever the distance of the vehicle behind, the image is always erect and diminished. A diminished image means a wider field of view is squeezed into the same small mirror, so the rider can see more of the road behind. The price paid is that the vehicles look smaller, and therefore farther away, than they really are — which is why such mirrors carry a warning that objects are closer than they appear.
A thin pencil of length 20 cm is kept along the principal axis of a concave mirror of curvature 30 cm. The nearest end of the pencil is 20 cm from the pole of the mirror. What will be the size of the image of the pencil?
Solution: Since the pencil lies along the axis, its two ends are at two different object distances. So we locate the image of each end separately and take the distance between those two image points as the length of the image.
\(R = 30\) cm
\[f = R/2 = -15 \ \text{cm} \ \ldots \text{(Concave mirror)}\]
\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
For the nearest end, \(u = u_1 = -20\) cm. Let the image distance be \(v_1\)
\[\therefore \frac{1}{-15} = \frac{1}{v_1} + \frac{1}{-20} \ \therefore v_1 = -60 \ \text{cm}\]
The nearest end is at 20 cm and the pencil itself is 20 cm long. Hence the farthest end is at \(20 + 20 = 40\) cm \(= -u_2\)
Let the image distance be \(v_2\)
\[\therefore \frac{1}{-15} = \frac{1}{v_2} + \frac{1}{-40} \ \therefore v_2 = -24 \ \text{cm}\]
\[\therefore \text{Length of the image} = 60 - 24 = 36 \ \text{cm}\]
Notice that the image (36 cm) is longer than the object (20 cm), even though the pencil is a single rigid object. This happens because the two ends are magnified by different amounts — a point worth remembering whenever an object is placed along the axis rather than perpendicular to it.
Defects or aberration of images
The theory of image formation by mirrors or lenses, and the formulae that we have used such as
\[f = \frac{R}{2} \quad \text{or} \quad \frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]
are based on the following assumptions:
Objects and images are situated close to the principal axis.
Rays diverging from the objects are confined to a cone of very small angle.
If there is a parallel beam of rays, it is paraxial, that is, parallel and close to the principal axis.
However, in reality these assumptions do not always hold good. Real objects have finite size, and real mirrors have finite aperture, so some rays are bound to travel far from the axis. This results in a distorted or defective image.
Commonly occurring defects are spherical aberration, coma, astigmatism, curvature and distortion. Except spherical aberration, all the others arise due to beams of rays inclined to the principal axis. These are not discussed here.
Spherical aberration: As mentioned earlier, the relation
\[f = \frac{R}{2}\]
giving a single point focus is applicable only for small aperture spherical mirrors and for paraxial rays. In reality, when the rays are farther from the principal axis, the focus gradually shifts towards the pole, as shown in Fig. 9.5. This phenomenon (defect) arises due to the spherical shape of the reflecting surface, hence it is called spherical aberration. It results in an unsharp (fuzzy) image with unclear boundaries.
Spherical aberration is the image defect in which rays striking a spherical mirror at different distances from the principal axis are focused at different points, so that no single sharp point image is obtained.
The distance between \(F_M\) and \(F_P\) (Fig. 9.5) is measured as the longitudinal spherical aberration. If there is no spherical aberration, we get a single point image on a screen placed perpendicular to the principal axis at that location, for a beam of incident rays parallel to the axis.
In the presence of spherical aberration, no such point is possible at any position of the screen, and the image is always a circle. At one particular location of the screen, the diameter of this circle is minimum. This is called the circle of least confusion. In the figure it is across AB. The radius of this circle is the transverse spherical aberration.
In the case of curved mirrors, this defect can be completely eliminated by using a parabolic mirror. Hence the surfaces of mirrors used in a search light, torch, headlight of a car, telescopes, etc., are parabolic and not spherical.
Q3. Why are the mirrors used in a search light, a torch or a car headlight made parabolic instead of spherical?
A. A spherical mirror suffers from spherical aberration: rays far from the principal axis are focused closer to the pole than paraxial rays, so there is no single sharp focus. A parabolic mirror has no such defect — every ray of a parallel beam reaches its focus after travelling exactly the same optical path, so all of them meet at one point. Run in reverse, this means a bulb kept at the focus of a parabolic mirror sends out a truly parallel, intense beam, which is exactly what a search light or a headlight needs.
Why does a parabolic mirror not have spherical aberration? A parabola is a geometrical shape drawn in such a way that every point on it is equidistant from a straight line and from a point. Figure 9.6 shows a parabola. Points A, B, C, … on it are equidistant from the line RS (called the directrix) and the point F (called the focus). Hence A′A = AF, B′B = BF, C′C = CF, and so on. If rays of equal optical path converge at a point, that point is the location of the real image corresponding to that beam of rays. Paths A″AA′, B″BB′, C″CC′, etc., are equal paths in the absence of the mirror. If the parabola ABC… is a mirror, then the respective optical paths will be A″AF, B″BF, C″CF, … and from the definition of a parabola these are also equal. Thus, F is the single point focus for the entire beam parallel to the axis, with NO spherical aberration.
A convex mirror used as a rear-view mirror has a radius of curvature of 2 m. A vehicle is 10 m behind the mirror’s pole. Using the Cartesian sign convention, find (i) the focal length of the mirror, (ii) the image distance, (iii) the lateral magnification, and (iv) the focal power of the mirror in dioptre. Comment on whether the image is erect and diminished, as expected for a convex mirror.
9.5 Refraction
Being an EM wave, the properties of light — speed, wavelength, direction of propagation, and so on — depend upon the medium through which it is travelling. Light is not an unchanging object that merely passes through matter; it adjusts itself to whatever medium it finds itself in.
So, if a ray of light comes to an interface between two media and enters into another medium of different refractive index, it changes itself suitable to that medium. This phenomenon is defined as refraction of light. The extent to which these properties change is decided by the index of refraction, ‘\(n\)’.
Refraction is the phenomenon in which light, on entering a medium of different refractive index across an interface, changes its speed, wavelength and (for oblique incidence) its direction of propagation.
- Logic behind the convention \(^1n_2\): The letter \(n\) is the symbol for refractive index, \(n_2\) corresponds to the refractive index of medium 2, and \(^1n_2\) indicates that it is with respect to medium 1. In this case light travels from medium 1 to medium 2, so we need to discuss medium 2 in the context of medium 1.
- Refraction is not the same as deviation. The dictionary meaning of the word refract is “to change the path”. However, in the context of Physics we should be more specific. We use the word deviate for changing the path. During refraction at normal incidence there is no change in path. Thus, there is refraction but no deviation. Deviation is associated with refraction only during oblique incidence. Deviation, or changing the path, or bending, is associated with many phenomena such as reflection, diffraction, scattering, and even gravitational bending of light due to a massive object.
Absolute refractive index
The absolute refractive index of a medium, which we met in Section 9.2, is defined formally as follows.
The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in the given medium.
\[n = \frac{c}{v}\]
where \(c\) and \(v\) are the respective speeds of light in vacuum and in the medium, both in m s\(^{-1}\). As \(n\) is the ratio of the same physical quantity, it is a unitless and dimensionless physical quantity.
For any material medium (including air), \(n > 1\), that is, light travels fastest in vacuum, faster than in any material medium. A medium having a greater value of \(n\) is called optically denser.
An optically denser medium need not be physically denser. For example, many oils are optically denser than water, but water is physically denser than them. So “denser” in optics is a statement about how much the medium slows light down, not about how much mass is packed into it.
Relative refractive index
The refractive index of medium 2 with respect to medium 1 is the ratio of the speed of light \(v_1\) in medium 1 to its speed \(v_2\) in medium 2.
\[^1n_2 = \frac{n_2}{n_1} = \frac{v_1}{v_2}\]
Here \(n_1\) and \(n_2\) are the absolute refractive indices of medium 1 and medium 2. This single quantity is what actually decides the bending at a boundary: Snell’s law can be rewritten as \(\sin\theta_1 / \sin\theta_2 = \ ^1n_2\), so it is the ratio of the two media, not either one alone, that matters.
Illustrations of refraction
1) The bottom of a water body appears to be raised. When seen from outside, the bottom of a pond or a bucket of water looks nearer to us than it really is. This is due to refraction at the plane surface of water. In this case,
\[n_{water} \cong \frac{Real \ depth}{apparent \ depth}\]
This relation holds good for a plane parallel transparent slab also, as shown below.
Figure 9.7 shows a plane parallel slab of a transparent medium of refractive index \(n\). A point object O at real depth \(R\) appears to be at I at apparent depth \(A\), when seen from outside (air). Incident rays OA (travelling undeviated, straight up along the normal) and OB (deviating along BC) are used to locate the image, since the image lies where these two emergent rays appear to meet.
By considering \(i\) and \(r\) to be small, we can write
\[\tan(r) = \frac{x}{A} \cong \sin(r) \quad \text{and} \quad \tan(i) = \frac{x}{R} \cong \sin(i)\]
where \(x\) is the small horizontal offset of the emerging ray from the normal, \(i\) is the angle of incidence inside the slab and \(r\) is the angle of refraction in air. Using Snell’s law in the form \(n = \sin r / \sin i\) for light going from the medium into air,
\[\therefore n = \frac{\sin(r)}{\sin(i)} \cong \frac{\left(\dfrac{x}{A}\right)}{\left(\dfrac{x}{R}\right)} = \frac{R}{A} = \frac{Real \ depth}{Apparent \ depth}\]
The common factor \(x\) cancels, which is why the result does not depend on where exactly the second ray strikes the surface — provided the angles stay small.
Small angle approximation: For small angles, expressed in radian, \(\sin\theta \cong \theta \cong \tan\theta\). For example, for \[\theta = 30^0 = \left(\frac{\pi}{6}\right)^c = 0.5236^c, \quad \text{we have} \ \sin\theta = 0.5\] In this case the error is \(0.5236 - 0.5 = 0.0236\) in 0.5, which is 4.72 %. For practical purposes we consider angles less than \(10^0\), where the error in using \(\sin\theta \cong \theta\) is less than 0.51 %. (Even for \(60^0\), it is still 15.7 %.) It is left to you to verify that this is almost equally valid for \(\tan\theta\) till \(20^0\) only.
2) A stick or pencil kept obliquely in a glass containing water appears broken, as its part in the water appears to be raised. The lower half is seen through the water surface and is shifted upwards by refraction, while the upper half in air is seen directly, so the two halves no longer line up.
Q2. Why does a pencil kept slanting in a glass of water look broken, and why does the bottom of a pond look shallower than it really is?
A. Light from the submerged part of the pencil bends away from the normal as it leaves the water and enters air. To our eye, which always assumes light has travelled in a straight line, the rays seem to come from a point higher up than the true one. So every submerged point appears raised, and the pencil seems to break at the water surface. For the same reason the bottom of the pond is seen at the apparent depth given by (real depth)/\(n\), which is smaller than the real depth — the pond looks shallower than it is.
The Sun is actually below the horizon at the moment we see it rise. The Earth’s atmosphere gets denser as we come down, so sunlight bends gradually while entering it, and the Sun appears raised — just as the bottom of a pond appears raised. This makes sunrise appear slightly earlier and sunset slightly later than they would be for an airless Earth, adding a few minutes of daylight each day.
A crane flying 6 m above a still, clear water lake sees a fish underwater. For the crane, the fish appears to be 6 cm below the water surface. How much deep should the crane immerse its beak to pick that fish? For the fish, how much above the water surface does the crane appear? Refractive index of water = 4/3.
Solution: For the crane, the apparent depth of the fish is 6 cm and the real depth is to be determined.
For the fish, the real depth (height, in this case) of the crane is 6 m and the apparent depth (height) is to be determined.
\[n = \frac{R}{A} = \frac{\text{Real depth}}{\text{Apparent depth}}\]
The only thing to be careful about is which medium is “medium 2” in each case — that is, whose refractive index we must use. The rule is that \(n\) here is the refractive index of the medium in which the real depth lies, taken with respect to the medium from which we are looking.
For the crane, it is water with respect to air, as the real depth is in water and the apparent depth is as seen from air.
\[\therefore n = \frac{4}{3} = \frac{R}{A} = \frac{R}{6} \ \therefore R = 8 \ \text{cm}\]
For the fish, it is air with respect to water, as the real height is in air and it is seen from water.
\[\therefore n = \frac{3}{4} = \frac{R}{A} = \frac{6}{A} \ \therefore A = 8 \ \text{m}\]
So the crane must dip its beak 8 cm into the water, while to the fish the crane appears to be 8 m above the surface, higher than it really is.
A coin lies at the bottom of a tank containing water (\(n = 4/3\)) up to a depth of 24 cm. Calculate the apparent depth of the coin as seen from directly above. If the water is replaced by a liquid of refractive index 1.6 up to the same depth, by how much does the apparent depth change?
9.6 Total Internal Reflection
Figure 9.8 shows refraction of light emerging from a denser medium into a rarer medium for various angles of incidence. Since the second medium is optically rarer, light speeds up on entering it and bends away from the normal. Hence the angles of refraction in the rarer medium are larger than the corresponding angles of incidence.
Now imagine slowly increasing the angle of incidence inside the denser medium. The refracted ray bends further and further away from the normal, and at some stage it must lie along the boundary itself. At a particular angle of incidence \(i_c\) in the denser medium, the corresponding angle of refraction in the rarer medium is exactly \(90^0\), as seen in Fig. 9.8.
For angles of incidence greater than \(i_c\), the angle of refraction would have to become larger than \(90^0\), which is geometrically impossible for a ray crossing into the second medium. The ray therefore does not enter the rarer medium at all, but is reflected totally back into the denser medium. This is called total internal reflection.
In general, there is always partial reflection and partial refraction at an interface: some of the light crosses over and some of it bounces back. During total internal reflection (TIR), it is total reflection and no refraction. The corresponding angle of incidence in the denser medium is greater than or equal to the critical angle.
Total internal reflection is the phenomenon in which light travelling in a denser medium and striking the boundary of a rarer medium at an angle of incidence greater than or equal to the critical angle is completely reflected back into the denser medium, with no refracted ray.
The critical angle for a pair of refracting media is that angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \(90°\).
Note both conditions carefully, since they are exam-critical: the light must be going from the denser medium to the rarer medium, and the angle of incidence must be at least the critical angle. If light travels from a rarer to a denser medium, TIR can never occur, no matter how large the angle of incidence.
In Physics the word critical is used when certain phenomena are not applicable, or when more than one phenomenon is applicable. Some examples are as follows. (i) In the case of total internal reflection, the phenomenon of reversibility of light is not applicable at the critical angle, and refraction is possible only for angles of incidence in the denser medium smaller than the critical angle. (ii) At the critical temperature, a substance coexists in all the three states: solid, liquid and gas. At all other temperatures, only two states are simultaneously possible. (iii) For liquids, streamline flow is possible till the critical velocity is achieved. At the critical velocity the flow can be either streamline or turbulent.
Expression for the critical angle
Let \(\mu\) be the relative refractive index of the denser medium with respect to the rarer medium. Applying Snell’s law at the critical angle of incidence, \(i_C\), with the angle of refraction equal to \(90^\circ\) in the rarer medium, we can write
\[(\mu)\sin (i_c) = (1) \sin 90°\]
\[\therefore \sin (i_c) = \frac{1}{\mu}\]
Here \(\mu\) is a ratio of two refractive indices and hence has no unit. The relation shows the physical logic clearly: the greater the value of \(\mu\) — that is, the more strongly the denser medium slows light down compared with its surroundings — the smaller is \(\sin(i_c)\), and hence the smaller is the critical angle. A medium with a small critical angle traps light very easily.
For commonly used glasses of \(\mu = 1.5\), \(i_c = 41° \ 49' \cong 42°\), and for water of \(\mu = \dfrac{4}{3}\), \(i_c = 48° \ 35'\) (both, with respect to air).
Q1. Why does light stay trapped inside a curving jet of water or inside a thin glass fibre, instead of leaking out through its sides?
A. Water and glass are optically denser than the air around them, so light inside them meets the side surface from the denser side. If the light is travelling nearly along the jet or the fibre, it strikes that surface at a large angle of incidence — larger than the critical angle (about \(48°\) for water and about \(42°\) for glass with respect to air). At such angles no refraction is possible, and the light is totally internally reflected back inside. The process repeats at every bounce, so the light is guided along the bending path instead of escaping.
Diamond sparkles far more than glass because it is optically very dense, with a refractive index of about 2.42. Its critical angle with respect to air is therefore only about \(24°\), so light entering a diamond meets almost every internal face beyond the critical angle and is totally internally reflected several times before it finally escapes. A gem cutter shapes the facets so that this trapped light emerges mainly through the top, which is what produces the brilliant glitter.
The refractive index of a certain transparent plastic with respect to air is 1.60. (i) Calculate its critical angle with respect to air. (ii) A ray inside this plastic strikes the plastic–air boundary at an angle of incidence of \(35°\). State, with reason, whether it will emerge into the air or be totally internally reflected.
9.6.1 Applications of Total Internal Reflection
(i) Optical fibre
Though a little costly for the initial set up, optic fibre communication is undoubtedly the most effective way of telecommunication by way of EM waves.
An optical fibre essentially consists of an extremely thin (slightly thicker than a human hair), transparent, flexible core, surrounded by an optically rarer (smaller refractive index), flexible cover called the cladding. Notice in Fig. 9.9 (a) that this system is further coated by a buffer and a jacket for protection. The entire thickness of the fibre is less than half a mm. A number of such fibres may be packed together in an outer cover.
The choice of a denser core inside a rarer cladding is the whole point of the design. It guarantees that light travelling nearly along the axis of the fibre always meets the core–cladding boundary from the denser side, and at an angle of incidence larger than the critical angle.
An optical signal (ray) entering the core therefore suffers multiple total internal reflections, as shown in Fig. 9.9 (b), and emerges after several kilometres with extremely low loss, travelling with the highest possible speed in that material (about 2,00,000 km/s for glass).
Some of the advantages of optic fibre communication are listed below.
Broad bandwidth (frequency range): For TV signals, a single optical fibre can carry over 90000 channels (independent signals).
Immune to EM interference: Being electrically non-conductive, it is not able to pick up nearby EM signals.
Low attenuation loss: The loss is lower than 0.2 dB/km, so that a single long cable can be used for several kilometres.
Electrical insulator: There is no issue with ground loops of metal wires or with lightning.
Theft prevention: It does not use copper or other expensive material.
Security of information: Internal damage is most unlikely.
An endoscope, used by doctors to look inside the stomach or the intestines, is a bundle of optical fibres. One set of fibres carries light down into the body to illuminate the region, and another set carries the image back out by repeated total internal reflection. Because the bundle is thin and flexible, it can follow the bends of the body without the light escaping through the sides, allowing a doctor to see inside a patient without surgery.
(ii) Prism binoculars
Binoculars using only two cylinders have a limitation of field of view, because the distance between the two cylinders cannot be greater than the distance between the two eyes. A wider separation of the two objectives would give a better sense of depth and a wider view, but the eyepieces must still end up in front of our eyes.
This limitation can be overcome by using two right-angled glass prisms (\(i_C \sim 42^0\)) used for total internal reflection, as shown in Fig. 9.10. The prisms fold the light path sideways, so the objectives can be set wider apart than the eyes while the eyepieces stay where they must be.
Total internal reflections occur inside the isosceles, right-angled prisms. Light strikes the hypotenuse face at \(45^\circ\), which is greater than the critical angle of about \(42^\circ\) for glass, so it is turned through a right angle without any silvering and with almost no loss of light.
(iii) Periscope
A periscope is used to see the objects on the surface of a water body from inside the water. For this, the rays of light should be reflected twice through a right angle: once at the top, to bring the light down the tube, and once at the bottom, to send it into the observer’s eye.
The reflections are similar to those in the binoculars (Fig. 9.11), and total internal reflections occur inside isosceles, right-angled prisms.
There is a tiny LED bulb at the centre of the bottom of a cylindrical vessel of diameter 6 cm. The height of the vessel is 4 cm. The beaker is filled completely with an optically dense liquid. The bulb is visible from any inclined position, but is just visible if seen along the edge of the beaker. Determine the refractive index of the liquid.
Solution: As seen from the accompanying figure, if the bulb is just visible from the edge, the angle of incidence in the liquid (at the edge) must be the critical angle of incidence, \(i_C\). “Just visible” means the emergent ray grazes along the surface, that is, the angle of refraction is \(90^\circ\) — which is exactly the condition defining the critical angle.
From the dimensions given, the horizontal distance from the centre of the bottom to the edge is half the diameter, that is 3 cm, and the vertical height is 4 cm.
\[\tan (i_c) = \frac{3}{4} \ \therefore \ \sin (i_c) = \frac{3}{5} \ \therefore \ n_{liquid} = \frac{\sin 90°}{\sin(i_c)} = \frac{5}{3}\]
The core of an optical fibre has a refractive index of 1.50 and its cladding has a refractive index of 1.45. Calculate the critical angle at the core–cladding boundary. Hence state the smallest angle of incidence (measured from the normal to that boundary) for which a ray inside the core will be guided along the fibre.
9.7 Refraction at a Spherical Surface and Lenses
In Section 9.5 we saw that due to refraction the bottom of a water body appears to be raised, and that
\[n_{water} = \frac{Real \ depth}{apparent \ depth}\]
However, this is valid only if we are dealing with refraction at a plane surface. In many cases — such as liquid drops, lenses, ellipsoid paper weights, and so on — curved surfaces are present, and the formula mentioned above may not be true.
In such cases we need to consider refraction at one or more spherical surfaces. This will involve parameters including the curvature, such as the radius of curvature, in addition to the refractive indices. The reason is easy to see: on a curved surface the normal points in a different direction at every point, so identical rays striking at different heights are bent by different amounts.
Lenses
Commonly used lenses can be visualized as consisting of the intersection of two spheres of radii of curvature \(R_1\) and \(R_2\), or of one sphere and a plane surface (\(R = \infty\)).
A lens is said to be convex if it is thicker in the middle and narrows towards the periphery. A lens is concave if it is thicker at the periphery and narrows down towards the centre.
A convex lens is visualized to be the internal cross section of two spheres (or of one sphere and a plane surface), while a concave lens is their external cross section, as seen in Figs. 9.12 (a) to 9.11 (f). Concavo-convex and convexo-concave lenses are commonly used for spectacles of positive and negative numbers, respectively.
For lenses of material optically denser than the medium in which those are kept, convex lenses have positive focal length (according to the Cartesian sign convention) and converge the incident beam, while concave lenses have negative focal length and diverge the incident beam.
Unless mentioned specifically, we assume lenses to be made up of optically denser material compared to the medium in which those are kept — for example, glass lenses in air or in water. As special cases we may consider lenses of rarer medium, such as an air lens in water or inside glass. A spherical hole inside a glass slab is also a lens of rarer medium. In such a case, physically (or geometrically, or shape-wise) a convex lens diverges the incident beam, while a concave lens converges the incident beam.
Thin lenses
For most of the applications of lenses, the maximum thickness of the lens is negligible — at least 50 times smaller — compared with all the other distances such as \(R_1\) and \(R_2\), \(u\), \(v\) and \(f\). Such a lens is called a thin lens, and the physical centre of such a lens can be assumed to be the common pole (or optical centre) for both its refracting surfaces.
This assumption is what makes lens problems manageable: instead of tracking the sideways shift of the ray as it crosses the thickness of the glass, we treat both surfaces as if they were at the same point.
For any thin lens,
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \qquad \text{--- (9.2)}\]
where \(u\) is the object distance, \(v\) the image distance and \(f\) the focal length, all measured from the optical centre with the proper sign, in metre in SI units.
Combination of thin lenses
If necessary, we can have a number of thin lenses in contact with each other, having a common principal axis. The focal power of such a combination is given by the algebraic addition (by considering \(\pm\) signs) of the individual focal powers.
\[\therefore \frac{1}{f} = \sum \left(\frac{1}{f_i}\right) = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + \ldots\]
\[= P_1 + P_2 + P_3 + \ldots = \sum P_i = P\]
Here each \(f_i\) is in metre and each \(P_i\) is in dioptre. Physically, this simply says that if one lens has already started bending the beam, the next lens in contact continues to bend it further, and the bending abilities add up.
For only two thin lenses, separated in air by a distance \(d\),
\[\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} = P_1 + P_2 - d P_1 P_2 = P\]
Note that the separation \(d\) (in metre) reduces the power of the combination. This extra term vanishes when \(d = 0\), and we return to the case of lenses in contact.
A convex lens of focal length 20 cm is kept in contact with a concave lens of focal length 30 cm. Calculate (i) the focal power of each lens in dioptre, (ii) the focal power of the combination, and (iii) the equivalent focal length of the combination. State whether the combination is converging or diverging.
Why f = R/2 fails for a lens
For lenses, the relations between \(u\), \(v\), \(R\) and \(f\) depend also upon the refractive index \(n\) of the material of the lens. The relation
\[f = \frac{R}{2}\]
does NOT hold good for lenses.
The reason is physical, not merely mathematical. A mirror bends light by reflection, and the law of reflection (\(i = r\)) does not involve the material at all — so only the shape of the mirror matters. A lens bends light by refraction, and refraction depends on how much the lens material slows light down compared with the surroundings. Two lenses of exactly the same shape but of different glasses will therefore have different focal lengths.
Below we shall derive the necessary relation by considering refraction at the two surfaces of a lens independently.
Refraction at a single spherical surface
Consider a spherical surface YPY′ of radius of curvature \(R\), separating two transparent media of refractive indices \(n_1\) and \(n_2\) respectively, with \(n_1 < n_2\). P is the pole and X′PX is the principal axis. A point object O is at an object distance \(-u\) from the pole, in the medium of refractive index \(n_1\).
Convexity or concavity of a surface is always decided with respect to the incident rays, that is, with respect to a real object. Hence in this case the surface is convex, as shown in Fig. 9.13.
To locate its image, and in order to minimize spherical aberration, we consider two paraxial rays.
The ray OP along the principal axis strikes the surface head-on, along the normal, and so travels undeviated along PX.
Another ray OA strikes the surface at A. CAN is the normal from the centre of curvature C of the surface at A — remember that for a spherical surface, the normal at any point is along the radius through that point. The angle of incidence in the medium \(n_1\) at A is \(i\).
As \(n_1 < n_2\), the ray deviates towards the normal, travels along AZ, and cuts the principal axis at I. Thus, a real image of the point object O is formed at I. The angle of refraction in medium \(n_2\) is \(r\).
According to Snell’s law,
\[n_1 \sin(i) = n_2 \sin(r) \qquad \text{--- (9.3)}\]
Let \(\alpha, \beta\) and \(\gamma\) be the angles subtended by the incident ray, the normal and the refracted ray with the principal axis.
From the exterior-angle property of the triangles formed at A,
\[\therefore i = \alpha + \beta \quad \text{and} \quad r = \beta - \gamma\]
For paraxial rays all these angles are small, and PA can be considered as an arc for \(\alpha, \beta\) and \(\gamma\).
\[\therefore \sin(i) \cong i \quad \text{and} \quad \sin(r) \cong r\]
Also, using the small-angle result that an angle equals (arc)/(radius),
\[\alpha \cong \frac{arc \ AP}{PO} = \frac{arc \ AP}{-u},\]
\[\beta = \frac{arc \ AP}{PC} = \frac{arc \ AP}{R} \quad \text{and}\]
\[\gamma \cong \frac{arc \ AP}{PI} = \frac{arc \ AP}{v}\]
Substituting the small-angle forms in Snell’s law,
\[\therefore n_1 i = n_2 r\]
\[\therefore n_1 (\alpha + \beta) = n_2 (\beta - \gamma)\]
\[\therefore (n_2 - n_1)\beta = n_2 \gamma + n_1 \alpha\]
Substituting \(\alpha, \beta\) and \(\gamma\), and cancelling the common factor ‘arc AP’, we get
\[\frac{n_2 - n_1}{R} = \frac{n_2}{v} - \frac{n_1}{u} \qquad \text{--- (9.4)}\]
This is the key result of the section. Notice that ‘arc AP’ cancels out completely, which is exactly why the position of the image does not depend on which paraxial ray we chose — all of them meet at the same point I.
The front surface of the human eye, the cornea, is a single curved refracting surface separating air from the transparent material of the eye. Because the change in refractive index there is large (from about 1 to about 1.376), most of the focusing done by the eye actually happens at this one spherical surface, and the eye lens behind it only makes fine adjustments. This is also why laser surgery works by reshaping the cornea — changing \(R\) in Eq. (9.4) changes where the image falls.
In this case (Fig. 9.13) we had considered the object placed in the rarer medium, a real image in the denser medium, and the surface facing the object to be convex. However, while deriving the relation, all the symbolic values (which could be numeric also) were substituted as per the Cartesian sign convention (e.g. ‘\(u\)’ as negative, etc.). Hence the final expression (Eq. 9.4) is applicable to any surface separating any two media, and to a real or virtual image, provided you substitute your values (symbolic or numerical) as per the Cartesian sign convention. The only restriction is that \(n_1\) is for the medium of the real object and \(n_2\) is the other medium (not necessarily the medium of the image). Only in the case of a real image will it be in medium \(n_2\). If virtual, it will be in the medium \(n_1\) (with image distance negative — how do you justify this?). We strongly suggest you do the derivations yourself for any other special case, such as an object placed in the denser medium, a virtual image, a concave surface, etc. It must be remembered that in any case you will land up with the same expression as in Eq. (9.4).
A glass paper-weight (\(n = 1.5\)) of radius 3 cm has a tiny air bubble trapped inside it. The closest distance of the bubble from the surface is 2 cm. Where will it appear when seen from the other end (from where it is farthest)?
Solution: The accompanying figure illustrates the location of the bubble. Since we are looking from the far side, the light from the bubble travels through the glass and leaves through the surface nearer to the observer. So the object (the bubble) lies in the glass, and the light emerges into air.
According to the symbols used in Eq. (9.4), we get,
\(n_1\) = refractive index of the medium of the real object (medium of incident rays) \(= 1.5\)
\(n_2\) = refractive index of the other medium \(= 1\)
The bubble is 2 cm from the near surface, so from the far surface (through which we view it) it is at \(6 - 2 = 4\) cm, and it lies to the left of that pole.
\(u = -4 \ cm\)
\(v = ?\)
\(R = -3 \ cm\)
\[\frac{n_2 - n_1}{R} = \frac{n_2}{v} - \frac{n_1}{u}\]
\[\therefore \frac{1 - 1.5}{-3} = \frac{1}{v} - \frac{1.5}{-4} \ \therefore \frac{1}{6} = \frac{1}{v} + \frac{3}{8} \ \therefore v = -4.8 \ \text{cm}\]
In this case the apparent depth is NOT less than the real depth. This is due to the curvature of the refracting surface. For a plane surface we would have expected the bubble to appear closer (apparent depth less than real depth); the curvature reverses that expectation, which is precisely why Eq. (9.4) is needed and the plane-surface formula cannot be used here.
Lens makers’ equation
Lens makers’ equation: This is the relation between the refractive index (\(n\)), the focal length (\(f\)) and the radii of curvature \(R_1\) and \(R_2\) for a thin lens.
Consider a lens of radii of curvature \(R_1\) and \(R_2\), kept in a medium such that \(n\) is the refractive index of the material of the lens with respect to the outside medium. Assuming the lens to be thin, P is the common pole for both the surfaces. O is a point object on the principal axis at a distance \(u\) from P. The first refracting surface of the lens, of radius of curvature \(R_1\), faces the object, as shown in Fig. 9.14.
The strategy of the derivation is to apply Eq. (9.4) twice — once for each surface — and to let the image formed by the first surface act as the object for the second.
The axial ray OP travels undeviated. The paraxial ray OA deviates towards the normal, and in the absence of the second refracting surface it would intersect the axis at \(I_1\). Here \(PI_1 = v_1\) is the image distance for the intermediate image \(I_1\).
Thus, the symbols to be used in Eq. (9.4) are
\[n_2 = n, \quad n_1 = 1, \quad R = R_1, \quad u = u, \quad v = v_1\]
\[\therefore \frac{n - 1}{R_1} = \frac{n}{v_1} - \frac{1}{(u)} \qquad \text{--- (9.5)}\]
(Note that, in this case, we are not substituting the algebraic values but just using different symbols.)
Before reaching \(I_1\), the ray \(PI_1\) is intercepted at B by the second refracting surface. In this case, the incident rays AB and OP are in the medium of refractive index \(n\) and are converging towards \(I_1\). Thus, \(I_1\) acts as a virtual object for the second surface of radius of curvature (\(R_2\)), and the object distance is \((u = v_1)\).
As the incident rays are in the medium of refractive index \(n\), this is the medium of the (virtual) object \(\therefore n_1 = n\), and the refractive index of the other medium is \(n_2 = 1\).
After refraction at the second surface, the ray bends away from the normal and intersects the principal axis at I, which is the real image of the object O formed due to the lens. \(\therefore PI = v\).
Substituting all these symbols in Eq. (9.4), we get
\[\frac{1 - n}{R_2} = \frac{n - 1}{-R_2} = \frac{1}{v} - \frac{n}{(v_1)} \qquad \text{--- (9.6)}\]
Adding Eq. (9.5) and (9.6), the intermediate image distance \(v_1\) cancels out — which is exactly why we could get away with never calculating it — and we get,
\[(n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{v} - \frac{1}{u}\]
For \(u = \infty, v = f\), since parallel rays from an infinitely distant object converge at the focus,
\[\therefore \frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \qquad \text{--- (9.7)}\]
For preparing spectacles, it is necessary to grind the glass (or acrylic, etc.) so as to have the desired radii of curvature. Equation (9.7) can be used to calculate the radii of curvature for the lens, and hence it is called the lens makers’ equation.
It should be remembered that while solving problems using equations 9.1, 9.2, 9.4, 9.7, etc., we will be substituting the values of the corresponding quantities. Hence this time it is algebraic substitution, that is, with the proper \(\pm\) sign.
Special cases: The most popular and most common special case is the one in which we have a thin, symmetric, double lens. In this case, \(R_1\) and \(R_2\) are numerically equal. (A) Thin, symmetric, double convex lens: \(R_1\) is positive, \(R_2\) is negative and numerically equal. Let \(\lvert R_1 \rvert = \lvert R_2 \rvert = R\). \[\therefore \frac{1}{f} = (n-1)\left(\frac{1}{R} - \frac{1}{-R}\right) = \frac{2(n-1)}{R}\] Further, for the popular variety of glasses, \(n \cong 1.5\). In such a case, \(f = R\). (B) Thin, symmetric, double concave lens: \(R_1\) is negative, \(R_2\) is positive and numerically equal. Let \(\lvert R_1 \rvert = \lvert R_2 \rvert = R\). \[\therefore \frac{1}{f} = (n-1)\left(\frac{1}{-R} - \frac{1}{R}\right) = \frac{2(n-1)}{-R}\] Further, if \(n \cong 1.5, \ f = -R\). (C) Thin, planoconvex lenses: One radius is \(R\) and the other is \(\infty\). \[\therefore \frac{1}{f} = \frac{n-1}{R}\] Further, if \(n \cong 1.5, \ f = 2\lvert R \rvert\).
A dense glass double convex lens (\(n = 2\)), designed to reduce spherical aberration, has \(\lvert R_1 \rvert : \lvert R_2 \rvert = 1 : 5\). If a point object is kept 15 cm in front of this lens, it produces its real image at 7.5 cm. Determine \(R_1\) and \(R_2\).
Solution: First find the focal length from the thin lens formula, then feed it into the lens makers’ equation.
\(u = -15\) cm, \(v = +7.5\) cm (real image is on the opposite side).
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \ \therefore \frac{1}{f} = \frac{1}{7.5} - \frac{1}{-15} \ \therefore f = +5 \ \text{cm}\]
The lens is double convex. Hence, \(R_1\) is positive and \(R_2\) is negative. Also, \(\lvert R_2 \rvert = 5 \lvert R_1 \rvert\) and \(n = 2\).
\[(n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{f}\]
\[\therefore (2-1)\left(\frac{1}{R_1} - \frac{1}{-(5R_1)}\right) = \frac{1}{5}\]
\[\therefore (1)\left(\frac{6}{5R_1}\right) = \frac{1}{5} \ \therefore R_1 = 6 \ \text{cm} \ \therefore R_2 = 30 \ \text{cm}\]
A thin, symmetric double convex lens is made of glass of refractive index 1.5, and each of its surfaces has a radius of curvature of magnitude 20 cm. Using the lens makers’ equation, calculate its focal length in air and its focal power in dioptre. Verify that your answer agrees with the special case \(f = R\) for \(n \cong 1.5\).
9.8 Dispersion of Light and Prisms
The colour of light that we see depends upon the frequency of that ray (wave). The refractive index of a material also depends upon the frequency of the wave, and it increases with frequency. Violet light has the highest frequency among visible colours and red the lowest, so glass slows violet down more than it slows red down.
Obviously, then, the refractive index of a material is different for different colours. As a result, for an obliquely incident ray, the angles of refraction are different for each colour, and they separate (disperse) as they travel along different directions. This phenomenon is called angular dispersion, as shown in Fig. 9.15.
Dispersion is the separation of a polychromatic beam into its component colours, caused by the refractive index of the medium being different for different frequencies (colours) of light.
In the 1660s, Isaac Newton let a narrow beam of sunlight pass through a glass prism in a darkened room and obtained a band of colours on the wall. The important part of his work was what he did next: he passed a single colour from that band through a second prism, and it did not split further. He also recombined the colours back into white light. This showed that white light is a mixture of colours and that the prism does not create colour but only separates what was already there.
Lateral dispersion through a slab
If a polychromatic beam of light (a bundle of rays of different colours) is obliquely incident upon a plane parallel transparent slab, the emergent beam consists of all the component colours separated out. However, in this case all those are parallel to each other, and also parallel to the initial direction.
This is lateral dispersion, which is measured as the perpendicular distance between the direction of the incident ray and the respective directions of the dispersed emergent rays (\(L_R\) and \(L_V\)), as shown in Fig. 9.16. For it to be easily detectable, the parallel surfaces must be separated over a very large distance, and \(i\) should be large.
The reason the emergent rays come out parallel to the incident ray is that the second surface of the slab undoes exactly what the first surface did — the two surfaces are parallel, so the bending at entry is cancelled by the bending at exit. Each colour is only shifted sideways, not turned. This is why a slab is useless for showing dispersion, as the next example makes clear numerically.
A fine beam of white light is incident upon the longer side of a plane parallel glass slab of breadth 5 cm at an angle of incidence of \(60^0\). Calculate the angular deviation of red and violet rays within the slab, and the lateral dispersion between them as they emerge from the opposite side. The refractive indices of the glass for red and violet are 1.51 and 1.53 respectively.
Solution: As shown in Fig. 9.16 above, VM \(= L_V\) and RT \(= L_R\) give the respective lateral deviations for violet and red colours, and \(L_{VR} = L_V - L_R\) is the lateral dispersion between these colours.
\(n_R = 1.51\), \(n_V = 1.53\) and \(i = 60°\)
First find the angle of refraction inside the slab for each colour, using Snell’s law.
\[\therefore \sin r_R = \frac{\sin i}{n_R} = \frac{\sin 60^0}{1.51} = 0.5735\]
\[\sin r_V = \frac{\sin i}{n_V} = \frac{\sin 60^0}{1.53} = 0.566\]
\[\therefore r_R = 35^0 \ \text{and} \ R_v = 34^0 28' \ \therefore \delta_{RV} = r_R - r_V = 32'\]
The angular deviation of each colour within the slab is the difference between the incident direction and the refracted direction, that is \((i - r)\).
\[\therefore i - r_R = 25^0, \quad i - r_V = 25^0 32'\]
Next find the path lengths inside the slab, using the breadth AN of the slab and the refracted angles.
\[\therefore \text{AR} = \frac{\text{AN}}{\cos r_R} = 6.104 \ \text{cm}\]
\[\text{AV} = \frac{\text{AN}}{\cos r_V} = 6.063 \ \text{cm}\]
The lateral shift of each colour is the component of its path perpendicular to the incident direction.
\[\therefore L_R = \text{RT} = \text{AR} \left(\sin \left[i - r_R\right]\right) = 2.58 \ \text{cm}\]
\[L_V = \text{VM} = \text{AV}\left(\sin\left[i - r_V\right]\right) = 2.58 \ \text{cm}\]
\[\therefore L_{VR} = L_V - L_R = 0.033 \ \text{cm} = 0.33 \ \text{mm}\]
It shows that the lateral dispersion is too small to detect.
Prisms
In order to have appreciable and observable dispersion, two parallel surfaces are not useful. In such a case we use prisms, in which two refracting surfaces inclined at an angle are used. Because the two surfaces are not parallel, the second refraction adds to the bending caused by the first instead of cancelling it — and the difference between colours therefore keeps growing.
The popular variety of prisms have three rectangular surfaces forming a triangle. At a time, two of these take part in the refraction. The one not involved in refraction is called the base of the prism, as shown in Fig. 9.17.
Any section of a prism perpendicular to the base is called the principal section of the prism. Usually we consider all the rays in this plane. Fig. 9.18 (a) and Fig. 9.18 (b) show refraction through a prism for monochromatic and white beams respectively. Angular dispersion is shown for the white beam.
Relations between the angles involved
Figure 9.19 shows the principal section ABC of a prism of absolute refractive index \(n\) kept in air. The refracting surfaces AB and AC are inclined at an angle \(A\), which is the refracting angle of the prism, or simply the ‘angle of prism’. Surface BC is the base.
A monochromatic ray PQ obliquely strikes the first refracting surface AB. The normal passing through the point of incidence Q is MQN. The angle of incidence at Q is \(i\). After refraction at Q, the ray deviates towards the normal (since it enters a denser medium) and strikes the second refracting surface AC at R, which is the point of emergence. MRN is the normal through R. The angles of refraction at Q and R are \(r_1\) and \(r_2\) respectively.
After R, the ray deviates away from the normal (since it now enters a rarer medium) and finally emerges along RS, making \(e\) as the angle of emergence. The incident ray PQ is extended as QT. The emergent ray RS meets QT at X if traced backward. The angle TXS is the angle of deviation \(\delta\) — the total turning suffered by the ray from entry to exit.
\[\angle \text{AQN} = \angle \text{ARN} = 90^0 \quad \ldots \ldots \text{(Angles at normal)}\]
\(\therefore\) From quadrilateral AQNR, whose angles must add up to \(360^0\),
\[A + \angle \text{QNR} = 180^0 \qquad \text{--- (9.8)}\]
From \(\Delta\) QNR,
\[r_1 + r_2 + \angle \text{QNR} = 180^0 \qquad \text{--- (9.9)}\]
\(\therefore\) From Eqs. (9.8) and (9.9),
\[A = r_1 + r_2 \qquad \text{--- (9.10)}\]
This is a neat result: whatever the angle of incidence, the two internal angles of refraction must always add up to the angle of the prism.
Angle \(\delta\) is the exterior angle for triangle XQR.
\[\therefore \angle \text{XQR} + \angle \text{XRQ} = \delta\]
\[\therefore (i - r_1) + (e - r_2) = \delta\]
\[\therefore (i + e) - (r_1 + r_2) = \delta\]
Hence, using Eq. (9.10), \((i + e) - (A) = \delta\)
\[\therefore i + e = A + \delta \qquad \text{--- (9.11)}\]
For a prism of refracting angle \(A\): \(A = r_1 + r_2\) and \(i + e = A + \delta\), where \(i\) is the angle of incidence, \(e\) the angle of emergence, \(r_1\) and \(r_2\) the angles of refraction at the two surfaces, and \(\delta\) the angle of deviation.
Deviation curve, minimum deviation and prism formula
From the relations (9.10) and (9.11), it is clear that \(\delta, e, r_1\) and \(r_2\) depend upon \(i\), \(A\) and \(n\).
After a certain minimum value of the angle of incidence, \(i_{min}\), the emergent ray is possible. This is because of the fact that for \(i < i_{min}\), we get \(r_2 > i_c\), and there is total internal reflection at the second surface, so there is no emergent ray at all. This will be shown later.
Then onwards, as \(i\) increases, \(r_1\) increases as \(\dfrac{\sin i}{\sin r_1} = n\), but \(r_2\) and \(e\) decrease — the last of these follows from \(A = r_1 + r_2\), since \(A\) is fixed. However, the variation in \(\delta\) with increasing \(i\) is different. It is as plotted in Fig. 9.20.
It shows that, with increasing values of \(i\), the angle of deviation \(\delta\) decreases initially to a certain minimum \((\delta_m)\) and then increases. It should also be noted that the curve is not a symmetric parabola, but the slope in the part after is less.
It is clear that except at \(\delta = \delta_m\) (the angle of minimum deviation), there are two values of \(i\) for any given \(\delta\). By applying the principle of reversibility of light to the path PQRS, it is obvious that if one of these values is \(i\), the other must be \(e\), and vice versa.
Thus at \(\delta = \delta_m\), we have \(i = e\). Also, in this case, \(r_1 = r_2\) and
\[A = r_1 + r_2 = 2r \ \therefore r = \frac{A}{2}\]
Only in this case is QR parallel to the base BC, and the figure is symmetric.
Using these in Eq. (9.11), we get,
\[i + i = A + \delta_m \ \therefore i = \frac{(A + \delta_m)}{2}\]
According to Snell’s law, applied at the first surface with \(i = (A+\delta_m)/2\) and \(r = A/2\),
\[\therefore n = \frac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)} \qquad \text{--- (9.12)}\]
Equation (9.12) is called the prism formula. It is the standard laboratory method of measuring the refractive index of a transparent material: shape it into a prism, measure \(A\) and \(\delta_m\), and substitute.
Prism formula: for a prism of refracting angle \(A\) in air, at minimum deviation \(\delta_m\), the refractive index is \(n = \dfrac{\sin\left(\dfrac{A+\delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}\).
(I) Grazing emergence and minimum angle of incidence: At the point of emergence, the ray travels from a denser medium into a rarer one (popular prisms are of denser material, kept in a rarer medium). Thus if \(r_2 = \sin^{-1}\left(\dfrac{1}{n}\right)\) is the critical angle, the angle of emergence is \(e = 90^0\). This is called grazing emergence, or we say that the ray just emerges. The angle of prism \(A\) is constant for a given prism, and \(A = r_1 + r_2\). Hence the corresponding \(r_1\) and \(i\) will have their minimum possible values. (II) For commonly used glass prisms, \[n = 1.5, \ \sin^{-1}\left(\frac{1}{n}\right) = \sin^{-1}\left(\frac{1}{1.5}\right) = 41^0 49' = (r_2)_{max}\] If the prism is symmetric (equilateral), \[A = 60^0 \ \therefore r_1 = 60^0 - 41^0 49' = 18^0 11'\] \[\therefore n = 1.5 = \frac{\sin(i_{min})}{\sin 18^0 11'} \ \therefore i_{min} = 27^0 55' \cong 28^0\] (III) For a symmetric (equilateral) prism, the prism formula can be written as \[n = \frac{\sin\left(\dfrac{60 + \delta_m}{2}\right)}{\sin\left(\dfrac{60}{2}\right)} = \frac{\sin\left(30 + \dfrac{\delta_m}{2}\right)}{\sin(30)} = 2\sin\left(30 + \frac{\delta_m}{2}\right)\] (IV) For a prism of denser material, kept in a rarer medium, the incident ray deviates towards the normal during the first refraction and away from the normal during the second refraction. However, during both the refractions it deviates towards the base only.
For a glass (\(n = 1.5\)) prism having a refracting angle of \(60^0\), determine the range of angle of incidence for which an emergent ray is possible from the opposite surface, and the corresponding angles of emergence. Also calculate the angle of incidence for which \(i = e\). How much is the corresponding angle of minimum deviation?
Solution: As shown in the box above,
\[i_{min} = 27^0 55'\]
The angle of emergence for this is
\[e_{max} = 90^0\]
From the principle of reversibility of light, the largest possible angle of incidence is the grazing one, and it pairs with the smallest emergence.
\[i_{max} = 90^0 \ and \ e_{min} = 27^0 55'\]
So the emergent ray is possible only for \(i\) between \(27^0 55'\) and \(90^0\).
Also, from the box above,
\[n = \frac{\sin\left(\dfrac{60 + \delta_m}{2}\right)}{\sin\left(\dfrac{60}{2}\right)} = \frac{\sin\left(30 + \dfrac{\delta_m}{2}\right)}{\sin(30)} = 2\sin\left(30 + \frac{\delta_m}{2}\right)\]
\[\therefore 1.5 = 2\sin\left(30 + \frac{\delta_m}{2}\right) \ \therefore 0.75 = \sin\left(30 + \frac{\delta_m}{2}\right)\]
\[\therefore \left(30 + \frac{\delta_m}{2}\right) = 48^0 35'\]
\[\therefore \frac{\delta_m}{2} = 18^0 35' \ \therefore \delta_m = 37^0 10'\]
\[i + e = A + \delta \quad \text{and} \quad i = e \ \text{for} \ \delta = \delta_m\]
\[\therefore i + i = 60 + 37° \ 10' = 97°10' \ \therefore i = 48°35'\]
Thin prisms
Prisms having a refracting angle less than \(10^0\) \((A < 10^0)\) are called thin prisms. For such prisms we can comfortably use \(\sin\theta \cong \theta\).
For such prisms to deviate the incident ray towards the base during both refractions, it is essential that \(i\) should also be less than \(10^0\), so that all the other angles will also be small.
Thus
\[n = \frac{\sin i}{\sin r_1} \cong \frac{i}{r_1} \quad \text{and} \quad n = \frac{\sin e}{\sin r_2} \cong \frac{e}{r_2}\]
\[\therefore i \cong n r_1 \quad \text{and} \quad e \cong n r_2\]
Using these in Eq. (9.11), we get,
\[i + e = n r_1 + n r_2 = n(r_1 + r_2) = nA = A + \delta\]
\[\therefore \delta = A(n-1) \qquad \text{--- (9.13)}\]
\(A\) and \(n\) are constant for a given prism. Thus, for a thin prism, for small angles of incidence, the angle of deviation is constant (independent of the angle of incidence). This is a considerable simplification — for a thin prism we no longer need to worry about how the prism is tilted.
A thin prism of refracting angle \(6^0\) is made of glass of refractive index 1.52. Calculate the angle of deviation produced by it for a small angle of incidence. If a second thin prism of the same angle but of refractive index 1.65 is used instead, by how much does the deviation increase?
Angular dispersion and mean deviation
As discussed earlier, if a polychromatic beam is incident upon a prism, the emergent beam consists of all the individual colours angularly separated. This is angular dispersion (Fig. 9.22).
It is measured for any two component colours.
\[\therefore \delta_{21} = \delta_2 - \delta_1\]
Normally we do it for extreme colours. For white light, violet and red are the extreme colours.
\[\therefore \delta_{VR} = \delta_V - \delta_R\]
Using the deviation for a thin prism (Eq. 9.13), we can write
\[\therefore \delta_{21} = \delta_2 - \delta_1 = A(n_2 - 1) - A(n_1 - 1)\]
\[= A(n_2 - n_1)\]
where \(n_1\) and \(n_2\) are the refractive indices for the two colours. Note that the angle of the prism \(A\) has not cancelled here — a bigger prism angle gives a bigger spread, which is why a prism disperses far better than a slab.
Also,
\[\delta_{VR} = \delta_V - \delta_R = A(n_V - 1) - A(n_R - 1)\]
\[= A(n_V - n_R) \qquad \text{--- (9.14)}\]
Yellow is practically chosen to be the mean colour for violet and red.
This gives the mean deviation
\[\delta_{VR} = \frac{\delta_V + \delta_R}{2} \cong \delta_Y = A(n_Y - 1) \qquad \text{--- (9.15)}\]
- If you see a rainbow widthwise, yellow appears to be centrally located. Hence the angular deviation of yellow is the average for the entire colour span. This may be the reason for choosing yellow as the mean colour. Remember, the red band is the widest, and violet is much thinner than blue.
- While obtaining the expression for \(\omega\), we have used the thin prism formula for \(\delta\). However, the expression for \(\omega\) (equation 9.16) is valid as well for equilateral prisms or right-angled prisms.
Dispersive power
Dispersive power: The ability of an optical material to disperse constituent colours is its dispersive power. It is measured, for any two colours, as the ratio of the angular dispersion to the mean deviation for those two colours. Thus, for the extreme colours of white light, the dispersive power is given by
\[\omega = \frac{\left[\delta_V - \delta_R\right]}{\left[\dfrac{\delta_V + \delta_R}{2}\right]} \cong \frac{\delta_V - \delta_R}{\delta_Y}\]
\[= \frac{A(n_V - n_R)}{A(n_Y - 1)} = \frac{n_V - n_R}{n_Y - 1} \qquad \text{--- (9.16)}\]
Dispersive power of a transparent material is the ratio of the angular dispersion between two colours to the mean deviation for those two colours. For the extreme colours of white light, \(\omega = \dfrac{n_V - n_R}{n_Y - 1}\).
As \(\omega\) is the ratio of the same physical quantities, it is a unitless and dimensionless quantity.
From the expression in terms of refractive indices it should be understood that dispersive power depends only upon the refractive index (hence upon the material only), and not upon the dimensions of the prism. The angle \(A\) has cancelled out of Eq. (9.16). So a small prism and a large prism of the same glass spread the colours by different absolute amounts, but they spread them by the same fraction of their mean deviation.
For commonly used glasses it is around 0.03.
For a dense flint glass prism of refracting angle \(10^0\), obtain the angular deviation for extreme colours and the dispersive power of dense flint glass. (\(n_{red} = 1.712\), \(n_{violet} = 1.792\))
Solution: Since the prism is thin (\(A = 10^0\)), we may use \(\delta = A(n-1)\) for each colour.
\[\delta_V = A(n_V - 1) = 10(1.792 - 1) = (7.92)°\]
\[\delta_R = A(n_R - 1) = 10(1.712 - 1) = (7.12)°\]
\[\therefore \text{Angular dispersion}, \ \delta_{VR} = \delta_V - \delta_R = (0.8)^0\]
Dispersive power, \(\omega =\)
\[= \frac{\delta_V - \delta_R}{\left(\dfrac{\delta_V + \delta_R}{2}\right)}\]
\[= 2\left(\frac{7.92 - 7.12}{7.92 + 7.12}\right)\]
\[= \frac{2 \times 0.8}{15.04} = 0.1064\]
(This is much higher than popular crown glass, whose dispersive power is around 0.03 — which is exactly why flint glass is chosen when strong dispersion is wanted, and crown glass when it is not.)
9.9 Some Natural Phenomena due to Sunlight
Mirage
On a hot, clear, sunny day, along a level road, a pond of water appears to be there ahead. However, if we physically reach that spot, there is nothing but the dry road — and the water pond again appears ahead. This illusion is called a mirage (Fig. 9.23).
On a hot day, the air in contact with the road is the hottest, and as we go up it gets gradually cooler. Hot air is less dense, so it is optically rarer. The refractive index of air thus increases with height.
This is exactly the situation in which our very first law of ray optics fails: the medium is no longer homogeneous. Its refractive index changes continuously from one layer to the next, so light cannot travel in a straight line through it.
As shown in Fig. 9.23, due to this gradual change in the refractive index, the ray of light coming from the top of an object becomes more and more horizontal as it almost touches the road. For some reason (mentioned in the note below) it bends above. Then onwards, upward bending continues due to the denser air higher up.
As a result, for an observer, the light appears to be coming from below the object, thereby giving an illusion of reflection from an (imaginary) water surface. What we take for a shining pool is really a bent-up image of the bright sky.
Q4. On a hot day, why does a shining “pool of water” appear far ahead on a dry road and disappear when we reach that spot?
A. The air just above the hot road is much hotter, and therefore optically rarer, than the air higher up. Light coming down from the sky or from the top of a distant object bends gradually as it passes through these layers, becomes almost horizontal near the road, and is then turned upwards into the observer’s eye. Since our eyes assume light travels straight, the sky appears to be coming from the road surface, which looks like a shimmering pool. As we approach, the geometry of the grazing rays no longer works for that spot, so the “pool” shifts further ahead and is never reached.
Possible reasons for the upward bending at the road during a mirage could be: (i) The angle of incidence at the road is glancing. At glancing incidence, the reflection coefficient is very large, which causes reflection. (ii) The air almost in contact with the road is not steady. The non-uniform motion of the air bends the ray upwards, and once it has bent upwards, it continues to do so. (iii) Using Maxwell’s equations for EM waves, a correct explanation is possible for the reflection. It may be pointed out that total internal reflection is NEVER possible here, because the relative refractive index is just less than 1, and hence the critical angle (discussed in Section 9.6) is also approaching \(90^0\).
The same physics works upside down over a cold sea. There the air near the water is the coldest and densest, so the refractive index decreases with height, and light from a distant ship curves downward. The ship then appears raised above the horizon — sometimes stretched, and sometimes visible even when it is geometrically below the horizon. Sailors call such superior mirages “looming”, and they can make land appear to float in the sky.
Rainbow
Undoubtedly, the rainbow is an eye-catching phenomenon occurring due to rain and Sunlight. It is most popular because it is observable from anywhere on the Earth.
A few lucky persons might have observed two rainbows simultaneously, one above the other. Some might have seen a complete circular rainbow from an aeroplane (of course, this time it is not a “bow”!). The optical phenomena discussed till now — refraction, dispersion, reflection and deviation — are sufficient to explain the formation of a rainbow.
The facts to be explained are:
It is seen during rains and on the opposite side of the Sun.
It is seen only during mornings and evenings, and not throughout the day.
In the commonly seen rainbow, the red arch is outside and violet is inside.
In the rarely occurring concentric secondary rainbow, the violet arch is outside and red is inside.
It is in the form of an arc of a circle.
A complete circle can be seen from a higher altitude, that is, from an aeroplane.
Total internal reflection is not possible in this case.
Conditions necessary for formation of a rainbow: A light shower with relatively large raindrops, morning or evening time, and enough Sunlight.
Optical phenomena involved: During the formation of a rainbow, the rays of Sunlight incident on water drops deviate and disperse during refraction, then internally (NOT total internally) reflect once (for the primary rainbow) or twice (for the secondary rainbow), and finally refract again into air. At all stages there is angular dispersion, which leads to a clear separation of the colours.
Read the phrase “NOT total internally” carefully — it is a favourite examination point, and the reason behind it is explained in the note that follows this discussion.
Primary rainbow
Figure 9.24 (a) shows the optical phenomena involved in the formation of a primary rainbow due to a spherical water drop.
The white ray AB from the Sun strikes the upper portion of a water drop at an incident angle \(i\). On entering into the water, it deviates and disperses into constituent colours. Extreme colours violet (V) and red (R) are shown.
The refracted rays BV and BR strike the opposite inner surface of the water drop and suffer internal (NOT total internal) reflection. These reflected rays finally emerge from V′ and R′, and can be seen by an observer on the ground. For the observer, they appear to be coming from the opposite side of the Sun — which is why we must always turn our back to the Sun to see a rainbow.
The minimum deviation rays of red and violet colour are inclined to the ground level at \(\theta_R = 42.8° \cong 43°\) and \(\theta_V = 40.8 \cong 41°\) respectively. As a result, in the ‘bow’ or arch, the red is above or outer, and violet is lower or inner.
Secondary rainbow
Figure 9.24 (b) shows some optical phenomena involved in the formation of a secondary rainbow due to a spherical water drop.
The white ray AB from the Sun strikes the lower portion of a water drop at an incident angle \(i\). On entering into the water, it deviates and disperses into constituent colours. Extreme colours violet (V) and red (R) are shown.
The refracted rays BV and BR finally emerge from the drop at V′ and R′ after suffering two internal reflections, and can be seen by an observer on the ground.
The minimum deviation rays of red and violet colour are inclined to the ground level at \(\theta_R \cong 51°\) and \(\theta_V \cong 53°\) respectively. As a result, in the ‘bow’ or arch, the violet is above or outer, and red is lower or inner — exactly the reverse of the primary rainbow. The extra internal reflection is what flips the order of the colours.
Since some light is lost at every reflection and refraction, the secondary rainbow is always fainter than the primary one, and it is seen only when the shower and the sunlight are strong.
Q5. In a rainbow, why is red always the outer arch and violet the inner one, and why do we see rainbows only in the morning or the evening?
A. In the common (primary) rainbow, the light entering a raindrop is dispersed and reflected once inside it. The minimum-deviation rays leave the drop inclined at about \(43°\) for red and about \(41°\) for violet with respect to the ground, so red reaches our eye from higher drops and violet from lower ones — putting red on the outside of the arch. The rainbow is seen only along cones of these fixed angles about the line joining the Sun to the observer. As the Sun climbs higher, the centre of these cones sinks below the horizon, so less and less of the arch remains visible; above a Sun elevation of about \(41°\) the violet arch of the primary rainbow disappears, and beyond about \(53°\) nothing is seen. Hence rainbows appear only when the Sun is low, that is, in the morning or in the evening.
(I) Why is total internal reflection not possible during the formation of a rainbow? The angle of incidence \(i\) in air, at the water drop, cannot be greater than \(90°\). As a result, the angle of refraction \(r\) in the water will always be less than the critical angle. From the accompanying figures (Fig. 9.25 and Fig. 9.26) and by simple geometry, it is clear that this \(r\) itself is the angle of incidence at any point for one or more internal reflections. Obviously, total internal reflection is not possible. (II) Why is a rainbow seen only for a definite angle range with respect to the ground? For clear visibility we must have a beam of enough intensity. From the deviation curve (Fig. 9.20) it is clear that near minimum deviation the curve is almost parallel to the \(x\)-axis, that is, for the majority of angles of incidence in this range, the angle of deviation is nearly the same, and those rays are almost parallel, forming a beam of enough intensity. Thus, the rays in the near vicinity of minimum deviation are almost parallel to each other. Rays beyond this range suffer wide angular dispersion and thus will not have enough intensity for visibility. By using simple geometry for Fig. 9.25 and Fig. 9.26, it can be shown that the angle of deviation between the final emergent ray and the incident ray is \(\delta = \pi + 2i - 4r\) during the primary rainbow, and \(\delta = 2\pi + 2i - 6r\) during the secondary rainbow. Using these relations and Snell’s law, \(\sin i = n \sin r\), we can obtain the derivatives of \(\delta\). The second derivative of \(\delta\) comes out to be negative, which shows that it is the minima condition. Equating the first derivative to zero, we can obtain the corresponding values of \(i\) and \(r\). Again, by using Fig. 9.25 and Fig. 9.26, we can obtain the corresponding angles \(\theta_R\) and \(\theta_V\) at the horizontal, which is the visible angular position for the rainbow. (III) Why is the rainbow a bow or an arch? Can we see a complete circular rainbow? Figure 9.27 illustrates the formation of primary and secondary rainbows with their common centre O, which is the point where the line joining the Sun and the observer meets the Earth when extended. P is the location of the observer. Different colours of the rainbows are seen on arches of cones of the respective angles described earlier. The smallest half angle refers to the cone of violet colour of the primary rainbow, which is \(41^0\). As the Sun rises, the common centre of the rainbows moves down. Hence, as the Sun comes up, a smaller and smaller part of the rainbows will be seen. If the Sun is above \(41^0\), the violet arch of the primary rainbow cannot be seen. Obviously, beyond \(53^0\), nothing will be seen. That is why rainbows are visible only during mornings and evenings. However, if the observer moves up (may be in an aeroplane), the line PO itself moves up, making the lower part of the arches visible. After a certain minimum elevation, the entire circle for all the cones can be visible. (IV) Size of water drops convenient for a rainbow: Water drops responsible for the formation of a rainbow should not be too small. For too small drops, the phenomenon of diffraction (redistribution of energy due to obstacles, discussed later) dominates, and a clear rainbow cannot be seen.
9.10 Defects of Lenses (Aberrations of Optical Images)
As mentioned in Section 9.4 for aberration of curved mirrors, while deriving the various relations we assume most of the rays to be paraxial, by using lenses of small aperture.
In reality, we have objects of finite sizes. Also, we need optical devices of large apertures — lenses and/or mirrors of size a few metres for telescopes, and so on. A large aperture is unavoidable there, because a big lens collects more light and shows fainter objects.
In such cases the beam of rays is no more paraxial, and quite often not parallel either. As a result, the spherical aberration discussed for spherical mirrors can occur for lenses also. That defect corresponds to a monochromatic beam of light, and it is the only such defect mentioned here.
But a lens suffers from one more defect that a mirror can never have, and we take that up first.
Chromatic aberration
In the case of mirrors, there is no dispersion of light due to refractive index — reflection does not depend on the material at all, so all colours are reflected identically.
However, lenses are prepared by using a transparent material medium having a different refractive index for different colours. Hence angular dispersion is present in a lens.
To see why, note that a convex lens can be approximated to two thin prisms connected base to base, and for a concave lens those are vertex to vertex, as shown in Fig. 9.28 (a) and Fig. 9.28 (b). Each half of the lens therefore disperses light exactly as a prism does.
If the lens is thick, this will result in notably different foci corresponding to each colour for a polychromatic beam, like white light. This defect is called chromatic aberration, with violet being focused closest to the pole, as it has the maximum deviation. See Fig. 9.29 (a) and Fig. 9.29 (b).
Chromatic aberration is the image defect in which a lens forms images of different colours at different points along the axis, because the refractive index of the lens material is different for different colours.
Longitudinal chromatic aberration, transverse chromatic aberration and the circle of least confusion are defined in the same manner as that of spherical aberration for spherical mirrors — that is, the axial separation of the extreme foci, the radius of the smallest blur circle, and that smallest blur circle itself.
Reducing or eliminating chromatic aberration
Eliminating chromatic aberration simultaneously for all the colours is impossible. We try to eliminate it for the extreme colours, which in turn reduces it for the other colours as well.
Convenient methods to do this use either a convex and a concave lens in contact, or two thin convex lenses with proper separation. Such a combination is called an achromatic combination.
The idea behind it is simple: a converging lens and a diverging lens spread the colours in opposite senses. If their dispersions are made equal and opposite, the colour spreading cancels while a net converging power still remains.
An achromatic combination is a combination of lenses so chosen that the resultant focal length is the same for the two extreme colours, so that chromatic aberration is eliminated for those colours.
Achromatic combination of two lenses in contact
Let \(\omega_1\) and \(\omega_2\) be the dispersive powers of the materials of the two component lenses used in contact for an achromatic combination. Their focal lengths \(f\) for violet, red and yellow (assumed to be the mean colour) are suffixed by the respective letters V, R and Y.
Also, let
\[K_1 = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)_1 \quad \text{for lens 1, and}\]
\[K_2 = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)_2 \quad \text{for lens 2.}\]
These \(K\) factors are purely geometrical — they depend only on the shape of each lens, not on the colour of the light.
For two thin lenses in contact,
\[\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \ \ldots\ldots\]
to be used separately for the respective colours.
For the combination to be achromatic, the resultant focal length of the combination must be the same for both the colours, that is,
\[f_V = f_R \quad \text{or} \quad \frac{1}{f_V} = \frac{1}{f_R}\]
\[\therefore \frac{1}{f_{1V}} + \frac{1}{f_{2V}} = \frac{1}{f_{1R}} + \frac{1}{f_{2R}}\]
Now use the lens makers’ equation (9.7) for each lens and each colour, writing \(1/f = (n-1)K\):
\[(n_{1V}-1) K_1 + (n_{2V}-1) K_2 = (n_{1R}-1) K_1 + (n_{2R}-1) K_2\]
\[\ldots\ldots \text{using lens makers' Eq. (9.7)}\]
Collecting terms in \(K_1\) and \(K_2\),
\[\therefore \frac{K_1}{K_2} = \frac{n_{2V} - n_{2R}}{n_{1V} - n_{1R}} \qquad \text{--- (9.17)}\]
For the mean colour, yellow,
\[\frac{1}{f_Y} = \frac{1}{f_{1Y}} + \frac{1}{f_{2Y}}\]
\[\text{with} \quad \frac{1}{f_{1Y}} = (n_{1Y}-1)K_1\]
\[\text{and} \quad \frac{1}{f_{2Y}} = (n_{2Y}-1)K_2\]
Dividing one by the other,
\[\therefore \frac{K_1}{K_2} = \left(\frac{n_{2Y}-1}{n_{1Y}-1}\right)\left(\frac{f_{2Y}}{f_{1Y}}\right) \qquad \text{--- (9.18)}\]
Equating the R.H.S. of (9.17) and (9.18) and rearranging, we can write
\[\frac{f_{2Y}}{f_{1Y}} = -\left(\frac{n_{2V} - n_{2R}}{n_{2Y}-1}\right) \div \left(\frac{n_{1V} - n_{1R}}{n_{1Y}-1}\right)\]
Each bracket is precisely the dispersive power of that material, as defined in Eq. (9.16). Hence
\[= -\frac{\omega_2}{\omega_1} \qquad \text{--- (9.19)}\]
Equation (9.19) is the condition for an achromatic combination of two lenses, in contact.
Condition for achromatism of two thin lenses in contact: \(\dfrac{f_{2Y}}{f_{1Y}} = -\dfrac{\omega_2}{\omega_1}\), where \(\omega_1\) and \(\omega_2\) are the dispersive powers of the two lens materials and \(f_{1Y}\), \(f_{2Y}\) are their focal lengths for the mean (yellow) colour.
Dispersive power \(\omega\) is always positive. Thus, the minus sign in Eq. (9.19) forces \(f_{1Y}\) and \(f_{2Y}\) to have opposite signs — that is, one of the lenses must be convex and the other concave.
If the second lens is concave, \(f_{2Y}\) is negative.
\[\therefore \frac{1}{f_Y} = \frac{1}{f_{1Y}} - \frac{1}{f_{2Y}}\]
For this combination to be converging (which is what we usually want, since the combination has to form an image), \(f_Y\) should be positive.
Hence, \(f_{1Y} < f_{2Y}\) and \(\omega_1 < \omega_2\)
Thus, for an achromatic combination, if there is a choice between flint glass (\(n = 1.655\)) and crown glass (\(n = 1.517\)), the convergent (convex) lens must be of crown glass and the divergent (concave) lens of flint glass. Crown glass has the smaller dispersive power, and by the condition just derived it must be the converging component.
The objective lens of a good telescope, a camera or a pair of quality binoculars is usually not a single lens but an achromatic doublet: a convex crown-glass lens cemented to a concave flint-glass lens. The pair is designed exactly by Eq. (9.19), so the red and violet images fall at the same place. Without it, every bright star would be ringed with colour and every photograph would show coloured fringes at sharp edges.
After a cataract operation, a person is recommended concavo-convex spectacles of curvatures 10 cm and 50 cm. Crown glass of refractive indices 1.51 for red and 1.53 for violet colours is used for this. Calculate the lateral chromatic aberration occurring due to these glasses.
Solution: For a concavo-convex lens, both the radii of curvature are either positive or both negative. If the convex shape faces the object, both will be positive. See the accompanying figure.
Diagram
Fig. 9.29 (c) – Concavo-convex lens with its convex face receiving the incident rays, showing centres of curvature C₁ and C₂ and both radii R₁ and R₂ positive.
\[\therefore R_1 = 10 \ \text{cm} \ \text{and} \ R_2 = 50 \ \text{cm}\]
\[\therefore \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \left(\frac{1}{+10} - \frac{1}{+50}\right) = 0.08 \ \text{cm}^{-1}\]
Now apply the lens makers’ equation separately for each colour, since only \(n\) changes.
\[\therefore \frac{1}{f_R} = (n_R - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]
\[= (1.51 - 1) \times 0.08 = 0.0408\]
\[\therefore f_R = 25.51 \ \text{cm}\]
and
\[\frac{1}{f_V} = (n_V - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]
\[= (1.53 - 1) \times 0.08 = 0.0424\]
\[\therefore f_V = 23.58 \ \text{cm}\]
\(\therefore\) Longitudinal chromatic aberration
\[= f_V - f_R = 25.51 - 23.58\]
\[= 1.93 \ \text{cm}, \ \ldots \text{(quite appreciable!)}\]
Verify that you get the same answer even if you consider the concave surface facing the incident rays.
A thin double convex lens has both radii of curvature of magnitude 24 cm. The refractive index of its glass is 1.60 for violet and 1.55 for red light. Calculate the focal length of the lens for each colour and hence the longitudinal chromatic aberration produced by it.
Spherical aberration in lenses
Spherical aberration: Longitudinal spherical aberration, transverse spherical aberration and the circle of least confusion are defined in the same manner as that for spherical mirrors. Notice in Fig. 9.30 (a) and Fig. 9.30 (b) that rays passing through the outer (marginal) parts of the lens are bent more strongly than the paraxial rays, so they cross the axis nearer the lens.
Methods to reduce or eliminate spherical aberration of lenses
(i) The cheapest method to reduce spherical aberration is to use a planoconvex or planoconcave lens with the curved side facing the incident rays (the real object). Reversing it increases the aberration appreciably. The reason is that the total bending is then shared roughly equally between the two surfaces, instead of being dumped almost entirely on one of them.
(ii) A certain ratio of radii of curvature for a given refractive index almost eliminates the spherical aberration. For \(n = 1.5\), the ratio is
\[\frac{R_1}{R_2} = \frac{1}{6}\]
and for \(n = 2\), it is \(\dfrac{1}{5}\).
(iii) The use of two thin converging lenses separated by a distance equal to the difference between their focal lengths, with the lens of larger focal length facing the incident rays, considerably reduces spherical aberration.
(iv) Spherical aberration of a convex lens is positive (for a real image), while that of a concave lens is negative. Thus, a suitable combination of them — preferably a double convex lens of smaller focal length and a planoconcave lens of greater focal length — can completely eliminate spherical aberration, since the two opposite aberrations cancel each other.
9.11 Optical Instruments
Introduction
Whether an object appears bigger or not does not necessarily depend upon its own size. Huge mountains far off may appear smaller than a small tree close to us.
This is because the angle subtended by the mountain at the eye from that distance — called the visual angle — is smaller than that subtended by the tree from its position. Hence, the apparent size of an object depends upon the visual angle subtended by the object from its position.
Obviously, then, for an object to appear bigger, we must bring it closer to us, or we should go closer to it. That is what we instinctively do when we cannot read small print.
However, due to the limitation in focusing of the eye lens, it is not possible to take an object closer than a certain distance. This distance is called the least distance of distinct vision, \(D\). For a normal, unaided human eye, \(D = 25\) cm. If an object is brought closer than this, we cannot see it clearly.
The least distance of distinct vision \(D\) is the smallest distance from the eye at which an object can be seen clearly. For a normal, unaided human eye, \(D = 25\) cm.
Now consider the two situations in which the naked eye fails us.
If an object is too small (like the legs of an ant), the corresponding visual angle from 25 cm is not enough to see it, and if we bring it closer than that, its image on the retina is blurred.
Also, the visual angle made by cosmic objects far away from us (such as stars) is too small to make out minor details, and we cannot bring those closer at all.
In such cases we need optical instruments — a microscope in the former case and a telescope in the latter. It means that microscopes and telescopes help us in increasing the visual angle. This is called angular magnification or magnifying power.
Magnifying power
Magnifying power: Angular magnification or magnifying power of an optical instrument is defined as the ratio of the visual angle made by the image formed by that optical instrument (\(\beta\)) to the visual angle subtended by the object when kept at the least distance of distinct vision (\(\alpha\)). See Fig. 9.31 (a) and Fig. 9.31 (b).
In the case of telescopes, \(\alpha\) is the angle subtended by the object from its own position, as it is not possible to get it closer.
Magnifying power (angular magnification) of an optical instrument is \(M = \dfrac{\beta}{\alpha}\), where \(\beta\) is the visual angle made by the image formed by the instrument, and \(\alpha\) is the visual angle made by the object kept at the least distance of distinct vision (or, for a telescope, at its own position).
Note carefully that magnifying power is not the same thing as lateral magnification \(m = v/u\). Lateral magnification compares the size of the image with the size of the object. Magnifying power compares the angles the two subtend at the eye — and it is the angle, not the size, that decides how big something looks.
Simple microscope or a reading glass
In order to read very small letters in a newspaper, sometimes we use a convex lens. You might have seen watch-makers using a special type of small convex lens while looking at very tiny parts of a wrist watch. The convex lens used for this purpose is a simple microscope.
Figure 9.31 (a) shows the visual angle \(\alpha\) made by an object when kept at the least distance of distinct vision \(D\). Without an optical instrument this is the greatest possible visual angle, as we cannot get the object closer than this.
Figure 9.31 (b) shows a convex lens forming an erect, virtual and magnified image of the same object, when the object is placed within the focus of the lens.
The visual angle \(\beta\) of the object and of the image in this case are the same, since the image lies along the same rays. However, this time the viewer is looking at the image, which is not closer than \(D\). Hence the same object is now at a distance smaller than \(D\). It makes \(\beta\) greater than \(\alpha\), and the same object appears bigger.
This is the entire trick of a simple microscope: the lens allows the object itself to be brought much closer than 25 cm, while placing its image far enough away for the eye to focus on comfortably.
Angular magnification or magnifying power, in this case, is given by
\[M = \frac{\text{Visual angle of the image}}{\text{Visual angle of the object, when at } D} = \frac{\beta}{\alpha}\]
For small angles \(\alpha\) and \(\beta\), we can write,
\[M = \frac{\beta}{\alpha} \cong \frac{\tan(\beta)}{\tan(\alpha)} = \frac{\left(\dfrac{BA}{PA}\right)}{\left(\dfrac{BA}{D}\right)} = \frac{D}{u} \quad \text{- (numerically)}\]
where \(u\) is the distance of the object from the lens and \(D = 25\) cm for a normal eye.
Limiting cases
(i) For maximum magnifying power, the image should be nearest possible, that is, at \(D\) — because the closer the image, the larger the angle it subtends.
For a thin lens,
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]
In this case, \(f = +f\), \(v = v_{min} = -D\), \(u = -u\) and \(M = M_{max}\)
\[\therefore \frac{1}{f} = \frac{1}{-D} - \frac{1}{-u} \ \therefore \frac{D}{f} = \frac{D}{-D} + \frac{D}{u}\]
\[\therefore M_{max} = \frac{D}{u} = 1 + \frac{D}{f}\]
(ii) For minimum magnifying power, \(v = \infty\), that is, \(u = f\) (numerically). The image is then at infinity and the eye is fully relaxed.
\[\therefore M_{min} = \frac{D}{u} = \frac{D}{f}\]
Thus the angular magnification by a lens of focal length \(f\) is between \(\left(\dfrac{D}{f}\right)\) and \(\left(1 + \dfrac{D}{f}\right)\) only.
For common human eyesight, \(D = 25\) cm. Thus, if \(f = 5\) cm,
\[M_{min} = \left(\frac{D}{f}\right) = 5 \quad \text{and} \quad M_{max} = \left(1 + \frac{D}{f}\right) = 6\]
Hence the image appears to be only 5 to 6 times bigger for a lens of focal length 5 cm.
For \(M_{min} = \left(\dfrac{D}{f}\right) = 5\), \(v = \infty\). \(\therefore m = \dfrac{v}{u} = \infty\).
Thus, the image size is infinite times that of the object, but it appears only 5 times bigger. This striking contrast is the clearest proof that size and apparent size are two different things — what the eye judges is the angle.
For
\[M_{max} = 1 + \left(\frac{D}{f}\right) = 6,\]
\(v = -25\) cm. Corresponding \(u = \dfrac{-25}{6}\) cm
\[\therefore m = \frac{v}{u} = 6\]
Thus, the image size is 6 times that of the object, and it appears also 6 times larger.
Q6. Why can a single magnifying glass never make letters look more than a few times bigger, and how do microscopes and telescopes get around this limit?
A. For a simple microscope the magnifying power lies only between \(D/f\) and \(1 + D/f\). With \(D = 25\) cm, even a fairly strong lens of \(f = 5\) cm gives just 5 to 6 times. To do better we would need a very small \(f\), but such a lens becomes thick and suffers badly from spherical and chromatic aberration. The way out is to use two lenses in series: an objective that first forms a magnified real image, and an eye lens that then magnifies that image. The two magnifications multiply, which is exactly how a compound microscope works. A telescope uses the same two-lens idea for the opposite problem — objects that cannot be brought closer at all.
A magnifying glass of focal length 10 cm is used to read letters of thickness 0.5 mm held 8 cm away from the lens. Calculate the image size. How big will the letters appear? Can you read the letters if held 5 cm away from the lens? If yes, of what size would the letters appear? If no, why not?
Solution: The object is within the focus (8 cm < 10 cm), so a virtual, erect, magnified image is expected.
\[f = +10 \ \text{cm}, \ u = -8 \ \text{cm}, \ v = ?\]
\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \ \therefore \frac{1}{10} = \frac{1}{v} - \frac{1}{-8} \ \therefore v = -40 \ \text{cm}\]
The negative sign confirms that the image is virtual and on the same side as the object.
\[m = \frac{v}{u} = \frac{\text{Image size } h_i}{\text{Object size } h_o} \ \therefore \frac{40}{8} = \frac{h_i}{0.5}\]
\[\therefore h_i = 2.5 \ \text{cm} \ (5 \ \text{times that of the object})\]
But how big it appears is decided by the magnifying power, not by \(m\):
\[M = \frac{D}{u} = \frac{25}{8} = 3.125\]
\(\therefore\) The image will appear to be 3.125 times bigger, i.e., \(3.125 \times 0.5 = 1.5625\) cm
For \(u = -5\) cm, \(v\) will be \(-10\) cm.
For an average human being to see clearly, the image must be at or beyond 25 cm. Here the image would be formed only 10 cm from the lens, which is inside the least distance of distinct vision. Thus, it will not be possible to read the letters if they are held 5 cm away from the lens.
Compound microscope
As seen above, the magnifying power of a simple microscope is inversely proportional to its focal length. However, if we need the focal length to be smaller and smaller, the corresponding lens becomes thicker and thicker. For such a lens both spherical as well as chromatic aberrations are dominant, so the image quality collapses even as the magnification rises.
Thus, if higher magnifying power is needed, we go for using more than one lens. The instrument is then called a compound microscope. It is used to view very small objects (sizes \(\sim 10^{-1}\) mm to \(10^{-3}\) mm). Also, whether the image is erect or inverted is immaterial here.
A compound microscope essentially uses two convex lenses of suitable focal lengths, fitted into a cylindrical tube with some adjustment possible for its length. The smaller lens (\(\sim\) 4 mm to 6 mm aperture) facing the object is called the objective. The other lens, with which the observer jams her/his eye, is a little larger and is called the eye lens (Fig. 9.32). During this discussion we consider the eye lens to be a single lens, but in practice it is an eyepiece, itself consisting of two planoconvex lenses.
As shown in Fig. 9.32, a tiny object AB is placed between \(f\) and \(2f\) of the objective, which produces its real, inverted and magnified image A′B′ in front of the eye lens.
The position of the eye lens is so adjusted that the (intermediate) image A′B′ is within its focus. Hence, for this object A′B′, the eye lens behaves as a simple microscope and produces its virtual and magnified image A″B″, which is inverted with respect to the original object AB.
Magnifying power of a compound microscope with two lenses: From its position, the final image A″B″ makes a visual angle \(\beta\) at the eye (jammed at the eye lens). The visual angle made by the object from distance \(D\) is \(\alpha\).
\[\therefore \tan\beta = \frac{\text{A''B''}}{v_e} = \frac{\text{A'B'}}{u_e}\]
\[\tan\alpha = \frac{\text{AB}}{D} \quad \text{(Fig. 9.31 (a))}\]
\(\therefore\) Angular magnification or magnifying power,
\[M = \frac{\beta}{\alpha} \cong \frac{\tan\beta}{\tan\alpha} = \left(\frac{\text{A'B'}}{u_e}\right) \times \left(\frac{D}{\text{AB}}\right)\]
\[= \left(\frac{\text{A'B'}}{\text{AB}}\right) \times \left(\frac{D}{u_e}\right)\]
\[\therefore M = m_o \times M_e\]
Where,
\[\left(\frac{\text{A'B'}}{\text{AB}}\right) = m_o = \frac{v_o}{u_o}\]
is the linear (lateral) magnification of the objective, and
\[\left(\frac{D}{u_e}\right) = M_e\]
is the angular magnification or magnifying power of the eye lens.
So the two stages simply multiply: the objective enlarges the object into a real image, and the eye lens then works on that image exactly as a simple microscope. This is why a compound microscope can go far beyond the 5-to-6 times limit of a single lens.
The length of the compound microscope then becomes
\[L = \text{distance between the two lenses} = v_0 + u_e\]
Remarks:
In order to increase \(m_o\), we need to decrease \(u_o\). Thereby, the object comes closer and closer to the focus of the objective. This increases \(v_0\) and hence the length of the microscope. Thus, \(m_o\) can be increased only within the limitation of the length of the microscope.
The minimum value of \(M_e\) is \(\left(\dfrac{D}{f_e}\right)\) for the final image at infinity, and the maximum value of \(M_e\) is \(\left(1 + \dfrac{D}{f_e}\right)\) for the final image at \(D\), respectively. \(M_e\) and \(m_o\) together decide the minimum and maximum magnifying power of the microscope.
The pocket microscope used by a student consists of an eye lens of focal length 6.25 cm and an objective of focal length 2 cm. At a microscope length of 15 cm, the final image appears biggest. Estimate the distance of the object from the objective and the magnifying power of the microscope.
Solution: “Final image appears biggest” means the final image is formed at the least distance of distinct vision, so \(v_e = 25\) cm.
\[f_e = 6.25 \ \text{cm}, \ f_o = 2 \ \text{cm}, \ L = \lvert v_o \rvert + \lvert u_e \rvert\]
\[= 15 \ \text{cm}, \ v_e = 25 \ \text{cm} \ (\text{Image appears largest})\]
Work backwards: first the eye lens, then the objective.
\[\frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \ \therefore \frac{1}{6.25} = \frac{4}{25}\]
\[= \frac{1}{-25} - \frac{1}{u_e} \ \therefore \lvert u_e \rvert = 5 \ \text{cm}\]
\[\therefore \lvert v_o \rvert = L - \lvert u_e \rvert = 15 - 5 = 10 \ \text{cm}\]
\[\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \ \therefore \frac{1}{2} = \frac{1}{+10} - \frac{1}{u_o} \ \therefore \lvert u_o \rvert = 2.5 \ \text{cm}\]
\[M = m_o \times M_e = \left(\frac{v_0}{u_o}\right)\left(\frac{D}{u_e}\right)\]
\[= \left(\frac{10}{2.5}\right)\left(\frac{25}{5}\right) = 4 \times 5 = 20\]
A compound microscope has an objective of focal length 1.5 cm and an eye lens of focal length 5 cm. An object is kept 1.8 cm from the objective, and the final image is formed at the least distance of distinct vision. Calculate (i) the image distance for the objective, (ii) the lateral magnification of the objective, (iii) the magnifying power of the eye lens, and (iv) the total magnifying power of the microscope.
Telescope
Telescope: Telescopes are used to see terrestrial or astronomical bodies. A telescope essentially uses two lenses (or one large parabolic mirror and a lens). The lens facing the object, called the objective, is of aperture as large as possible, so that it collects as much light as possible from a faint, distant object. For Newtonian telescopes, a large parabolic mirror faces the object.
For terrestrial telescopes, the objects to be seen are on the Earth — like mountains, trees, players playing a match in a stadium, and so on. In such a case, the final image must be erect. The eye lens used for this purpose must be concave, and such a telescope is popularly called a binocular. A variety of binoculars use three convex lenses with proper separation; the third lens again inverts the second intermediate image and makes the final image erect with respect to the object. In this text we shall be discussing the astronomical telescope.
For an astronomical telescope, the objects to be seen are planets, stars, galaxies, etc. In this case there is no necessity of an erect image — a star looks the same either way. Such telescopes use a convex lens as the eye lens (Fig. 9.33).
In 1609–1610, Galileo Galilei built telescopes of his own and turned them towards the sky. He saw craters and mountains on the Moon, and four small points of light beside Jupiter that changed position from night to night — the moons of Jupiter, orbiting a body other than the Earth. A modest instrument of two lenses, magnifying only about twenty times, thus changed our picture of the Solar System.
Magnifying power of a telescope: The objects to be seen through a telescope cannot be brought to a distance \(D\) from the objective, as they can in microscopes. Hence, for telescopes, \(\alpha\) is the visual angle of the object from its own position, which is practically at infinity.
The visual angle of the final image is \(\beta\), and its position can be adjusted to be at \(D\). However, under normal adjustment, the final image is also at infinity, but making a greater visual angle than that of the object. (If the image is really at infinity, there will not be any parallax at the cross wires.)
The beam of incident rays is now inclined at an angle \(\alpha\) with the principal axis, while the emergent beam is inclined at a greater angle \(\beta\) with the principal axis, causing angular magnification (Fig. 9.33).
The objective of focal length \(f_o\) focusses the parallel incident beam at a distance \(f_o\) from the objective, giving an inverted image AB. For normal adjustment, the eye lens is so adjusted that the intermediate image AB happens to be at the focus of the eye lens. Rays refracted beyond the eye lens then form a parallel beam inclined at an angle \(\beta\) with the principal axis, resulting in the final image also at infinity.
\(\therefore\) Angular magnification or magnifying power,
\[M = \frac{\beta}{\alpha} \cong \frac{\tan\beta}{\tan\alpha} = \frac{\left(\dfrac{\text{BA}}{\text{P}_e\text{B}}\right)}{\left(\dfrac{\text{BA}}{\text{P}_o\text{B}}\right)} = \frac{\left(\dfrac{\text{BA}}{f_e}\right)}{\left(\dfrac{\text{BA}}{f_o}\right)}\]
\[\therefore M = \frac{f_o}{f_e}\]
The common intermediate image size BA cancels, leaving a result of pleasing simplicity: the magnifying power of a telescope is just the ratio of the two focal lengths.
The length of the telescope for normal adjustment is
\[L = f_o + f_e\]
Under the allowed limit of length, an objective of maximum possible focal length \(f_o\) and an eye lens of minimum possible focal length \(f_e\) can be chosen for maximum magnifying power.
The focal length of the objective of an astronomical telescope is 1 m. Under normal adjustment, the length of the telescope is 1.05 m. Calculate the focal length of the eyepiece and the magnifying power under normal adjustment.
Solution: For an astronomical telescope, under normal adjustment the length is simply the sum of the two focal lengths.
\[L = f_0 + f_e \ \therefore 1.05 = 1 + f_0 \ \therefore f_e = 0.05 \ \text{m} = 5 \ \text{cm}\]
Under normal adjustments,
\[M = \frac{f_o}{f_e} = \frac{1}{0.05} = 20\]
9.12 Summary
Nature of light
- Optics is the science of light. Light is a form of energy that travels as an electromagnetic (EM) wave and also behaves as a stream of energy-carrying particles called photons. Light thus has a dual nature.
- In vacuum, light travels with speed \(c = 299792458 \ \text{m s}^{-1}\), taken as \(c = 3 \times 10^{8} \ \text{m s}^{-1}\) for practical purposes. By Einstein’s special theory of relativity, this is the maximum possible speed for any object.
- In a material medium, the speed of EM waves is \(c = \sqrt{\dfrac{1}{\epsilon \mu}}\), where \(\epsilon\) is the permittivity and \(\mu\) the permeability of that medium.
- Phenomena of light are studied in three ways: ray (geometrical) optics — reflection, refraction, total internal reflection; wave (physical) optics — interference, diffraction, polarization, Doppler effect; and the particle (quantum) nature of light — photoelectric effect, spectral lines, Compton effect.
Foundations of ray optics
- A ray of light is a particular direction of propagation of energy from a source.
- The four fundamental laws are: (i) light travels in a straight line in a homogeneous and isotropic medium; (ii) rays intersect without affecting each other’s paths; (iii) laws of reflection — incident ray, reflected ray and normal are coplanar with the rays on either side of the normal, and \(i = r\); (iv) laws of refraction — incident ray, refracted ray and normal are coplanar with the rays on either side of the normal, and Snell’s law, \((n_1)\sin\theta_1 = (n_2)\sin\theta_2\).
- All four laws follow from Fermat’s principle: a ray of light travelling from one point to another by one or more reflections or refractions chooses the path of extreme (least or greatest) time.
Cartesian sign convention
- All distances are measured from the pole (optical centre), taken as origin, with incident light travelling from left to right.
- Distances to the right of the pole are positive, to the left negative. Distances above the principal axis are positive, below it negative.
- A diverging incident beam corresponds to a real object (to the left of the pole); a converging incident beam corresponds to a virtual object (to the right of the pole); a parallel beam corresponds to an object at infinity.
- Unless stated otherwise, objects are taken to be real.
Reflection
- A plane mirror forms a virtual, laterally inverted image of the same size, at the same distance behind the mirror.
- For two plane mirrors inclined at angle \(\theta\), calculate \(n = \dfrac{360}{\theta}\). If \(n\) is an even integer, \(N = (n-1)\) for any object position. If \(n\) is an odd integer, \(N = (n-1)\) for an object on the angle bisector and \(N = n\) for an object off it. If \(n\) is not an integer, \(N\) is the integral part of \(n\).
- For a spherical mirror, \(f = \dfrac{R}{2}\). With incident light from the left, the focal length of a convex mirror is positive and that of a concave mirror is negative.
- Mirror formula: \(\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\) (Eq. 9.1), valid for a small mirror whose aperture is at least ten times smaller than \(u\), \(v\) and \(f\).
- Focal power: \(P = \dfrac{1}{f}\), with \(f\) in metre; SI unit is the dioptre, \(1 \ D = 1 \ m^{-1}\).
- Lateral magnification: \(m = \dfrac{v}{u}\).
- A convex mirror always forms a virtual, erect, diminished image (\(m < 1\)). For a concave mirror, the image depends on the object position (see Table 9.2).
- Spherical aberration: rays far from the axis focus nearer the pole than paraxial rays, giving a fuzzy image. The axial separation \(F_M F_P\) is the longitudinal spherical aberration; the smallest blur circle is the circle of least confusion, and its radius is the transverse spherical aberration.
- A parabolic mirror has a single point focus and hence no spherical aberration, since all parallel rays travel equal optical paths to the focus. Search lights, torches, headlights and telescope mirrors are therefore parabolic.
Refraction
- Refraction is the change in speed, wavelength and (for oblique incidence) direction of light on entering a medium of different refractive index. At normal incidence there is refraction but no deviation.
- Absolute refractive index: \(n = \dfrac{c}{v}\); unitless and dimensionless. For every material medium \(n > 1\); a medium of greater \(n\) is optically denser (which need not mean physically denser).
- Relative refractive index: \(^1n_2 = \dfrac{n_2}{n_1} = \dfrac{v_1}{v_2}\).
- For refraction at a plane surface, \(n \cong \dfrac{Real \ depth}{Apparent \ depth}\), obtained using the small-angle approximation \(\sin\theta \cong \theta \cong \tan\theta\) (valid within about 0.51 % error for \(\theta < 10^0\)).
Total internal reflection
- Total internal reflection (TIR) occurs when light travelling in a denser medium strikes the boundary of a rarer medium at an angle of incidence greater than or equal to the critical angle. There is then total reflection and no refraction.
- The critical angle is that angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \(90°\): \((\mu)\sin(i_c) = (1)\sin 90°\), so \(\sin(i_c) = \dfrac{1}{\mu}\).
- For glass, \(\mu = 1.5\) gives \(i_c = 41°49' \cong 42°\); for water, \(\mu = \dfrac{4}{3}\) gives \(i_c = 48°35'\) (both with respect to air).
- Applications: optical fibre (denser core, rarer cladding; broad bandwidth, immunity to EM interference, attenuation below 0.2 dB/km, electrical insulation, theft prevention, security), prism binoculars, and the periscope.
Refraction at a spherical surface and lenses
- Refraction at a single spherical surface: \(\dfrac{n_2 - n_1}{R} = \dfrac{n_2}{v} - \dfrac{n_1}{u}\) (Eq. 9.4), where \(n_1\) is the refractive index of the medium of the real object and \(n_2\) that of the other medium. It holds for any such surface, provided all values are substituted with Cartesian signs.
- Thin lens formula: \(\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\) (Eq. 9.2). A lens is thin if its thickness is at least 50 times smaller than \(R_1\), \(R_2\), \(u\), \(v\) and \(f\).
- Lens makers’ equation: \(\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\) (Eq. 9.7). Note that \(f = R/2\) does NOT hold for a lens.
- Special cases (thin, symmetric lenses with \(\lvert R_1 \rvert = \lvert R_2 \rvert = R\)): double convex, \(\dfrac{1}{f} = \dfrac{2(n-1)}{R}\), giving \(f = R\) if \(n \cong 1.5\); double concave, \(\dfrac{1}{f} = \dfrac{2(n-1)}{-R}\), giving \(f = -R\) if \(n \cong 1.5\); planoconvex, \(\dfrac{1}{f} = \dfrac{n-1}{R}\), giving \(f = 2\lvert R \rvert\) if \(n \cong 1.5\).
- Thin lenses in contact: \(\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \dfrac{1}{f_3} + \ldots = \sum P_i = P\).
- Two thin lenses separated in air by distance \(d\): \(\dfrac{1}{f} = \dfrac{1}{f_1} + \dfrac{1}{f_2} - \dfrac{d}{f_1 f_2} = P_1 + P_2 - dP_1P_2 = P\).
- For lenses denser than the surroundings, a convex lens has positive focal length and converges light, a concave lens has negative focal length and diverges it. For a lens of rarer material (e.g. an air lens in water), the behaviour reverses.
Dispersion and prisms
- The refractive index of a material increases with frequency, so violet is refracted most and red least. This separation of colours is dispersion.
- A plane parallel slab gives only lateral dispersion, which is too small to detect; prisms are used for appreciable angular dispersion.
- Prism relations: \(A = r_1 + r_2\) (Eq. 9.10) and \(i + e = A + \delta\) (Eq. 9.11).
- The deviation curve shows that \(\delta\) falls to a minimum \(\delta_m\) and then rises. At \(\delta = \delta_m\), \(i = e\) and \(r_1 = r_2 = \dfrac{A}{2}\).
- Prism formula: \(n = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}\) (Eq. 9.12). For an equilateral prism, \(n = 2\sin\left(30 + \dfrac{\delta_m}{2}\right)\).
- Grazing emergence (\(e = 90°\)) occurs when \(r_2\) equals the critical angle; this fixes \(i_{min}\). For an equilateral glass prism of \(n = 1.5\), \((r_2)_{max} = 41°49'\), \(r_1 = 18°11'\) and \(i_{min} = 27°55' \cong 28^0\).
- Thin prism (\(A < 10^0\)): \(\delta = A(n-1)\) (Eq. 9.13), independent of the angle of incidence.
- Angular dispersion between two colours: \(\delta_{21} = A(n_2 - n_1)\); for extreme colours, \(\delta_{VR} = A(n_V - n_R)\) (Eq. 9.14). Mean (yellow) deviation: \(\delta_{VR} = \dfrac{\delta_V + \delta_R}{2} \cong \delta_Y = A(n_Y - 1)\) (Eq. 9.15).
- Dispersive power: \(\omega = \dfrac{\delta_V - \delta_R}{\delta_Y} = \dfrac{n_V - n_R}{n_Y - 1}\) (Eq. 9.16). It is unitless, depends only on the material, and is around 0.03 for common glasses.
Natural phenomena due to Sunlight
- Mirage: hot air near the road is optically rarer, so the refractive index of air increases with height; rays from a distant object bend gradually, are turned upwards near the road, and reach the eye from below, giving the illusion of a water surface. TIR is never possible here, since the relative refractive index is just less than 1 and the critical angle approaches \(90^0\).
- Rainbow: sunlight incident on raindrops refracts and disperses, reflects internally (NOT total internally) once for the primary and twice for the secondary rainbow, and refracts out again.
- Primary rainbow: minimum-deviation rays leave at \(\theta_R = 42.8° \cong 43°\) and \(\theta_V = 40.8 \cong 41°\) to the ground; red is outer and violet inner.
- Secondary rainbow: after two internal reflections, \(\theta_R \cong 51°\) and \(\theta_V \cong 53°\); violet is outer and red inner.
- Rainbows need a light shower with fairly large drops, enough sunlight, and a low Sun (mornings or evenings). Seen as arcs of circles about the line joining the Sun and the observer; the full circle can be seen from a sufficient height.
Defects of lenses
- Chromatic aberration: a lens focuses different colours at different points, violet closest to the pole. It cannot occur in mirrors.
- An achromatic combination brings the extreme colours to a common focus. For two thin lenses in contact the condition is \(\dfrac{f_{2Y}}{f_{1Y}} = -\dfrac{\omega_2}{\omega_1}\) (Eq. 9.19), so one lens must be convex and the other concave. For a converging achromat, \(f_{1Y} < f_{2Y}\) and \(\omega_1 < \omega_2\); hence the convex lens is of crown glass (\(n = 1.517\)) and the concave lens of flint glass (\(n = 1.655\)).
- Spherical aberration in lenses is reduced by: using a planoconvex/planoconcave lens with the curved face towards the incident rays; choosing the radii ratio \(\dfrac{R_1}{R_2} = \dfrac{1}{6}\) for \(n = 1.5\) (and \(\dfrac{1}{5}\) for \(n = 2\)); using two thin converging lenses separated by the difference of their focal lengths, with the longer-focus lens facing the light; or combining a convex lens (positive aberration) with a concave lens (negative aberration).
Optical instruments
- The apparent size of an object depends on the visual angle it subtends. The least distance of distinct vision for a normal eye is \(D = 25\) cm.
- Magnifying power: \(M = \dfrac{\beta}{\alpha}\), where \(\beta\) is the visual angle of the image formed by the instrument and \(\alpha\) that of the object at \(D\) (or, for a telescope, at its own position).
- Simple microscope: \(M = \dfrac{D}{u}\), with \(M_{min} = \dfrac{D}{f}\) (image at infinity) and \(M_{max} = 1 + \dfrac{D}{f}\) (image at \(D\)). For \(f = 5\) cm and \(D = 25\) cm, \(M\) lies only between 5 and 6.
- Compound microscope: \(M = m_o \times M_e\), where \(m_o = \dfrac{v_o}{u_o}\) is the lateral magnification of the objective and \(M_e = \dfrac{D}{u_e}\) is the magnifying power of the eye lens. Length \(L = v_0 + u_e\). Increasing \(m_o\) increases \(v_0\) and hence the length of the microscope.
- Astronomical telescope (normal adjustment): \(M = \dfrac{f_o}{f_e}\) and length \(L = f_o + f_e\). For maximum magnifying power, choose the largest permissible \(f_o\) and the smallest practical \(f_e\).
- Terrestrial telescopes need an erect final image and use a concave eye lens (binocular) or three convex lenses.
9.13 Check Your Concepts
Light is described both as an EM wave and as a stream of photons. Which category of optical phenomena requires each description, and why is ray optics still a valid way to study mirrors and lenses?
Why does the first law of ray optics specify a “homogeneous and isotropic” medium? Name one everyday phenomenon that occurs precisely because this condition is violated, and explain how.
A ray strikes a glass slab along the normal. Has refraction occurred? Has deviation occurred? Justify your answer using Snell’s law and using the definition of refraction.
Two students substitute the object distance for a mirror as \(u = +20\) cm and \(u = -20\) cm respectively, for the same real object placed in front of the mirror. Which one is correct, and what physical information does the sign carry?
What is a virtual object? Under what circumstances does a lens or mirror receive one? Why is its distance taken as positive?
Two plane mirrors are inclined at \(72^0\). Why does the number of images depend on where exactly the object is kept, whereas at \(60^0\) it does not?
Why does a convex mirror always form a virtual, erect and diminished image, whatever the position of the object? Relate your answer to its use as a rear-view mirror.
A spherical mirror and a parabolic mirror of the same aperture are each used to make a search light. Which one gives a truly parallel beam, and why does the other fail?
An oil is optically denser than water, yet water is physically denser than the oil. Explain how both statements can be true at once, and state what “optically denser” actually measures.
Why is total internal reflection impossible when light travels from a rarer medium into a denser medium, no matter how large the angle of incidence?
In an optical fibre, why must the cladding be optically rarer than the core? What would happen to a signal if the cladding were made of a material denser than the core?
Prism binoculars and periscopes both use right-angled isosceles prisms rather than plane mirrors to turn the light. State two advantages of this choice and justify them physically.
Why does the relation \(f = R/2\) hold for a spherical mirror but fail for a lens? Your answer should refer to the physical difference between reflection and refraction.
An air bubble inside a glass slab behaves as a lens. Explain why a bubble that is geometrically convex actually diverges the incident beam.
A white beam passes obliquely through a thick plane parallel glass slab. The colours separate inside the slab, yet the emergent beam appears white to the eye. Explain, and state what change to the slab or to the angle of incidence would make the dispersion detectable.
Why must \(i = e\) at the angle of minimum deviation for a prism? Use the deviation curve and the principle of reversibility of light in your answer.
For a thin prism, the angle of deviation is said to be independent of the angle of incidence. State the approximations that make this true, and explain why the statement fails for an equilateral prism.
Dispersive power is independent of the dimensions of the prism, while angular dispersion is not. Show how both facts follow from the corresponding expressions, and explain what this means for a prism designer.
During the formation of a rainbow, the light suffers internal reflection but never total internal reflection. Explain why total internal reflection is impossible inside a raindrop.
In the primary rainbow, red is the outer arch, while in the secondary rainbow, violet is the outer arch. What single difference between the two causes this reversal?
Why can a rainbow not be seen at noon on a clear day, even if a light shower is falling? Under what circumstance could an observer see a complete circular rainbow?
Why can a mirror never suffer from chromatic aberration, while a lens of the same shape does? What consequence does this have for the design of very large astronomical telescopes?
In an achromatic combination of two lenses in contact, one lens must be convex and the other concave. Show how this requirement follows from the condition for achromatism, and explain why the convex lens is made of crown glass rather than flint glass.
A planoconvex lens is used with its curved side facing the incident light. What happens to the spherical aberration if the lens is reversed, and why?
A simple microscope produces an image of infinite size when the object is at its focus, yet the magnifying power is then only \(D/f\). Explain this apparent contradiction, distinguishing clearly between lateral magnification and magnifying power.
Why is a compound microscope preferred over a simple microscope of very small focal length for obtaining high magnification?
For an astronomical telescope, \(M = f_o / f_e\) and \(L = f_o + f_e\). Explain why these two relations together place a practical limit on the magnifying power, and state how a designer chooses \(f_o\) and \(f_e\) within a fixed tube length.
In a telescope, why is the visual angle \(\alpha\) taken as the angle subtended by the object at its own position, whereas in a microscope it is taken at the least distance of distinct vision?
As per recent development, what is the nature of light? Wave optics and the particle nature of light are used to explain which phenomena of light, respectively?
Which phenomena can be satisfactorily explained using ray optics? State the assumptions on which ray optics is based.
What is the focal power of a spherical mirror or of a lens? What may be the reason for using \(P = \dfrac{1}{f}\) as its expression?
At which positions of the objects do spherical mirrors produce (i) a diminished image, (ii) a magnified image?
State the restrictions for having images produced by spherical mirrors to be appreciably clear.
Explain spherical aberration for spherical mirrors. How can it be minimized? Can it be eliminated by some curved mirrors?
Define absolute refractive index and relative refractive index. Explain in brief, with an illustration for each.
Explain ‘mirage’ as an illustration of refraction.
Under what conditions is total internal reflection possible? Explain it with a suitable example. Define the critical angle of incidence and obtain an expression for it.
Describe the construction and working of an optical fibre. What are the advantages of optical fibre communication over electronic communication?
Why are prism binoculars preferred over traditional binoculars? Describe their working in brief.
A spherical surface separates two transparent media. Derive an expression that relates the object and image distances with the radius of curvature for a point object. Clearly state the assumptions, if any.
Derive the lens makers’ equation. Why is it called so? Under which conditions are the focal length \(f\) and the radii of curvature \(R\) numerically equal for a lens?
What are the different types of dispersion of light? Why do they occur?
Define angular dispersion for a prism. Obtain its expression for a thin prism. Relate it with the refractive indices of the material of the prism for the corresponding colours.
Explain and define the dispersive power of a transparent material. Obtain its expressions in terms of angles of deviation and refractive indices.
Answer the following about the rainbow:
- State the conditions under which a rainbow can be seen.
- Explain the formation of a primary rainbow. For which angular range with the horizontal is it visible?
- Explain the formation of a secondary rainbow. For which angular range with the horizontal is it visible?
- Is it possible to see the primary and secondary rainbow simultaneously? Under what conditions?
Answer the following about aberration in lenses:
- Explain chromatic aberration for spherical lenses. State a method to minimize or eliminate it.
- What is achromatism? Derive a condition to achieve achromatism for a lens combination. State the conditions for it to be converging.
Describe spherical aberration for spherical lenses. What are the different ways to minimize or eliminate it?
Define and describe the magnifying power of an optical instrument. How does it differ from linear or lateral magnification?
Derive an expression for the magnifying power of a simple microscope. Obtain its minimum and maximum values in terms of its focal length.
Derive the expressions for the magnifying power and the length of a compound microscope using two convex lenses.
What is a terrestrial telescope and an astronomical telescope?
Obtain the expressions for the magnifying power and the length of an astronomical telescope under normal adjustment.
What are the limitations in increasing the magnifying powers of (i) a simple microscope, (ii) a compound microscope, (iii) an astronomical telescope?
9.14 Practice with Numericals
The speed of light in a certain glass is \(1.875 \times 10^{8} \ \text{m s}^{-1}\). Calculate the absolute refractive index of the glass. How long will light take to cross a glass block of thickness 6 mm made of this material? Express the answer with a suitable prefix.
A ray of light travelling in air strikes the surface of a liquid of refractive index 1.4 at an angle of incidence of \(45^0\). Calculate the angle of refraction in the liquid, and the speed of light inside it.
A tiny object is kept symmetrically between two plane mirrors inclined at \(45^0\). Determine the number of images seen. If the angle is now reduced to \(36^0\), how many images will be seen?
An object is placed 30 cm in front of a concave mirror of focal length 20 cm. Determine the image distance, the lateral magnification, and the nature of the image. Also calculate the focal power of the mirror in dioptre.
A pencil of length 10 cm is kept along the principal axis of a concave mirror of radius of curvature 40 cm, with its nearest end 30 cm from the pole. Determine the length of the image of the pencil.
A convex mirror of radius of curvature 3 m is fixed at a turning of a road. A motorcycle 1.8 m long is 12 m away from the mirror. Calculate the image distance and the size of the image.
A tank contains a layer of an oil of refractive index 1.5 and thickness 9 cm floating over water of refractive index 4/3 and depth 16 cm. Calculate the apparent depth of the bottom of the tank when seen from vertically above.
The critical angle of a certain transparent medium with respect to air is \(37^0\). Calculate the refractive index of the medium, and the speed of light inside it. [Take \(\sin 37^0 = 0.6\).]
A tiny bulb is fixed at the centre of the bottom of a cylindrical vessel of diameter 16 cm filled with a liquid of refractive index 5/3 up to a height of 6 cm. Determine whether light from the bulb can emerge from the liquid surface at the edge of the vessel, by comparing the angle of incidence there with the critical angle.
The core of an optical fibre has a refractive index of 1.52 and the cladding a refractive index of 1.48. Calculate the critical angle at the core–cladding boundary.
A point object is kept in air, 20 cm in front of a convex spherical surface of radius of curvature 10 cm, separating air from glass of refractive index 1.5. Determine the position and nature of the image.
A glass sphere (\(n = 1.5\)) of radius 4 cm has a small air bubble trapped inside it, at a distance of 1 cm from the centre. Determine where the bubble will appear when seen from the nearer side, along the line joining the bubble and the centre.
A thin planoconvex lens is made of glass of refractive index 1.5, with the curved surface of radius of curvature 15 cm. Calculate its focal length and focal power. Verify the special-case result \(f = 2\lvert R \rvert\).
A convex lens of focal length 15 cm is kept in contact with a concave lens of focal length 25 cm. Calculate the focal power and the equivalent focal length of the combination, and state whether it is converging or diverging.
Two thin convex lenses of focal lengths 12 cm and 18 cm are kept in air, separated by a distance of 6 cm, with a common principal axis. Calculate the focal power and the equivalent focal length of the combination.
An object is placed 40 cm from a thin convex lens of focal length 25 cm. Determine the image distance and the lateral magnification. State the nature of the image.
An equilateral glass prism of refractive index 1.5 is used at minimum deviation. Calculate the angle of minimum deviation and the corresponding angle of incidence.
A prism of refracting angle \(60^0\) produces a minimum deviation of \(40^0\) for a certain colour. Determine the refractive index of the material of the prism for that colour.
A thin prism of refracting angle \(8^0\) is made of glass having refractive indices 1.53 for violet and 1.51 for red, with a mean (yellow) refractive index of 1.52. Calculate the angular deviation of each extreme colour, the angular dispersion, the mean deviation, and the dispersive power of the glass.
For a certain prism, the angles of deviation for red and violet are \(5.2°\) and \(5.6°\) respectively. Calculate the angular dispersion and the dispersive power of the material of the prism.
A thin double convex lens has both faces of radius of curvature 30 cm. The refractive index of its material is 1.66 for violet and 1.60 for red. Calculate the focal length for each colour and the longitudinal chromatic aberration produced by the lens.
A simple microscope of focal length 4 cm is used by a person of normal eyesight. Determine the minimum and maximum magnifying power obtainable. If the object is kept 3.5 cm from the lens, where is the image formed, and can the person see it clearly?
A compound microscope has an objective of focal length 2 cm and an eye lens of focal length 5 cm, separated by 20 cm. The final image is formed at the least distance of distinct vision. Determine the object distance from the objective and the magnifying power of the microscope.
An astronomical telescope has an objective of focal length 120 cm and an eyepiece of focal length 4 cm. Under normal adjustment, calculate the magnifying power and the length of the telescope.
The length of an astronomical telescope under normal adjustment is 90 cm and its magnifying power is 17. Determine the focal lengths of the objective and the eyepiece.
A monochromatic ray of light strikes the water (\(n = 4/3\)) surface in a cylindrical vessel at an angle of incidence of \(53^0\). The depth of water is 36 cm. After striking the water surface, how long will the light take to reach the bottom of the vessel? [The angles of the most popular Pythagorean triangle, of sides in the ratio 3 : 4 : 5, are nearly \(37^0\), \(53^0\) and \(90^0\).]
Estimate the number of images produced if a tiny object is kept in between two plane mirrors inclined at \(35^0\), \(36^0\), \(40^0\) and \(45^0\).
A rectangular sheet of length 30 cm and breadth 3 cm is kept on the principal axis of a concave mirror of focal length 30 cm. Draw the image formed by the mirror on the same ray diagram, as far as possible on scale.
A car uses a convex mirror of curvature 1.2 m as its rear-view mirror. A minibus of cross section 2.4 m × 2.4 m is 6.6 m away from the mirror. Estimate the image size.
A glass slab of thickness 2.5 cm having refractive index 5/3 is kept on an ink spot. A transparent beaker of very thin bottom, containing water of refractive index 4/3 up to 8 cm, is kept on the glass block. Calculate the apparent depth of the ink spot when seen from the outside air.
A convex lens held some distance above a 6 cm long pencil produces its image of SOME size. On shifting the lens by a distance equal to its focal length, it again produces the image of the SAME size as earlier. Determine the image size.
The figure below shows the section ABCD of a transparent slab. There is a tiny green LED light source at the bottom left corner B. A certain ray of light from B suffers total internal reflection at the nearest point P on the surface AD and strikes the surface CD at point Q. Determine the refractive index of the material of the slab and the distance DQ. At Q, will the ray PQ suffer partial or total internal reflection? [You may use the approximation given in Q 39 above.]
A point object is kept 10 cm away from one of the surfaces of a thick double convex lens of refractive index 1.5 and radii of curvature 10 cm and 8 cm. The central thickness of the lens is 2 cm. Determine the location of the final image, considering paraxial rays only. Hint: The single spherical surface formula is to be used twice.
A monochromatic ray of light is incident at \(37^0\) on an equilateral prism of refractive index 3/2. Determine the angle of emergence and the angle of deviation. If the angle of prism is adjustable, what should its value be for the emergent ray to be just possible for the same angle of incidence?
From the given data set, determine the angular dispersion by the prism and the dispersive power of its material for extreme colours. \(n_R = 1.62\), \(n_V = 1.66\), \(\delta_R = 3.1°\).
The refractive index of a flint glass varies from 1.60 to 1.66 for the visible range. The radii of curvature of a thin convex lens are 10 cm and 15 cm. Calculate the chromatic aberration between extreme colours.
A person uses spectacles of ‘number’ 2.00 for reading. Determine the range of magnifying power (angular magnification) possible. It is a concavo-convex lens (\(n = 1.5\)) having the curvature of one of its surfaces to be 10 cm. Estimate that of the other.
The focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be used for a maximum magnifying power (angular magnification) of at least 12.5. The packing instructions demand that the length of the microscope should be 25 cm. Determine the minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of \(n = 1.5\)?
9.15 Objective Questions
- As per recent understanding, light consists of
- rays
- waves
- corpuscles
- photons obeying the rules of waves
- Consider optically denser lenses P, Q, R and S drawn below. According to the Cartesian sign convention, which of these have positive focal length?
- Only P
- Only P and Q
- Only P and R
- Only Q and S
- Two plane mirrors are inclined at an angle of \(40^0\) between them. The number of images seen of a tiny object kept between them is
- Only 8
- Only 9
- 8 or 9
- 9 or 10
- A concave mirror of curvature 40 cm, used for shaving purposes, produces an image of double the size as that of the object. The object distance must be
- 10 cm only
- 20 cm only
- 30 cm only
- 10 cm or 30 cm
- Which of the following aberrations will NOT occur for spherical mirrors?
- Chromatic aberration
- Coma
- Distortion
- Spherical aberration
- There are different fish, monkeys and water on the habitable planet of the star Proxima b. A fish swimming underwater feels that there is a monkey 2.5 m above, on the top of a tree. The same monkey feels that the fish is 1.6 m below the water surface. Interestingly, the height of the tree and the depth at which the fish is swimming are exactly the same. The refractive index of that water must be
- 6/5
- 5/4
- 4/3
- 7/5
- Consider the following phenomena/applications: P) Mirage, Q) rainbow, R) Optical fibre and S) glittering of a diamond. Total internal reflection is involved in
- Only R and S
- Only R
- Only P, R and S
- all the four
- A student uses spectacles of number −2 for seeing distant objects. The lenses commonly used for her/his spectacles are
- bi-concave
- double concave
- concavo-convex
- convexo-concave
- A spherical marble of refractive index 1.5 and curvature 1.5 cm contains a tiny air bubble at its centre. Where will it appear when seen from outside?
- 1 cm inside
- at the centre
- 5/3 cm inside
- 2 cm inside
- Select the WRONG statement.
- A smaller angle of prism is recommended for greater angular dispersion.
- A right-angled isosceles glass prism is commonly used for total internal reflection.
- The angle of deviation is practically constant for thin prisms.
- For an emergent ray to be possible from the second refracting surface, a certain minimum angle of incidence is necessary from the first surface.
- Angles of deviation for extreme colours are given for different prisms. Select the one having the maximum dispersive power of its material.
- 7°, 10°
- 8°, 11°
- 12°, 16°
- 10°, 14°
- Which of the following is not involved in the formation of a rainbow?
- refraction
- angular dispersion
- angular deviation
- total internal reflection
- Consider the following statements regarding a simple microscope: (P) It allows us to keep the object within the least distance of distinct vision. (Q) The image appears to be biggest if the object is at the focus. (R) It is simply a convex lens.
- Only (P) is correct
- Only (P) and (Q) are correct
- Only (Q) and (R) are correct
- Only (P) and (R) are correct