← Back to Subjects ororo.world

Chapter 5: Refraction Through A Lens

In the 1670s, a cloth merchant in the Dutch town of Delft began sending strange letters to the Royal Society in London. His name was Antonie van Leeuwenhoek, and he had no training as a scientist. What he had was a skill borrowed from his trade: cloth merchants used small glass lenses to count the threads in a piece of fabric, and Leeuwenhoek had taught himself to make better ones than anybody else.

His lenses were tiny — single beads of glass, some no larger than a pinhead, each ground and polished by hand and clamped between two brass plates. He would hold the plate almost touching his eye, bring a drop of rainwater up to the other side, and look.

What he reported seemed impossible. In a drop of ordinary water there were creatures, swimming. In the scrapings from his own teeth there were more, wriggling. He called them animalcules, and drew them carefully. For years, few believed him, until other members of the Society built lenses good enough to see the same thing.

A whole living world had been sitting in plain sight, too small for the eye alone. One curved piece of glass, smaller than a pea, was enough to reveal it. Fig. 5.0-1 shows the kind of instrument he used.

Diagram

Fig. 5.0-1 — Leeuwenhoek’s Single-Lens Microscope Simple line drawing of a small rectangular brass plate held upright, with a tiny bead-shaped glass lens set in a hole near the top of the plate; a fine adjustable screw and a pointed specimen pin on the back of the plate hold a droplet of water in front of the lens; show an eye placed very close to the lens on the other side of the plate. Label the glass bead lens, the brass plate, the specimen pin, the adjusting screw, and the position of the eye.

NoteCuriosity Corner

Q1. Leeuwenhoek’s best lenses were the smallest and most steeply curved ones. Why should a more sharply curved piece of glass magnify more strongly than a large, gently curved one? Q2. He had to hold the lens almost touching his eye, and the water drop very close to the lens. Why would moving the drop further away have spoiled the view? Q3. However bright the image of an animalcule looked, it could never have been caught on a screen or a photographic plate. What does that tell you about the kind of image a magnifying glass makes? Q4. The very same kind of curved glass, held out in sunlight, can scorch a hole in a sheet of paper. How can one piece of glass do two such different jobs? Q5. A lens that is thin in the middle and thick at the edges is no use at all as a magnifier. What must such a lens be doing to the light that a magnifying lens does not?

5.1 Lens

5.1.1 What A Lens Is

You already use lenses every day, often without noticing. They sit in spectacles, in the camera of a phone, in a hand magnifier, in a projector. Each of them is a shaped piece of glass or clear plastic that light can pass through, and that bends the light in a useful way as it passes.

A lens is a transparent refracting medium bounded by two surfaces, of which at least one is curved. Usually both surfaces are parts of spheres. Sometimes one surface is a part of a sphere and the other is flat.

A flat surface may look like an exception, but it is not really one. A sphere of a larger and larger radius is flatter and flatter over any small patch of it, so a plane surface can be treated as a spherical surface of infinite radius of curvature. On that view, every lens surface is spherical.

NoteDefinition

Lens A transparent refracting medium bounded either by two spherical surfaces, or by one spherical surface and one plane surface.

Lenses are of two kinds, and the difference between them is simply a matter of shape. A lens may be thicker in the middle than at its edge, or thinner in the middle than at its edge. That single difference decides everything the lens does to light.

The word lens comes from the Latin name for a lentil, the small flat seed used in dal. A double-convex lens has exactly that shape — rounded on both faces and tapering to a rim.

5.1.2 Convex Or Converging Lens

A convex lens is thick in its middle and thin at its periphery. In other words, a lens that bulges outwards in the middle is a convex lens. A beam of light comes closer together after passing through such a lens, so it is also called a converging lens.

A convex lens may be of three kinds, and the shapes of all three are drawn in Fig. 5.1-1.

  • bi-convex, also called double-convex or equi-convex
  • plano-convex
  • concavo-convex

Diagram

Fig. 5.1-1 — The Three Kinds Of Convex Lens Three lens cross-sections drawn side by side in outline, each shaded lightly. First: bi-convex, both faces bulging outwards, thick centre tapering to a thin rim. Second: plano-convex, left face flat and vertical, right face bulging outwards. Third: concavo-convex, both faces curving the same way to the right, the left face hollow and the right face bulging, drawn so that the centre is clearly thicker than the rim. Label the three shapes BI-CONVEX, PLANO-CONVEX and CONCAVO-CONVEX.

A bi-convex lens has both its surfaces convex. A plano-convex lens has one surface plane and the other surface convex. A concavo-convex lens has one surface convex and the other surface concave, shaped so that it is still thicker in the middle than at its periphery.

5.1.3 Concave Or Diverging Lens

A concave lens is thick at its periphery and thin in the middle. In other words, a lens that is hollowed inwards in the middle is a concave lens. Such a lens spreads out the light rays falling on it, so it is also called a diverging lens.

A concave lens too may be of three kinds, shown in Fig. 5.1-2.

  • bi-concave, also called double-concave or equi-concave
  • plano-concave
  • convexo-concave

Diagram

Fig. 5.1-2 — The Three Kinds Of Concave Lens Three lens cross-sections drawn side by side in outline, each shaded lightly. First: bi-concave, both faces hollowed inwards, thin waist and thick rim. Second: plano-concave, left face flat and vertical, right face hollowed inwards. Third: convexo-concave, both faces curving the same way to the right, the left face bulging and the right face hollow, drawn so that the centre is clearly thinner than the rim. Label the three shapes BI-CONCAVE, PLANO-CONCAVE and CONVEXO-CONCAVE.

A bi-concave lens has both its surfaces concave. A plano-concave lens has one surface plane and the other surface concave. A convexo-concave lens has one surface concave and the other surface convex, shaped so that it is thinner in the middle than at its periphery.

The last shape in each set is worth a second look. A concavo-convex lens and a convexo-concave lens both have one convex surface and one concave surface, so on a list of surfaces they sound identical. They are not. The concavo-convex lens is thicker in the middle, and therefore has a converging action on a beam of light. The convexo-concave lens is thinner in the middle, and therefore has a diverging action. What decides the action of a lens is never the names of its two surfaces, but where the glass is thicker.

Concavo-convex and convexo-concave lenses. Both have one convex and one concave face. The concavo-convex lens is thicker at the centre, so it converges light. The convexo-concave lens is thinner at the centre, so it diverges light.

Q. A lens is thin at its middle and thick at its rim. Without touching it, can you say whether it converges or diverges light?

A. It diverges light. Thin in the middle means a concave, or diverging, lens.

Collect three or four clear objects with curved faces — a marble, a smooth glass paperweight, a spectacle lens borrowed from a family member, the base of a glass tumbler. Run a fingertip across each from rim to centre and decide, by thickness alone, whether it should converge or diverge light.

5.2 Action Of A Lens As A Set Of Prisms

You have already studied the refraction of light through a prism. The result you need here is a single one: a ray of light passing through a prism always bends towards the base of the prism. That one fact is enough to explain everything a lens does.

Refraction of light through a lens becomes easy to picture if we imagine the lens cut into a stack of prisms, as drawn in Fig. 5.2-1. Near the top and the bottom of the lens the glass is sharply wedge-shaped, and each wedge behaves like a small prism.

Diagram

Fig. 5.2-1 — A Lens Treated As A Set Of Prisms Two outline drawings side by side. (a) Convex lens: the lens outline divided by horizontal lines into a stack of pieces — a sharp prism at the top with its base downwards, then a wider prism, a nearly rectangular piece across the middle, then prisms below with their bases upwards, tapering to a sharp prism at the bottom. (b) Concave lens: the same stacking for a concave outline, with the top prisms having their bases upwards and the bottom prisms their bases downwards. Label the two drawings (a) Convex lens and (b) Concave lens.

To make the picture simpler still, the several prisms in the central portion may be replaced by a single rectangular glass slab. The whole lens is then just one rectangular slab at the centre with one prism above it and one prism below it, as shown in Fig. 5.2-2.

Diagram

Fig. 5.2-2 — A Lens As A Slab With One Prism On Either Side Two outline drawings side by side. (a) Convex lens: a vertical rectangular slab across the middle of the lens outline, a triangular prism above it with its base downwards resting on the slab, and a triangular prism below it with its base upwards resting on the slab. (b) Concave lens: a vertical rectangular slab across the middle, a prism above it with its base upwards, and a prism below it with its base downwards. Label the two drawings (a) Convex lens and (b) Concave lens.

Read the two parts of Fig. 5.2-2 carefully, because the whole difference between the two kinds of lens is contained in them. A convex lens has, in its upper part, a prism with its base downwards, and in its lower part a prism with its base upwards — both bases point towards the middle. A concave lens has, in its upper part, a prism with its base upwards, and in its lower part a prism with its base downwards — both bases point away from the middle.

The stack of prisms is a model, not a description. A real lens has smoothly curved faces and no corners at all. The model is used only because it makes the direction of bending easy to see.

5.2.1 Convergent Action Of A Convex Lens

Consider three rays of light, A, B and C, travelling parallel to one another and falling on the upper, central and lower parts of a convex lens, as in Fig. 5.2-3.

The prism in the upper part has its base downwards, so it bends ray A downwards. The prism in the lower part has its base upwards, so it bends ray C upwards. Ray B strikes the central parallel-sided slab normally and passes straight on without any bending at all.

The three rays therefore end up closer together than they began, and they meet at a point F beyond the lens. The set of prisms that makes up a convex lens converges the parallel rays, or brings them closer. This is why a convex lens is said to have a converging action on the incident light rays.

Diagram

Fig. 5.2-3 — Convergent Action Of A Convex Lens The slab-and-two-prisms outline of a convex lens as in Fig. 5.2-2(a). Three horizontal rays travel from the left with arrowheads: ray A entering the upper prism, ray B entering the central slab along the axis, and ray C entering the lower prism. Show A bent downwards on emerging, C bent upwards on emerging, and B continuing straight and undeviated. All three emergent rays cross at a single point to the right of the lens, marked F. Label the incident rays A, B and C, and label the point F.

Q. Why does ray B pass through the middle of the lens without bending?

A. The central portion behaves as a parallel-sided glass slab, and B falls on it normally, so it goes straight through undeviated.

5.2.2 Divergent Action Of A Concave Lens

Now send the same three parallel rays A, B and C on to a concave lens, as in Fig. 5.2-4.

Here the prism in the upper part has its base upwards, so it bends ray A upwards — once again towards the base. The prism in the lower part has its base downwards, so it bends ray C downwards. The central slab passes the normally incident ray B undeviated, exactly as before.

The three emergent rays now travel further apart than they began. Produced backwards, they all appear to start from a common point F, which lies on the same side of the lens as the incident rays. The set of prisms that makes up a concave lens therefore diverges, or spreads, the parallel rays, and a concave lens is said to have a diverging action on the incident light rays.

Diagram

Fig. 5.2-4 — Divergent Action Of A Concave Lens The slab-and-two-prisms outline of a concave lens as in Fig. 5.2-2(b). Three horizontal rays travel from the left with arrowheads: ray A entering the upper prism, ray B entering the central slab along the axis, and ray C entering the lower prism. Show A bent upwards on emerging, C bent downwards on emerging, and B continuing straight. Draw the emergent rays A and C produced backwards as dotted lines that meet the axis at a single point F on the left of the lens, on the same side as the incident rays. Label the incident rays A, B and C, and label the point F.

Notice the difference in the two figures. For the convex lens the rays really pass through F. For the concave lens no light ever reaches F — the rays only appear to come from it. That distinction will matter a great deal later in this chapter.

Redraw Fig. 5.2-2(b) for a plano-concave lens, with its flat face towards the incoming light. Mark the base of the upper prism and the base of the lower prism, and use them to predict the path of a ray striking the upper half of the lens.

5.4 Refraction Of Light Through An Equi-Convex And An Equi-Concave Lens

The prism model of Section 5.2 told us which way a lens bends light. Now we look at what actually happens at the glass itself, surface by surface.

Light entering a lens crosses two boundaries, not one. It goes from air into glass at the first surface, and back from glass into air at the second. Recall the rule for each crossing: on entering a denser medium a ray bends towards the normal, and on entering a rarer medium it bends away from the normal. At a curved surface the normal at any point is simply the straight line joining that point to the centre of curvature of the surface.

Fig. 5.4-1 shows the refraction of a single ray of light at the two surfaces of an equi-convex lens and of an equi-concave lens. The two surfaces are marked I and II.

Diagram

Fig. 5.4-1 — Refraction Through A Lens At Its Two Surfaces Two drawings side by side, each of a fairly thick lens outline shaded to show glass, with the region outside marked AIR on both sides and the region inside marked GLASS. (a) Convex lens: an equi-convex outline, its left face labelled I and its right face labelled II. A horizontal ray AB with an arrowhead arrives from the left and meets surface I at the point B. Draw the normal N₁B at B as a dotted line along the radius, and show the refracted ray BC inside the glass bent towards this normal, sloping down to the right to meet surface II at C. Draw the normal N₂C at C as a dotted line, and show the emergent ray CD bent away from this normal, leaving the lens sloping further downwards to the right. Label A, B, C, D, N₁ and N₂. (b) Concave lens: an equi-concave outline with faces labelled I and II, a horizontal ray AB arriving at B on surface I, the normal N₁ at B, the refracted ray BC bent towards the normal inside the glass, the normal N₂ at C on surface II, and the emergent ray CD bent away from that normal so that it leaves sloping upwards, away from the axis. Label A, B, C, D, N₁ and N₂.

Follow the convex lens first. The ray AB suffers refraction at surface I, passing from air to glass, so it bends towards the normal \(\text{N}_1\text{B}\) at the point B. The refracted ray BC then falls on surface II, where it passes from glass to air, so it bends away from the normal \(\text{N}_2\text{C}\) at the point C and emerges out of the lens as CD.

The important thing is that the ray bends the same way at both surfaces. Neither bend undoes the other. The total deviation produced by the lens is therefore the sum of the deviations produced at its two surfaces.

Trace the concave lens in the same way and you will find the same pattern, with the bending the other way about. It is clear from Fig. 5.4-1 that a convex lens bends the ray of light towards its middle, that is, it converges the light, while a concave lens bends a ray of light towards its edges, or away from its middle, that is, it diverges the light.

One sentence covers both cases: in each lens, the deviation is towards the thicker portion of the lens.

Draw a bold arrow on a sheet of paper and stand a clear glass of water in front of it, a few centimetres away. Look through the water at the arrow and move the paper slowly further back. Watch the arrow flip. A cylinder of water is thick in the middle, so it acts as a converging lens.

How Lenses Are Drawn From Here On

Drawing both bends every time would clutter every ray diagram in the rest of this chapter, and would gain us nothing. From here on we shall consider the lens to be thin, and we shall not show the bending of the ray of light at each of the two surfaces separately. Instead we shall show the net bending produced by the lens by drawing a straight vertical line passing through the optical centre, and bending each ray once, at that line. Fig. 5.4-2 compares the two ways of drawing the same ray.

Diagram

Fig. 5.4-2 — The Thin-Lens Way Of Drawing Refraction Two drawings side by side of the same convex lens on a horizontal principal axis, with the same horizontal incident ray arriving from the left, parallel to and above the axis. (a) Labelled ‘bending at each surface’: a thick lens outline showing the ray bending once as it enters the front face and once as it leaves the back face, then crossing the axis. (b) Labelled ‘net bending at the lens line’: a thin lens outline with a single straight vertical line ruled through the optical centre O; the incident ray runs straight up to this vertical line and bends once there, leaving along the same final direction as in (a) and crossing the axis at the same point. Label O in both drawings and mark the point where each emergent ray cuts the axis as F₂.

A ray of light travelling parallel to the principal axis strikes the upper half of an equi-concave lens. Copy the outline of the lens, mark the point of incidence, draw the normal there, and mark clearly at which surface the ray bends towards the normal and at which surface it bends away from it.

5.5 Principal (Or Construction) Rays For Ray Diagrams

The position, size and nature of the image of an object formed by a lens can all be found by drawing a ray diagram. To draw one, we need at least two rays starting from the same point on the object.

Any two rays would do in principle, but most rays are useless to us, because we do not know where they go after the lens. The rays we choose are those for which the path after refraction from the lens is already known. There are three such rays, and they are called the principal rays or construction rays.

5.5.1 The Three Principal Rays

The three rays are listed below, and drawn for a convex lens in Fig. 5.5-1 and for a concave lens in Fig. 5.5-2.

  1. A ray of light incident at the optical centre O of the lens passes undeviated through the lens. You met this ray in Fig. 5.3-3; it is the reason the optical centre was defined at all.
  2. A ray of light incident parallel to the principal axis of the lens, after refraction, passes through the second focus \(\text{F}_2\) in a convex lens, or appears to come from the second focus \(\text{F}_2\) in a concave lens.
  3. A ray of light passing through the first focus \(\text{F}_1\) in a convex lens, or directed towards the first focus \(\text{F}_1\) in a concave lens, emerges parallel to the principal axis after refraction.

Diagram

Fig. 5.5-1 — The Three Principal Rays For A Convex Lens A thin bi-convex lens drawn as an outline with a straight vertical line ruled through its optical centre O, on a horizontal principal axis; mark F₁ on the axis to the left of the lens and F₂ to the right, at equal distances from O. Draw three separate incident rays coming from the left, each with arrowheads, in a different line style so they can be told apart. Ray 1: an oblique ray aimed at O, passing straight through the lens undeviated and continuing in the same direction. Ray 2: a horizontal ray above the axis, bending at the lens line and crossing the axis exactly at F₂, continuing beyond it. Ray 3: a ray passing through F₁ before reaching the lens, bending at the lens line and emerging horizontal, parallel to the principal axis. Label the rays 1, 2 and 3, and label O, F₁ and F₂.

Diagram

Fig. 5.5-2 — The Three Principal Rays For A Concave Lens A thin bi-concave lens drawn as an outline with a straight vertical line ruled through its optical centre O, on a horizontal principal axis; for a concave lens mark F₂ on the axis to the left of the lens and F₁ to the right, at equal distances from O. Draw three separate incident rays coming from the left, each in a different line style, with arrowheads. Ray 1: an oblique ray aimed at O, passing straight through undeviated. Ray 2: a horizontal ray above the axis, bending upwards and away at the lens line; produce this emergent ray backwards as a dotted line to meet the axis at F₂. Ray 3: a ray directed towards F₁ on the far side of the lens, drawn as a solid line up to the lens and as a dotted line beyond it towards F₁, which after refraction emerges horizontal, parallel to the principal axis. Label the rays 1, 2 and 3, and label O, F₁ and F₂.

From a point of the object an infinite number of rays travel out in all possible directions. We choose only two convenient rays out of the three listed above. The point where these two rays meet, or appear to meet, after refraction from the lens is the image of that point of the object.

The image of every other point of the object is found in exactly the same way, and all these image points together give the full image of the object. In practice we need only trace the rays from the top point of the object, since the foot of the object usually lies on the principal axis and its image lies there too.

Two rays fix an image point; the third is a check. If all three cross at one place, the drawing is right. Draftsmen use the third ray for exactly this reason.

5.5.2 Kinds Of Images

The images formed by a lens can be of two kinds: real and virtual. Which kind you get depends on whether light actually arrives where the image seems to be.

5.5.2.1 Real Image

If the rays from a point of an object, after refraction through a lens, actually meet at a point, the image is real. Light energy is genuinely delivered to that point.

Because the light really is there, a screen placed at this point catches the image, and a real image can be obtained on a screen. A real image formed by a lens is always inverted with respect to the object.

5.5.2.2 Virtual Image

If the rays from a point of an object, after refraction through a lens, do not actually meet at a point, but only appear to diverge from a point when they are produced backwards, the image is virtual. No light ever reaches that point, so a screen placed there will show nothing at all — a virtual image cannot be obtained on a screen.

Yet you can see a virtual image perfectly well, and this is worth understanding. An eye kept in the path of the diverging rays receives them just as it would receive rays from a real object at that place. The eye lens, being convex, converges these diverging rays and forms a real image of them on the retina. A virtual image is always erect, or upright, with respect to the object.

Q3. However bright the image of an animalcule looked, it could never have been caught on a screen or a photographic plate. What does that tell you about the kind of image a magnifying glass makes?

A. It is a virtual image — the light only appears to come from it, so no screen can catch it, though the eye placed in the diverging rays sees it clearly.

Q. A screen is held where a lens seems to be forming an image, but the screen stays blank. What kind of image is it?

A. Virtual. Light never reaches that point, so there is nothing for the screen to catch.

5.5.3 Distinction Between A Real And A Virtual Image

The two kinds of image differ in how they are formed, in whether they can be caught, and in which way up they are. Table 5.2 sets the three differences side by side.

Table 5.2 — Real Image Compared With Virtual Image | Property | Real image | Virtual image | |—|—|—| | How it is formed | Formed by the actual intersection of the rays refracted by the lens | Formed where the rays refracted by the lens appear to meet when they are produced backwards | | On a screen | Can be obtained on a screen | Cannot be obtained on a screen | | Orientation | Inverted with respect to the object | Erect with respect to the object | | Example | The image of a distant object formed by a convex lens on a screen | The image of an object formed by a concave lens as seen by the eye |

On a bright day, stand a few metres inside a room facing a window and hold a hand magnifier so that it casts a small picture of the window on a sheet of white paper. That is a real image. Now bring the magnifier close to a fingertip and look through it. That image cannot be put on paper at all.

Copy Fig. 5.5-2. From a point on the incident ray 2, draw the ray that would reach an eye placed to the right of the concave lens, and mark with a cross the point from which that eye judges the light to be coming. State whether the point you have marked is a real or a virtual image point.

5.6 Construction Of Ray Diagram For A Lens

Knowing the three principal rays is one thing; laying them down on paper in the right order is another. This section sets out the drawing itself, step by step.

5.6.1 Steps For A Convex Lens

A ray diagram to determine the position and the characteristics of the image formed by a convex lens is drawn as follows.

  1. Draw a line PQ on a paper and name it the principal axis. Choose a point O on the principal axis as the optical centre, and draw a convex lens with its principal section shown by a vertical line passing through O, as in Fig. 5.6-1. For a thin lens we shall show the light rays bending at this line, instead of bending them separately at the two surfaces.
  2. Choose a proper scale and mark the two principal foci \(\text{F}_1\) and \(\text{F}_2\) on the principal axis PQ. For a convex lens \(\text{F}_1\) is on the left of O and \(\text{F}_2\) is on the right of O, at an equal distance, that is \(\text{OF}_1 = \text{OF}_2\), since there is the same medium — air — on both sides of the lens.
  3. Draw a straight line AB standing on the principal axis to represent the vertical linear object, of the given height and at the given position, according to the scale chosen (Fig. 5.6-2).
  4. Draw one ray of light AO from the top point A of the object, passing straight through the optical centre O of the lens without any deviation (Fig. 5.6-3).
  5. Draw a second ray of light AD from the same top point A, parallel to the principal axis PQ up to the lens. After refraction this ray passes through the focus \(\text{F}_2\), and is drawn onward as \(\text{DA}'\) (Fig. 5.6-4).
  6. The point \(\text{A}'\) where the two refracted rays meet, or appear to meet, is the image of the point A of the object. Drop a perpendicular \(\text{A}'\text{B}'\) from \(\text{A}'\) on to the line PQ, and \(\text{A}'\text{B}'\) is the image of the object AB (Fig. 5.6-6).
  7. Measure the height \(\text{A}'\text{B}'\) and the distance \(\text{OB}'\) of the image, and convert them on the chosen scale to get the actual height or size of the image and the actual position or distance of the image from the lens.

Diagram

Fig. 5.6-1 — Setting Up The Axis, The Lens And The Foci A long horizontal line ruled across the page, labelled P at its left end and Q at its right end. At the middle, the outline of a thin bi-convex lens with a straight vertical line ruled through it; mark the point where this vertical line cuts the horizontal line as O. Mark a point F₁ on the axis to the left of O and a point F₂ to the right of O, at clearly equal distances from O, each marked with a short cross-stroke.

Diagram

Fig. 5.6-2 — Placing The Object On The Axis The same set-up as Fig. 5.6-1, with a short vertical arrow drawn standing up from the axis to represent the object; its foot on the axis is labelled B and its tip is labelled A. Place B on the axis between F₁ and the left-hand end P, closer to F₁ than to P. Label P, Q, O, F₁, F₂, A and B.

Diagram

Fig. 5.6-3 — Drawing The First Ray, Through The Optical Centre The same drawing as Fig. 5.6-2, with one ray added: a straight line with arrowheads from the tip A of the object to the optical centre O, continuing straight on beyond the lens, sloping downwards to the right and crossing the principal axis at O. Label the ray AO.

Diagram

Fig. 5.6-4 — Drawing The Second Ray, Parallel To The Axis The drawing of Fig. 5.6-3 with a second ray added: a horizontal line with arrowheads from the tip A of the object to the lens line, meeting it at a point D above the axis. From D the refracted ray is drawn sloping downwards to the right, passing exactly through F₂ and continuing beyond it, until it crosses the first ray at a point A′ below the axis. Label D, F₂ and A′.

For the second ray, there is another choice available. Instead of the ray parallel to the axis, we may draw a ray \(\text{AD}'\) from the top point A up to the lens, passing through the first focus \(\text{F}_1\) on the way. After refraction this ray becomes parallel to the principal axis PQ and is drawn onward as \(\text{D}'\text{A}'\), as in Fig. 5.6-5. Either choice leads to the same point \(\text{A}'\).

Diagram

Fig. 5.6-5 — The Alternative Second Ray, Through The First Focus The drawing of Fig. 5.6-3 with a different second ray added: a straight line with arrowheads from the tip A of the object, sloping downwards to the right so that it passes exactly through F₁, and meeting the lens line at a point D′ below the axis. From D′ the refracted ray is drawn horizontal, parallel to the principal axis, running to the right until it crosses the first ray at the point A′ below the axis. Label D′, F₁ and A′.

Fig. 5.6-6 shows all three construction rays drawn on one diagram for a convex lens, together with the completed image.

Diagram

Fig. 5.6-6 — Three Construction Rays In A Convex Lens A thin bi-convex lens with a vertical lens line through O, on the principal axis PQ, with F₁ and 2F₁ marked to the left and F₂ and 2F₂ marked to the right. The object AB is a vertical arrow standing on the axis between F₁ and 2F₁, with tip A. From A draw all three rays: the ray AO through the optical centre continuing straight on; the ray AD parallel to the axis, refracted at D through F₂; and the ray AD′ through F₁, refracted at D′ to emerge parallel to the axis. Show all three refracted rays crossing at a single point A′ below the axis, beyond 2F₂. Draw a perpendicular from A′ down to the axis meeting it at B′, and draw the image as an inverted arrow A′B′. Label A, B, D, D′, O, F₁, F₂, 2F₁, 2F₂, A′ and B′.

Where the first focus sits. For a convex lens \(\text{F}_1\) is on the left of the optical centre and \(\text{F}_2\) on the right. For a concave lens it is reversed: \(\text{F}_1\) on the right, \(\text{F}_2\) on the left.

A scale must be chosen before anything is drawn, and stated on the diagram — for example, 1 cm on paper standing for 10 cm in reality. Distances and heights are then measured on the drawing and converted back at the end.

5.6.2 Ray Diagram For A Concave Lens

The same procedure is adopted to draw a ray diagram for a concave lens, with one change made at step 2. In a concave lens the focus \(\text{F}_1\) is on the right of the optical centre O, and the focus \(\text{F}_2\) is on its left.

In Fig. 5.6-7, the ray of light AO incident at the optical centre passes undeviated as AOX. The ray of light AD incident parallel to the principal axis appears to diverge as DY from \(\text{F}_2\) after refraction. The ray \(\text{AD}'\) directed towards the focus \(\text{F}_1\) becomes parallel to the principal axis as \(\text{D}'\text{Z}\) after refraction.

The refracted rays DY and \(\text{D}'\text{Z}\) do not meet on the right of the lens at all. When produced backwards, they meet at \(\text{A}'\). Thus \(\text{A}'\) is the virtual image of the point A of the object, and for the whole object AB the virtual image is \(\text{A}'\text{B}'\).

Diagram

Fig. 5.6-7 — Three Construction Rays In A Concave Lens A thin bi-concave lens with a vertical lens line through O, on the principal axis PQ, with F₂ and 2F₂ marked on the left of O and F₁ marked on the right. The object AB is a vertical arrow standing on the axis to the left of 2F₂, with tip A. From A draw three rays. Ray AO through the optical centre, continuing straight on to the lower right and labelled AOX at its far end. Ray AD, horizontal and parallel to the axis, meeting the lens at D and emerging sloping upwards to the right as DY; produce DY backwards as a dotted line to F₂. Ray AD′ directed towards F₁ on the far side, meeting the lens at D′ and emerging horizontal as D′Z; produce D′Z backwards as a dotted line. Show the backward productions meeting at a point A′ above the axis, between F₂ and the lens. Draw a perpendicular from A′ to the axis at B′, and draw the erect image A′B′ as a short upright arrow. Label A, B, D, D′, O, F₁, F₂, 2F₂, X, Y, Z, A′ and B′.

Compare Fig. 5.6-6 with Fig. 5.6-7 before going on. In the convex lens the rays themselves cross, and the image is drawn where they cross. In the concave lens only the dotted backward productions cross, and the image is drawn there — which is why it is virtual, erect, and smaller than the object.

Using a scale of 1 cm to 5 cm, draw a ray diagram for an object 10 cm tall standing on the principal axis at 40 cm from a convex lens of focal length 10 cm. Use the ray through the optical centre and the ray parallel to the axis, and measure the distance of the image from the lens on your drawing.

5.7 Characteristics And Location Of Images For A Convex Lens

We shall now use the construction of Section 5.6 to find the image formed by a convex lens of focal length \(f\) for six different positions of the object. Let \(u\) denote the distance of the object from the lens.

Read the six cases as one story rather than six separate results. The object begins very far away and walks steadily towards the lens, and at each stage we ask where its image has gone, how big it is, and whether it is real or virtual.

5.7.1 Object At Infinity

When the object is at infinity, that is \(u = \infty\), the rays coming from it — the sun, for example — are parallel to each other and are incident on the convex lens parallel to its principal axis. After refraction they pass through the second focus \(\text{F}_2\) of the lens, as in Fig. 5.7-1.

Diagram

Fig. 5.7-1 — Image Formation By A Convex Lens For An Object At Infinity A thin bi-convex lens with a vertical lens line through O on a horizontal principal axis, with F₁ marked to the left and F₂ to the right. Four horizontal rays with arrowheads arrive from the left, all parallel to the principal axis and evenly spaced above and below it. After refraction at the lens line all four converge and cross the axis at the single point F₂, continuing beyond it as diverging rays. Label O, F₁ and F₂.

Thus a real, inverted and diminished image — diminished almost to a point — is formed at the second focus, on the other side of the lens.

An Object At A Far Distance

A real object is never truly at infinity, so it is worth seeing what happens when the object AB is merely very far away, that is \(u \gg f\). The light rays reaching the lens from any one point of the object, such as the top of a distant tree, are almost parallel to each other. But rays coming from different points of the object arrive inclined at different angles to the principal axis.

In Fig. 5.7-2 the object AB is too far away to be shown. The rays from its top point A are obliquely incident on the convex lens. One of these rays passes through the optical centre O and travels undeviated through the lens as \(\text{OA}'\), while another ray from the same point A passes through \(\text{F}_1\) and, after refraction, becomes parallel to the principal axis as \(\text{DA}'\).

These two refracted rays meet at the point \(\text{A}'\), which is therefore the real image of the object point A. It lies in the focal plane of the lens passing through the second focus \(\text{F}_2\). The bottom point B of the object lies on the principal axis, so its image \(\text{B}'\) is at \(\text{F}_2\) itself. Thus \(\text{B}'\text{A}'\) is the real, inverted and highly diminished image of the very distant object AB, formed in the second focal plane of the lens.

Diagram

Fig. 5.7-2 — Image Formation By A Convex Lens For A Very Distant Object A thin bi-convex lens with a vertical lens line through O on a horizontal principal axis; mark 2F₁ and F₁ to the left of the lens and F₂ and 2F₂ to the right. The object is not shown. Two oblique rays arrive from the far upper left, sloping downwards to the right: one passing straight through the optical centre O and continuing undeviated to the lower right, and one passing through F₁ before reaching the lens at D and emerging horizontal, parallel to the principal axis. Show the two refracted rays crossing at a point A′ just below the axis and a little beyond F₂. Mark B′ on the axis at F₂ and draw the tiny inverted image B′A′ between them. Label O, F₁, F₂, 2F₁, 2F₂, D, A′ and B′.

The image is at the focus, or in the focal plane, on the other side of the lens. It is real, inverted, and highly diminished.

Both these arrangements are put to use. A convex lens is used in this manner as a burning glass, and as a camera lens. To use the convex lens as a burning glass, the rays of light from the sun, which is at infinity, are brought to focus on a piece of paper kept in the second focal plane of the lens; the sun’s heat is concentrated into so small a patch that the paper burns. In a camera the object lies very far from the lens, and the image is formed at the camera film or sensor, which is placed at the second focus.

Q4. The very same kind of curved glass, held out in sunlight, can scorch a hole in a sheet of paper. How can one piece of glass do two such different jobs?

A. Because what the lens does depends entirely on where the object sits. Sunlight comes from infinity, so it is gathered to a tiny hot point at the focus; a printed letter placed nearer than the focus gives instead a large virtual image, as Section 5.7.6 shows.

5.7.2 Object Beyond 2F₁

In Fig. 5.7-3, AB is an object placed on the principal axis of a convex lens at a point beyond \(2\text{F}_1\). The situation is very similar to that of a far object, except that now a scale drawing can be used for the object and its distance.

From the point A of the object, the ray AD incident parallel to the principal axis passes, after refraction through the lens, through the second focus \(\text{F}_2\) as \(\text{DA}'\). The other ray AO incident at the optical centre O of the lens travels undeviated as \(\text{OA}'\). The two refracted rays \(\text{DA}'\) and \(\text{OA}'\) meet at a point \(\text{A}'\), which is the real image of the point A of the object.

The image is formed in the same way for all other points of the object between A and B. For the point B of the object, the image formed is at \(\text{B}'\). Thus \(\text{A}'\text{B}'\) is the real, inverted and diminished image of the object AB.

Diagram

Fig. 5.7-3 — Image Formation By A Convex Lens For An Object Beyond 2F₁ A thin bi-convex lens with a vertical lens line through O on a horizontal principal axis; mark 2F₁ and F₁ to the left and F₂ and 2F₂ to the right. The object AB is drawn as an upright arrow standing on the axis clearly to the left of 2F₁, foot B on the axis and tip A above it. From A draw the ray AD parallel to the axis, refracted at D through F₂ and continuing beyond; and the ray AO through the optical centre, continuing straight on. The two refracted rays cross at A′ below the axis, between F₂ and 2F₂. Drop a perpendicular from A′ to the axis at B′ and draw the inverted image A′B′, clearly shorter than the object. Label A, B, D, O, F₁, F₂, 2F₁, 2F₂, A′ and B′.

The image is between \(\text{F}_2\) and \(2\text{F}_2\) on the other side of the lens. It is real, inverted, and diminished. A convex lens is used in this manner as a camera lens, when the object to be photographed is not very far away.

5.7.3 Object At 2F₁

In Fig. 5.7-4, AB is an object placed on the principal axis of a convex lens at a distance equal to twice the focal length of the lens, that is, at \(2\text{F}_1\).

From the point A of the object, the ray AD incident parallel to the principal axis passes after refraction through the second focus \(\text{F}_2\) as \(\text{DA}'\). The other ray AO incident through the optical centre O travels undeviated as \(\text{OA}'\). The two refracted rays meet at \(\text{A}'\), the real image of the point A. Similarly the image is formed for all other points between A and B, and for the point B the image is at \(\text{B}'\). Thus \(\text{A}'\text{B}'\) is the real and inverted image of the object AB, and this image is of the same size as the object.

Diagram

Fig. 5.7-4 — Image Formation By A Convex Lens For An Object At 2F₁ A thin bi-convex lens with a vertical lens line through O on a horizontal principal axis; mark 2F₁ and F₁ to the left and F₂ and 2F₂ to the right. The object AB is an upright arrow standing exactly at 2F₁, foot B on the axis, tip A above. From A draw the ray AD parallel to the axis, refracted at D through F₂; and the ray AO through the optical centre, continuing straight on. The two refracted rays cross at A′ lying exactly at 2F₂, below the axis. Draw the perpendicular from A′ to the axis at B′ and draw the inverted image A′B′, drawn exactly as long as the object. Label A, B, D, O, F₁, F₂, 2F₁, 2F₂, A′ and B′.

The image is at \(2\text{F}_2\) on the other side of the lens. It is real, inverted, and of the same size as the object. A convex lens is used in this manner in a terrestrial telescope, for erecting an inverted image formed by the objective lens.

5.7.4 Object Between F₁ And 2F₁

In Fig. 5.7-5, AB is an object placed on the principal axis of a convex lens at a point between \(\text{F}_1\) and \(2\text{F}_1\).

From the point A of the object, a ray AD incident parallel to the principal axis passes after refraction through the second focus \(\text{F}_2\) as \(\text{DA}'\). The other ray AO incident towards the optical centre O passes undeviated through the lens as \(\text{OA}'\). The two refracted rays \(\text{DA}'\) and \(\text{OA}'\) meet at a point \(\text{A}'\), which is the real image of the point A. For the point B of the object the image is at \(\text{B}'\). Thus \(\text{A}'\text{B}'\) is the real, inverted and magnified image of the object AB.

Diagram

Fig. 5.7-5 — Image Formation By A Convex Lens For An Object Between F₁ And 2F₁ A thin bi-convex lens with a vertical lens line through O on a horizontal principal axis; mark 2F₁ and F₁ to the left and F₂ and 2F₂ to the right. The object AB is an upright arrow standing on the axis between F₁ and 2F₁, foot B, tip A. From A draw the ray AD parallel to the axis, refracted at D through F₂ and continuing well beyond; and the ray AO through the optical centre, continuing straight on. The two refracted rays cross at A′ below the axis, clearly beyond 2F₂. Drop the perpendicular from A′ to the axis at B′ and draw the inverted image A′B′, drawn clearly taller than the object. Label A, B, D, O, F₁, F₂, 2F₁, 2F₂, A′ and B′.

The image is beyond \(2\text{F}_2\) on the other side of the lens. It is real, inverted, and magnified.

A convex lens is used in this manner in cinema projectors and slide projectors. The magnified image is obtained on a screen placed at a large distance on the other side of the lens. Care is taken to put the slide, that is the illuminated object, in front of the lens just beyond its focus and in an inverted position, so that an erect and magnified image is obtained on the screen.

Two magnified images. An object between \(\text{F}_1\) and \(2\text{F}_1\) gives a magnified image that is real and inverted, and lands on a screen. An object nearer than \(\text{F}_1\) gives a magnified image that is virtual and erect, and no screen will catch it.

5.7.5 Object At F₁

In Fig. 5.7-6, AB is an object placed at the focus \(\text{F}_1\) on the principal axis of a convex lens.

From the point A of the object, a ray AD incident parallel to the principal axis passes after refraction through the second focus \(\text{F}_2\) of the lens as \(\text{DF}_2\). The other ray AO incident towards the optical centre O passes undeviated through it as \(\text{OA}'\). These two refracted rays are parallel to each other, and so they do not converge at any point at a finite distance. The same is true for the rays from every other point between A and B. For the point B, the image will be at \(\text{B}'\) at infinity on the principal axis. Thus a highly enlarged image \(\text{A}'\text{B}'\) is formed at infinity.

Diagram

Fig. 5.7-6 — Image Formation By A Convex Lens For An Object At F₁ A thin bi-convex lens with a vertical lens line through O on a horizontal principal axis; mark 2F₁ and F₁ to the left and F₂ and 2F₂ to the right. The object AB is an upright arrow standing exactly at F₁, foot B, tip A. From A draw the ray AD parallel to the axis, refracted at D so that it passes through F₂ and continues to the lower right; and the ray AO through the optical centre, continuing straight on to the lower right. Draw the two emergent rays clearly parallel to each other, with no crossing point shown, and letter along one of them TO INFINITY. Label A, B, D, O, F₁, F₂, 2F₁ and 2F₂.

The image is at infinity, that is, at a very far distance, on the other side of the lens. It is real, inverted, and highly magnified. A convex lens is used in this manner in the collimator of a spectrometer, to obtain a parallel beam of light by placing the source of light at the focus of a convex lens.

The lighthouse lens. A lighthouse works this case backwards. Its lamp is set at the focus of a huge lens, so the light leaves as a parallel beam that can still be seen far out at sea instead of spreading uselessly in all directions. Casting such a lens in solid glass would make it unliftably heavy, so Augustin-Jean Fresnel, in the 1820s, replaced the thick glass with a set of thin concentric rings, each ring shaped like the surface of the lens at that radius. The rings bend the light exactly as the full lens would, at a fraction of the weight. The same ringed design is used today in the flat plastic magnifiers sold as page readers.

5.7.6 Object Between The Lens And F₁

In Fig. 5.7-7, AB is an object placed on the principal axis of a convex lens between its optical centre O and the first focus \(\text{F}_1\), that is \(u < f\).

From the point A of the object, a ray AD incident parallel to the principal axis passes after refraction through the lens through the second focus \(\text{F}_2\) as \(\text{DF}_2\). The other ray AO incident at the optical centre O passes undeviated as \(\text{OO}'\).

These two refracted rays \(\text{DF}_2\) and \(\text{OO}'\) do not meet each other at all. But when they are produced backwards, they meet at a point \(\text{A}'\). Thus \(\text{A}'\) is the virtual image of the point A of the object. For all other points of the object between A and B the virtual images are formed between \(\text{A}'\) and \(\text{B}'\), and for the point B the virtual image is \(\text{B}'\).

Thus \(\text{A}'\text{B}'\) is the virtual, erect and magnified image of the object AB, formed on the same side of the lens as the object and behind it. The image can be distinctly seen by keeping the eye at the position shown in Fig. 5.7-7. The eye lens converges the diverging rays emerging from the lens to form a real image on the retina of the eye.

Diagram

Fig. 5.7-7 — Image Formation By A Convex Lens For An Object Between The Optical Centre And The Focus A thin bi-convex lens with a vertical lens line through O on a horizontal principal axis; mark 2F₁ and F₁ to the left and F₂ and 2F₂ to the right. The object AB is a short upright arrow standing on the axis between O and F₁, foot B, tip A. From A draw the ray AD parallel to the axis, refracted at D to pass through F₂ and continue to the lower right; and the ray AO through the optical centre, continuing straight on to the lower right and labelled OO′. Show an eye symbol at the far lower right in the path of these two diverging rays. Produce both emergent rays backwards as dotted lines to meet at a point A′ far above the axis on the left, beyond 2F₁; drop a perpendicular from A′ to the axis at B′ and draw the erect image A′B′ as a tall upright arrow, clearly larger than the object. Label A, B, D, O, O′, F₁, F₂, 2F₁, 2F₂, A′ and B′.

The image is on the same side of the lens as the object, and behind it. It is virtual, erect or upright, and magnified. A convex lens is used in this manner as a reading lens, that is, as a magnifying glass or a simple microscope, to form a magnified virtual image of a tiny object such as a small letter in a book or a small part of a watch.

Cast the image of a bright window on to a sheet of paper with a hand magnifier, and hold the sharpest image you can. Now cover the lower half of the lens with a piece of card. Watch what happens to the size of the image, to its position, and to its brightness.

5.7.7 Relative Positions Of Object And Image In A Convex Lens

From all six cases we notice that both real as well as virtual images are formed by a convex lens, depending upon the position of the object relative to the lens. The size of the image — magnified, diminished or the same — also depends on the position of the object. The pattern can be stated in five steps.

  1. When an object is very far off from the convex lens, a real, inverted and diminished image is formed at the focus.
  2. As the object moves towards the lens up to 2F, the image on the other side of the lens moves away from the focus of the lens up to 2F. The image formed remains real, inverted and diminished, but its size gradually increases.
  3. As the object comes at the position 2F, the image is real, inverted and of the same size. It is obtained at the position 2F on the other side of the lens.
  4. On further moving the object from 2F towards its focus F, the image remains real and inverted, but it becomes magnified and is formed beyond 2F on the other side of the lens. When the object comes at the focus F, the image moves to infinity.
  5. On further moving the object from F towards the lens, that is, if the object is brought between the focus and the lens, the image becomes magnified, erect and virtual, on the same side as the object. It is formed behind the object.

These observations will be the same if the lens is moved towards the object instead of the object being moved towards the lens. Only the separation between the two matters.

Table 5.3 gathers the position, size and nature of the image formed by a convex lens, together with the application, for each position of the object.

Table 5.3 — Relative Positions Of The Object And Image In A Convex Lens

Position of object Position of image Size of image Nature of image Application
At infinity At \(\text{F}_2\) Highly diminished Real and inverted As a burning glass
Beyond \(2\text{F}_1\) Between \(\text{F}_2\) and \(2\text{F}_2\) Diminished Real and inverted In a camera lens
At \(2\text{F}_1\) At \(2\text{F}_2\) Same size Real and inverted In a terrestrial telescope
Between \(\text{F}_1\) and \(2\text{F}_1\) Beyond \(2\text{F}_2\) Magnified Real and inverted In a slide projector
At \(\text{F}_1\) At infinity Highly magnified Real and inverted In a collimator of a spectrometer
Between the lens and \(\text{F}_1\) On the same side, behind the object Magnified Virtual and upright As a magnifying glass

Example 5.1 — Reading The Case From The Image

  1. What is given, and what is asked. A convex lens of focal length 12 cm forms a real, inverted image that is larger than the object and is caught on a screen. We are asked between which two positions the object must lie, and where the screen must be placed.

  2. The relation used. No formula is needed here; the pattern of Table 5.3 decides it. A real, inverted, magnified image is produced in one case only — the object between \(\text{F}_1\) and \(2\text{F}_1\).

  3. The working. For this lens \(f = 12\ \text{cm}\), so \(\text{F}_1\) is at 12 cm from the lens and \(2\text{F}_1\) is at \(2 \times 12\ \text{cm} = 24\ \text{cm}\). The object must therefore lie somewhere between 12 cm and 24 cm in front of the lens. For that case the image forms beyond \(2\text{F}_2\), that is, at more than 24 cm behind the lens.

  4. The answer. The object lies between 12 cm and 24 cm from the lens, and the screen must be placed more than 24 cm behind the lens. The answer is sensible: this is exactly how a slide projector is set up, with the slide just outside the focus and the screen far away.

A convex lens of focal length 8 cm is used to form the image of a small object placed 5 cm in front of it. State, with a reason drawn from Table 5.3, whether the image can be received on a screen, and whether it is erect or inverted. Support your answer with a ray diagram.

5.8 Characteristics And Location Of Images For A Concave Lens

Let us now determine the position and characteristics of the image formed by a concave lens, by drawing ray diagrams for different positions of an object. After the six cases of the convex lens this will seem a short section, and for a good reason: a concave lens gives the same kind of image wherever the object is put.

5.8.1 Object At Infinity

The light rays coming from an object at infinity are parallel to each other, and as shown in Fig. 5.8-1 they are incident parallel to the principal axis of the concave lens, the face of the lens being parallel to the object. After refraction from the concave lens, the rays appear to diverge from the second focus \(\text{F}_2\).

Diagram

Fig. 5.8-1 — Ray Diagram For Image Formation By A Concave Lens When The Object Is At Infinity A thin bi-concave lens with a vertical lens line through O on a horizontal principal axis; mark F₂ on the axis to the left of the lens and F₁ to the right. Three horizontal rays with arrowheads arrive from the left, parallel to the principal axis and evenly spaced above and below it. After refraction all three emerge diverging away from the axis. Produce each emergent ray backwards as a dotted line so that all three dotted lines meet at the single point F₂ on the axis, on the same side as the incoming light. Label O, F₁ and F₂.

Thus a virtual, erect and highly diminished, or point, image is formed at the second focus, on the side of the object, in front of the concave lens.

An Object At A Far Distance

If the object is at a far distance on the principal axis of a concave lens, and is too far off to be shown in the diagram, the rays reaching the lens from its top point A are parallel to each other but are incident obliquely on the lens. Fig. 5.8-2 traces two of them.

The ray PD, incident towards the first focus \(\text{F}_1\) of the lens, becomes parallel to the principal axis after refraction, and is drawn as \(\text{DD}'\). The other ray QO from the same point A, incident at the optical centre O of the lens, passes undeviated through the lens as \(\text{OO}'\).

The refracted rays \(\text{DD}'\) and \(\text{OO}'\) do not meet each other. When produced backwards, they meet at a point \(\text{A}'\), which is the virtual image of the point A of the object. For all the other points of the object between A and B, the virtual images are formed between \(\text{A}'\) and \(\text{B}'\). For the bottom point B of the object, which lies on the principal axis, the virtual image is formed at \(\text{B}'\) at the second focus \(\text{F}_2\) of the lens.

Diagram

Fig. 5.8-2 — Ray Diagram For Image Formation By A Concave Lens When The Object Is At A Very Large Distance A thin bi-concave lens with a vertical lens line through O on a horizontal principal axis; mark F₂ to the left of the lens and F₁ to the right. The object is not shown. Two parallel oblique rays arrive from the far upper left, sloping downwards to the right, and are labelled PARALLEL RAYS: ray PD, aimed towards F₁ on the far side, meeting the lens at D and emerging horizontal as DD′; and ray QO, passing through the optical centre O and continuing undeviated to the lower right as OO′. Produce both emergent rays backwards as dotted lines to meet at a point A′ just above the axis and close to F₂; drop a perpendicular from A′ to the axis at B′, placed at F₂, and draw the tiny erect image A′B′. Label P, Q, D, D′, O, O′, F₁, F₂, A′ and B′.

Thus \(\text{A}'\text{B}'\) is the virtual, erect and highly diminished image of the object AB, formed in front of the lens in the focal plane at the second focus \(\text{F}_2\) of the lens. Note that light rays coming from different points of the same object are parallel to one another, but are incident at different angles on the lens, which is why the image has a size at all.

The image is at the second focus \(\text{F}_2\), or in the focal plane at \(\text{F}_2\), on the side of the object. It is virtual, erect, and highly diminished. A concave lens is used in this manner in a Galilean telescope.

Q5. A lens that is thin in the middle and thick at the edges is no use at all as a magnifier. What must such a lens be doing to the light that a magnifying lens does not?

A. It spreads the rays apart instead of bringing them together, so the rays only ever appear to come from a point nearer the lens than the object. The image is always upright and always smaller than the object.

5.8.2 Object At Any Finite Distance

Let AB be an object placed at a point on the principal axis at any finite distance in front of a concave lens, as in Fig. 5.8-3. Two rays AD and AO from the point A of the object diverge from each other after refraction through the lens.

The ray AD, incident parallel to the principal axis, is refracted as \(\text{DD}'\) so as to appear to come from the second focus \(\text{F}_2\) of the lens. The ray AO, incident towards the optical centre O, passes undeviated after refraction as \(\text{OO}'\).

The two refracted rays \(\text{DD}'\) and \(\text{OO}'\) do not meet at any point on the far side, but on being produced backwards they meet at a point \(\text{A}'\), which is the virtual image of the point A. For all other points of the object AB, the virtual images are formed between \(\text{A}'\) and \(\text{B}'\), and \(\text{B}'\) is the virtual image of the point B.

Diagram

Fig. 5.8-3 — Ray Diagram For Image Formation By A Concave Lens When The Object Is Between Infinity And The Lens A thin bi-concave lens with a vertical lens line through O on a horizontal principal axis; mark F₂ and 2F₂ to the left of the lens and F₁ and 2F₁ to the right. The object AB is drawn as an upright arrow standing on the axis to the left of 2F₂, foot B on the axis and tip A above it. From A draw the ray AD parallel to the axis, meeting the lens at D and emerging sloping upwards to the right as DD′; and the ray AO through the optical centre, continuing undeviated to the lower right as OO′. Produce both emergent rays backwards as dotted lines, meeting at a point A′ above the axis between F₂ and the lens; drop a perpendicular from A′ to the axis at B′ and draw the erect image A′B′, clearly shorter than the object. Label A, B, D, D′, O, O′, F₁, F₂, 2F₂, A′ and B′.

Thus \(\text{A}'\text{B}'\) is the virtual, erect and diminished image of the object AB, formed in front of the lens, between the optical centre and the second focus of the lens.

The image is between the lens and the focus, on the side of the object. It is virtual, erect, and diminished. A concave lens is used in spectacles for short-sighted persons in this manner.

Two virtual images. A concave lens always gives a virtual, erect image that is smaller than the object. A convex lens gives a virtual, erect image only when the object is inside the focus, and that one is larger than the object.

Example 5.2 — Naming The Lens From The Image

  1. What is given, and what is asked. A lens forms an image of an object that is erect, virtual, and one-third the size of the object. The object is 30 cm from the lens. We are asked to name the kind of lens and to say on which side of the lens the image lies.

  2. The relation used. Again the pattern of the two tables decides it, not a formula. An erect and virtual image is given by a concave lens for every object position, and by a convex lens only for an object inside the focus. But a convex lens in that position always magnifies, while this image is diminished.

  3. The working. Since the image is virtual, erect and diminished, the lens must be concave. For a concave lens the image always lies on the same side of the lens as the object, between the optical centre and the second focus.

  4. The answer. The lens is concave, and the image lies on the same side as the object, somewhere within 30 cm of the lens and closer to it than the focus. The answer is sensible: a diminished upright view is exactly what you see through the spectacles of a short-sighted person.

5.8.3 Relative Positions Of Object And Image In A Concave Lens

From the two cases above, we notice that irrespective of the position of the object, the image formed by a thin concave, or divergent, lens is always virtual, upright and diminished, and it is situated on the side of the object between the focus and the lens. Two further points complete the picture.

  1. When the object moves from a large distance towards the lens, the image shifts from the focus towards the optical centre of the lens, and its size gradually increases, but it always remains smaller than the object.
  2. When the object is at a distance equal to the focal length of the lens, the image is exactly at the mid-point between the optical centre and the second focus of the lens.

Table 5.4 gathers the position, size and nature of the image formed by a concave lens for the different positions of the object.

Table 5.4 — Relative Positions Of The Object And Image In A Concave Lens

Position of the object Position of the image Nature of the image Size of the image Application
At infinity At the focus, on the same side of the lens as the object Virtual and upright Highly diminished In a Galilean telescope
At any position between infinity and the optical centre Between the focus and the optical centre, on the same side of the lens as the object Virtual and upright Diminished In making spectacles for a myopic eye

The clip-on wide-angle lens. The tiny lens you can clip over a phone camera to fit a whole room into one photograph is a diverging lens. It forms a diminished virtual image of the scene, which the camera’s own lens then photographs, so far more of the room is squeezed on to the same sensor. The price is the familiar bulge at the centre of such pictures, and the stretching of anything near the corners.

An object 6 cm tall is placed 20 cm in front of a concave lens of focal length 20 cm. Using a scale of 1 cm to 5 cm, draw the ray diagram and mark the image. State, from your drawing, whether the image lies nearer to the lens than the focus, and measure how tall it is.

5.9 Difference Between The Image Formed By A Convex And A Concave Lens

Sections 5.7 and 5.8 dealt with the two lenses separately. Setting their results side by side brings out the real difference between them, which is not about shape at all but about how much variety each lens offers.

A convex lens can do many things, depending on where the object is placed. A concave lens does only one thing, wherever the object is placed. Table 5.5 sets out the three differences that follow from this.

Table 5.5 — Image By A Convex Lens Compared With Image By A Concave Lens | Property | Image by a convex lens | Image by a concave lens | |—|—|—| | Real or virtual | Can be real as well as virtual. It is real if the object lies beyond the focus, and virtual if the object lies within the focal distance | Always virtual, for all positions of the object | | Size | Can be magnified, of the same size, or diminished. It is magnified if the object lies between the focus and 2F, of the same size if the object is at 2F, and diminished if the object is beyond 2F | Always diminished | | Erect or inverted | Can be inverted as well as erect. It is inverted if the object is at a distance beyond the focus, and erect if the object is within the focal distance | Always erect |

The three rows are linked, and it is worth seeing how. Whenever a convex lens forms a real image, that image is inverted; whenever it forms a virtual image, that image is erect and magnified. For a concave lens all three entries are fixed together, because the rays leaving it always diverge and can never meet.

Q. A lens forms an inverted image of a candle flame on a wall. Which kind of lens is it, and why can you say so at once?

A. A convex lens. The image is on a wall, so it is real, and only a convex lens can form a real image.

A lens forms an image of a small object placed 10 cm in front of it. The image is on the same side as the object and is the same size as the object. State, with a reason, whether such a lens can exist among the two kinds you have studied.

5.10 Sign Convention In The Measurement Of Distances

Up to now we have located images by drawing. From here on we shall locate them by calculation, and calculation needs numbers that carry direction as well as size. An image 20 cm behind a lens and an image 20 cm in front of it are quite different things, and the arithmetic must be able to tell them apart.

That is what a sign convention is for. We follow the cartesian sign convention to measure the distances in a lens, according to which:

  1. The optical centre of the lens is chosen as the origin of the coordinate system.
  2. The object is considered to be placed on the principal axis to the left of the lens. Light therefore always travels from left to right in our diagrams.
  3. All distances are measured along the principal axis from the optical centre of the lens. The distance of an object from the lens is denoted by \(u\), the distance of the image by \(v\), and the distance of the second focus, that is the focal length of the lens, by \(f\). All three are lengths, measured in metre (m) or in centimetre (cm).
  4. Distances measured in the direction of the incident ray are taken positive, while distances measured opposite to the direction of the incident ray are taken negative.
  5. The length above the principal axis is taken positive, while the length below the principal axis is taken negative. This rule is used for the heights of the object and the image.

Diagram

Fig. 5.10-1 — Sign Convention For A Lens Two drawings, one above the other. (a) Convex lens: a thin bi-convex lens with a vertical lens line through O on a horizontal principal axis labelled X′ at the left end and X at the right end; mark F₁ to the left and F₂ to the right. Draw a short upright arrow to the left of the lens labelled OBJECT, with a horizontal arrow beside it pointing right labelled INCIDENT LIGHT. To the right of O along the axis draw an arrow pointing right labelled POSITIVE and beneath it DIRECTION OF INCIDENT LIGHT; to the left of O draw an arrow pointing left labelled NEGATIVE and beneath it DIRECTION OPPOSITE TO INCIDENT LIGHT. Above the lens draw a vertical arrow pointing up labelled POSITIVE and below the lens a vertical arrow pointing down labelled NEGATIVE. (b) Concave lens: exactly the same labelling for a thin bi-concave lens, with F₂ marked to the left of O and F₁ to the right.

Fig. 5.10-1 shows the distances for convex and concave lenses according to this convention. Three consequences of it are used in every numerical problem, and they are worth stating separately.

The focal length, which is \(\text{OF}_2\), of a convex lens is positive, and that of a concave lens is negative. This follows from rule 4: the second focus of a convex lens lies on the far side, in the direction the light travels, while that of a concave lens lies on the near side.

The distance of the object \(u\) in front of the lens is always negative, because the object is always placed to the left of the lens, against the direction of the incident light.

The distance of the image \(v\) is positive if the image is real and formed behind the lens, while it is negative if the image is virtual and formed in front of the lens.

\(u\), \(v\) and \(f\) are all measured from the optical centre, in cm or m. Measured along the direction of the incident light they count positive, against it negative. So \(u\) is always negative; \(f\) is \(+\) for convex and \(-\) for concave; \(v\) is \(+\) for a real image and \(-\) for a virtual one.

Notice that these signs are not bookkeeping. Each one is a statement about where the light actually goes. A negative \(v\) says, before any drawing is made, that no light reaches the image and that a screen placed there would stay dark.

Q. An object stands 25 cm in front of a concave lens of focal length 15 cm. What are \(u\) and \(f\)?

A. \(u = -25\ \text{cm}\), since the object is always in front of the lens, and \(f = -15\ \text{cm}\), since the lens is concave.

An object stands 8 cm in front of a convex lens of focal length 12 cm, and the lens forms a virtual image 24 cm from the lens on the same side as the object. Write down the values of \(u\), \(v\) and \(f\) that you would substitute in a calculation, each with its correct sign and unit.

The rule is borrowed from graph work. The optical centre is the origin, the principal axis is the x-axis pointing along the light, and the y-axis points upwards — which is exactly why heights above the axis count as positive.

5.11 Lens Formula

A ray diagram will always tell you where the image is, but it takes a sheet of paper, a ruler and a chosen scale. There is a quicker route. The three distances that matter — where the object is, where the image is, and how strong the lens is — are tied together by one equation.

5.11.1 The Relation Between u, v And f

Say it in words first. The nearer an object comes to a lens, the further away the lens throws its image; and the more strongly a lens bends light, the closer in it pulls that image. The equation that ties these three distances together is called the lens formula, and it is the same for both the convex and the concave lens.

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]

Here \(u\) is the distance of the object from the optical centre in cm or m, \(v\) is the distance of the image from the optical centre in the same unit, and \(f\) is the focal length of the lens, again in the same unit. All three are measured along the principal axis from the optical centre, and each carries the sign fixed by the convention of Section 5.10. The relation holds for a thin lens only, and only when the same medium — air — lies on both sides of it.

Where The Formula Comes From

Fig. 5.11-1 shows a convex lens forming the image \(\text{A}_1\text{B}_1\) of a linear object AB. Two pairs of similar triangles do all the work.

Diagram

Fig. 5.11-1 — Obtaining The Lens Formula For A Convex Lens A thin convex lens outline with a vertical line ruled through its optical centre O, on a horizontal principal axis. Mark F₁ and 2F₁ on the axis to the left and F₂ and 2F₂ to the right. Stand an upright object arrow AB on the axis beyond 2F₁, foot B on the axis, tip A above it. Draw the horizontal ray from A meeting the lens line at D and bending to pass through F₂, and the ray from A straight through O; show the two refracted rays crossing at A₁ below the axis, between F₂ and 2F₂, and drop a perpendicular from A₁ to the axis at B₁ so that A₁B₁ is the inverted image. Beneath the axis mark three double-headed measuring arrows: from B to O labelled (−) u, from O to F₂ labelled (+) f, and from O to B₁ labelled (+) v. Label A, B, D, O, F₁, F₂, 2F₁, 2F₂, A₁ and B₁.

Triangles AOB and \(\text{A}_1\text{OB}_1\) are similar, because AB and \(\text{A}_1\text{B}_1\) are both perpendicular to the axis and the ray AO passes straight through O. Therefore

\[\frac{\text{BA}}{\text{B}_1\text{A}_1} = \frac{\text{OB}}{\text{OB}_1} \qquad \dots\text{(i)}\]

Triangles \(\text{DOF}_2\) and \(\text{A}_1\text{B}_1\text{F}_2\) are also similar. Therefore

\[\frac{\text{OD}}{\text{B}_1\text{A}_1} = \frac{\text{OF}_2}{\text{F}_2\text{B}_1}\]

But OD and BA are opposite sides of a rectangle, so \(\text{BA} = \text{OD}\), which lets us write

\[\frac{\text{BA}}{\text{B}_1\text{A}_1} = \frac{\text{OF}_2}{\text{F}_2\text{B}_1} \qquad \dots\text{(ii)}\]

Comparing (i) and (ii), the left-hand sides are the same, so the right-hand sides must be equal:

\[\frac{\text{OB}}{\text{OB}_1} = \frac{\text{OF}_2}{\text{F}_2\text{B}_1} \qquad \dots\text{(iii)}\]

Now put in the signs, reading them off Fig. 5.11-1. The object lies in front of the lens, so \(\text{OB} = u\) is negative and the length OB itself is \(-u\). The lens is convex, so \(\text{OF}_2 = f\) is positive. The image is real and behind the lens, so \(\text{OB}_1 = v\) is positive, and the length \(\text{F}_2\text{B}_1 = \text{OB}_1 - \text{OF}_2 = v - f\), which is positive here. Substituting in (iii),

\[\frac{-u}{v} = \frac{f}{v - f}\]

Cross-multiplying, \(-u(v - f) = vf\), that is \(-uv + uf = vf\), which rearranges to

\[uf - vf = uv\]

Divide every term by \(uvf\). The first term gives \(1/v\), the second gives \(1/u\), and the right-hand side gives \(1/f\):

\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]

Repeating the same work for a concave lens, using Fig. 5.11-2, leads to exactly the same equation. There \(\text{OB} = u\), \(\text{OF}_2 = f\) and \(\text{OB}_1 = v\) are all negative, and \(\text{F}_2\text{B}_1 = \text{OF}_2 - \text{OB}_1 = f - v\); substituting these in (iii) and clearing the fractions gives \(uf - vf = uv\) once more. This is why one formula serves both lenses.

Diagram

Fig. 5.11-2 — Obtaining The Lens Formula For A Concave Lens A thin concave lens outline with a vertical line ruled through its optical centre O, on a horizontal principal axis, with F₂ marked to the left of O and F₁ to the right. Stand an upright object arrow AB on the axis well to the left of F₂. Draw the horizontal ray from A meeting the lens line at D and emerging upward and away, with its backward continuation drawn dotted through F₂, and the ray from A straight through O continuing to the lower right. Show the dotted backward continuations meeting at A₁ between F₂ and the lens, above the axis, and drop a perpendicular from A₁ to the axis at B₁ so that A₁B₁ is the small upright virtual image. Beneath the axis mark three double-headed measuring arrows: from B to O labelled (−) u, from O to F₂ labelled (−) f, and from O to B₁ labelled (−) v. Label A, B, D, O, F₁, F₂, A₁ and B₁.

1/v − 1/u = 1/f The image distance and the object distance, each turned upside down and subtracted, give the focal length turned upside down. Only for a thin lens, with every distance signed and measured from the optical centre.

5.11.2 Using The Lens Formula In Numerical Problems

In numerical problems the known values are substituted with their proper signs, and the unknown quantity is then obtained by calculation. The sign of the calculated quantity is not a leftover to be discarded — it tells you where on the principal axis the quantity lies, and it must be read.

Four rules cover almost every problem you will meet.

  • For a convex lens, \(u\) is always negative and \(f\) is always positive.
  • For a convex lens, \(v\) is positive for a real image and negative for a virtual image.
  • For a concave lens, \(u\), \(v\) and \(f\) are all negative.
  • For a concave lens, the numerical value of \(u\) is always greater than that of \(v\), because the image is always formed between the lens and the focus.

One practical caution before any substitution: all three quantities must be in the same unit. If the focal length is given in metre and the object distance in centimetre, convert one of them before you begin, and state the unit of your answer.

The lens and mirror formulae. For a lens it is \(1/v - 1/u = 1/f\), with a minus sign. For a spherical mirror it is \(1/v + 1/u = 1/f\), with a plus. Writing the mirror form for a lens will place the image on the wrong side.

Example 5.3 — Locating The Image Formed By A Convex Lens

  1. Given: focal length \(f = +15\ \text{cm}\), positive because the lens is convex; object distance \(u = -25\ \text{cm}\), negative because the object is in front of the lens. To find: the image distance \(v\) in cm, and the nature of the image.

  2. Relation used: the lens formula \(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\), rearranged as \(\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}\), since \(f\) and \(u\) are known and \(v\) is wanted.

  3. Working: \(\dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{-25} = \dfrac{5 - 3}{75} = \dfrac{2}{75}\), so \(v = \dfrac{75}{2} = +37.5\ \text{cm}\).

  4. Answer: the image is formed \(37.5\ \text{cm}\) behind the lens. The positive sign means it is a real image, and a real image formed by a lens is inverted. Check: the object at \(25\ \text{cm}\) lies between \(f\) and \(2f\), that is between \(15\ \text{cm}\) and \(30\ \text{cm}\), so by Section 5.7.4 the image should be real and beyond \(2f = 30\ \text{cm}\) — and \(37.5\ \text{cm}\) is indeed beyond it.

Example 5.4 — Locating The Image Formed By A Concave Lens

  1. Given: focal length \(f = -20\ \text{cm}\), negative because the lens is concave; object distance \(u = -30\ \text{cm}\). To find: the image distance \(v\) in cm, and the nature of the image.

  2. Relation used: the lens formula in the form \(\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}\), which holds for a concave lens exactly as it does for a convex one.

  3. Working: \(\dfrac{1}{v} = \dfrac{1}{-20} + \dfrac{1}{-30} = -\dfrac{3 + 2}{60} = -\dfrac{5}{60} = -\dfrac{1}{12}\), so \(v = -12\ \text{cm}\).

  4. Answer: the image is formed \(12\ \text{cm}\) from the lens, on the same side as the object. The negative sign means it is virtual, and therefore erect. Check: \(12\ \text{cm}\) is less than the object distance of \(30\ \text{cm}\) and less than the focal length of \(20\ \text{cm}\), so the image lies between the lens and the focus, exactly where Section 5.8 said a concave lens always puts it.

Example 5.5 — Finding The Focal Length From A Measurement

  1. Given: an object placed \(20\ \text{cm}\) in front of a lens, so \(u = -20\ \text{cm}\); a sharp image caught on a screen \(60\ \text{cm}\) behind the lens, so \(v = +60\ \text{cm}\), positive because the image is real. To find: the focal length \(f\) in cm, and the kind of lens.

  2. Relation used: the lens formula \(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\), used as it stands, since both \(u\) and \(v\) are known.

  3. Working: \(\dfrac{1}{f} = \dfrac{1}{60} - \dfrac{1}{-20} = \dfrac{1}{60} + \dfrac{1}{20} = \dfrac{1 + 3}{60} = \dfrac{4}{60} = \dfrac{1}{15}\), so \(f = +15\ \text{cm}\).

  4. Answer: the focal length is \(15\ \text{cm}\), and the positive sign shows the lens is convex. Check: the image was caught on a screen, so it had to be real, and only a convex lens can form a real image.

An object is placed \(12\ \text{cm}\) in front of a convex lens of focal length \(18\ \text{cm}\). Find the position of the image and state its nature.

5.12 Linear Magnification

5.12.1 Meaning Of Linear Magnification

It is not sufficient to know the position of the image of an object formed by a lens. It is also important to know its size. When the position of the object changes, there is a change in the position as well as in the size of the image, as every case in Section 5.7 showed.

Size is best given as a comparison. The ratio of the height of the image I, measured perpendicular to the principal axis, to the height of the object O, is called the linear magnification of the lens. It is generally denoted by the letter \(m\).

NoteDefinition

Linear Magnification The ratio of the height of the image, measured perpendicular to the principal axis, to the height of the object. It is a ratio of two lengths, so it has no unit.

Heights are hard to measure and distances are easy, so it is useful that the same ratio can be written in terms of the two distances we already know:

\[m = \frac{\text{height of image (I)}}{\text{height of object (O)}} = \frac{v}{u}\]

where I and O are the two heights in the same unit, \(v\) is the image distance in cm or m and \(u\) is the object distance in the same unit. Both heights and both distances carry the signs of the convention of Section 5.10, and the relation holds for a thin convex lens and a thin concave lens alike.

Where The Relation Comes From

Fig. 5.12-1 shows the image \(\text{A}_1\text{B}_1\) of an object AB formed by a convex lens, and Fig. 5.12-2 shows the same for a concave lens. In each figure the ray through the optical centre makes triangles ABO and \(\text{A}_1\text{B}_1\text{O}\) similar, so that

\[\frac{\text{B}_1\text{A}_1}{\text{BA}} = \frac{\text{OB}_1}{\text{OB}} \qquad \dots\text{(i)}\]

Diagram

Fig. 5.12-1 — Magnification By A Convex Lens A thin convex lens outline with a vertical line ruled through its optical centre O, on a horizontal principal axis, with F₁ and 2F₁ marked to the left and F₂ and 2F₂ to the right. An upright object arrow AB stands on the axis beyond 2F₁, its height marked with a small double-headed arrow labelled (+) O. Draw the horizontal ray from A bending at the lens line through F₂, and the ray from A straight through O; the refracted rays meet at A₁ below the axis, and a perpendicular from A₁ meets the axis at B₁ giving the inverted image arrow A₁B₁, whose height is marked with a small double-headed arrow labelled (−) I. Label A, B, O, F₁, F₂, 2F₁, 2F₂, A₁ and B₁.

For the convex lens of Fig. 5.12-1, the image is real and inverted, so its height \(\text{B}_1\text{A}_1 = \text{I}\) is negative, while the object height \(\text{BA} = \text{O}\) is positive. The image distance \(\text{OB}_1 = v\) is positive and the object distance \(\text{OB} = u\) is negative. Putting these into (i),

\[\frac{-\text{I}}{\text{O}} = \frac{v}{-u} \qquad \text{or} \qquad \frac{\text{I}}{\text{O}} = \frac{v}{u}\]

Diagram

Fig. 5.12-2 — Magnification By A Concave Lens A thin concave lens outline with a vertical line ruled through its optical centre O, on a horizontal principal axis, with F₂ marked to the left of O and F₁ to the right. An upright object arrow AB stands on the axis to the left of F₂, its height marked with a small double-headed arrow labelled (+) O. Draw the horizontal ray from A meeting the lens line and emerging upward and away, with its backward continuation dotted through F₂, and the ray from A straight through O; the dotted backward continuations meet at A₁ between F₂ and the lens, above the axis. A perpendicular from A₁ meets the axis at B₁, giving the small upright image arrow A₁B₁, whose height is marked with a small double-headed arrow labelled (+) I. Label A, B, O, F₁, F₂, A₁ and B₁.

For the concave lens of Fig. 5.12-2, the image is virtual and erect, so \(\text{B}_1\text{A}_1 = \text{I}\) is positive and \(\text{BA} = \text{O}\) is positive, while \(\text{OB}_1 = v\) and \(\text{OB} = u\) are both negative. Putting these into (i),

\[\frac{\text{I}}{\text{O}} = \frac{-v}{-u} = \frac{v}{u}\]

The two results are identical, so the expression for magnification is the same for the convex and the concave lens.

m = v/u How many times taller the image is than the object equals how many times further the image is from the lens than the object. A minus sign means the image is upside down; a plus sign means it is the right way up.

5.12.2 What The Sign And Size Of m Tell You

A single value of \(m\) carries two separate pieces of information at once. Its sign tells you which way up the image is, and its numerical value tells you how big it is. Read them separately.

For a real image, which is inverted, the magnification \(m\) is negative. For a virtual image, which is erect, the magnification \(m\) is positive. Since a convex lens can form real images as well as virtual ones, it can have a positive as well as a negative value of \(m\); a concave lens always forms a virtual erect image, so it always has a positive value of \(m\).

Now the numerical value, ignoring the sign. It is greater than 1 if the image is magnified, exactly 1 for an image of the same size as the object, and less than 1 for a diminished image. The numerical value of \(m\) is always less than 1 for a concave lens, while for a convex lens it can be greater than, equal to, or less than 1, depending on where the object is placed.

One more reading is worth having. A convex lens forms a real image on the other side of the lens, so there \(u\) and \(v\) have opposite signs. But for a convex lens as well as for a concave lens, a virtual image is always on the same side as the object, so there \(u\) and \(v\) have the same signs — which is exactly why \(m = v/u\) comes out positive for every virtual image.

The sign of \(m\) and the size of \(m\). The sign says which way up: minus for inverted, plus for erect. The size says how big: above 1 magnified, below 1 diminished. An \(m\) of \(-3\) is inverted and three times larger, not smaller.

Q. A lens gives \(m = +0.4\). What kind of lens must it be?

A. The image is erect and diminished, so the lens is concave — a convex lens can give an erect image only if it is magnified.

Example 5.6 — Position And Size Of An Image In A Convex Lens

  1. Given: object height \(\text{O} = 3.0\ \text{cm}\), object distance \(u = -30\ \text{cm}\), focal length \(f = +10\ \text{cm}\) for a convex lens. To find: the image distance \(v\) in cm and the image height I in cm.

  2. Relations used: the lens formula \(\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}\) to fix the position, then \(m = \dfrac{\text{I}}{\text{O}} = \dfrac{v}{u}\) to fix the size.

  3. Working: \(\dfrac{1}{v} = \dfrac{1}{10} + \dfrac{1}{-30} = \dfrac{3 - 1}{30} = \dfrac{2}{30}\), so \(v = +15\ \text{cm}\). Then \(m = \dfrac{v}{u} = \dfrac{+15\ \text{cm}}{-30\ \text{cm}} = -0.5\), and \(\text{I} = m \times \text{O} = -0.5 \times 3.0\ \text{cm} = -1.5\ \text{cm}\).

  4. Answer: the image is formed \(15\ \text{cm}\) behind the lens and is \(1.5\ \text{cm}\) tall. The minus signs say it is real and inverted, standing below the principal axis. Check: the object at \(30\ \text{cm}\) lies beyond \(2f = 20\ \text{cm}\), so a real, inverted, diminished image between \(\text{F}_2\) and \(2\text{F}_2\) is exactly what Section 5.7.2 predicts.

Example 5.7 — Size Of An Image In A Concave Lens

  1. Given: object height \(\text{O} = 5.0\ \text{cm}\), object distance \(u = -24\ \text{cm}\), focal length \(f = -12\ \text{cm}\) for a concave lens. To find: the image distance \(v\) in cm and the image height I in cm.

  2. Relations used: the lens formula \(\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}\), then \(m = \dfrac{v}{u}\), both of which apply unchanged to a concave lens.

  3. Working: \(\dfrac{1}{v} = \dfrac{1}{-12} + \dfrac{1}{-24} = -\dfrac{2 + 1}{24} = -\dfrac{3}{24} = -\dfrac{1}{8}\), so \(v = -8\ \text{cm}\). Then \(m = \dfrac{-8\ \text{cm}}{-24\ \text{cm}} = +\dfrac{1}{3}\), and \(\text{I} = \dfrac{1}{3} \times 5.0\ \text{cm} = 1.7\ \text{cm}\), to two significant figures.

  4. Answer: the image is \(8\ \text{cm}\) from the lens on the same side as the object, is \(1.7\ \text{cm}\) tall, and is virtual and erect. Check: \(m\) is positive and less than 1, as it must be for every concave lens.

Example 5.8 — Working Backwards From A Given Magnification

  1. Given: a convex lens of focal length \(f = +20\ \text{cm}\) forms a real image twice the size of the object, so \(m = -2\), the minus sign because a real image is inverted. To find: the object distance \(u\) and the image distance \(v\), both in cm.

  2. Relations used: \(m = \dfrac{v}{u}\) gives \(v = -2u\); this is then substituted into the lens formula \(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\), so that only one unknown is left.

  3. Working: \(\dfrac{1}{-2u} - \dfrac{1}{u} = \dfrac{1}{20}\), that is \(\dfrac{-1 - 2}{2u} = \dfrac{1}{20}\), so \(\dfrac{-3}{2u} = \dfrac{1}{20}\) and \(2u = -60\ \text{cm}\), giving \(u = -30\ \text{cm}\). Then \(v = -2u = -2 \times (-30\ \text{cm}) = +60\ \text{cm}\).

  4. Answer: the object must be placed \(30\ \text{cm}\) in front of the lens, and the image is then formed \(60\ \text{cm}\) behind it. Check: \(30\ \text{cm}\) lies between \(f = 20\ \text{cm}\) and \(2f = 40\ \text{cm}\), and Section 5.7.4 says such an object gives a real, inverted, magnified image beyond \(2\text{F}_2 = 40\ \text{cm}\) — and \(60\ \text{cm}\) is beyond it.

An object \(2.0\ \text{cm}\) tall is placed \(15\ \text{cm}\) in front of a convex lens of focal length \(10\ \text{cm}\). Find the position of the image, its magnification and its height, and state whether it is erect or inverted.

5.13 Power Of A Lens

5.13.1 Power And Focal Length

When a beam of light passes through a lens, it gets deviated from its path. Some lenses bend that beam sharply; others barely bend it at all. The deviation in the path of light produced by the lens is expressed in terms of its power, and a lens which produces more deviation has more power.

NoteDefinition

Power Of A Lens The deviation of the incident light rays produced by a lens on refraction through it is a measure of its power. It is measured in dioptre (D).

Now connect this to something we can measure with a metre rule. A thick lens — one having surfaces of more curvature — is of short focal length, and it deviates the rays more. A thin lens — one having surfaces of less curvature — is of large focal length, and it deviates the rays less. Fig. 5.13-1 puts the two side by side.

Diagram

Fig. 5.13-1 — More Deviation Means More Power Two drawings side by side, each on its own horizontal principal axis, with the same three horizontal parallel rays entering from the left at the same heights above and below the axis. (a) A steeply curved, strongly bulging convex lens: the rays bend sharply at the lens line and cross the axis close to the lens at a point marked F₂; mark the distance from the lens to F₂ with a short double-headed arrow labelled small f, and caption the drawing ‘large power’. (b) A gently curved, nearly flat convex lens of the same diameter: the same three rays bend only slightly and cross the axis far from the lens at a point marked F₂; mark the distance with a long double-headed arrow labelled large f, and caption the drawing ‘small power’. Label O and F₂ in both.

So a short focal length goes with a large deviation, and a long focal length with a small one. The power of a lens is therefore expressed, or measured, in terms of the reciprocal of its focal length:

\[P = \frac{1}{f}\]

where \(P\) is the power of the lens in dioptre (D) and \(f\) is its focal length in metre (m). The unit metre is part of the formula, not a detail: a focal length given in centimetre must be converted to metre before it is used here.

A lens is of the power 1 dioptre, or 1 D, if its focal length is 1 m, that is 100 cm. While giving a prescription to a patient, an optician does not quote the focal length of the lens, but quotes its power instead.

Why bother with a reciprocal at all? Because powers simply add up when lenses are put together, while focal lengths do not. One addition replaces a page of fractions.

5.13.2 Sign Of The Power Of A Lens

The power of a lens carries a sign, and the sign is a statement about which way the lens bends light. Depending on the direction in which a lens deviates the light ray, its power is either positive or negative.

If a lens deviates a ray towards its centre, the power is positive. If it deviates the ray away from its centre, the power is negative. Therefore the power of a convex lens is positive and that of a concave lens is negative — which agrees with the sign convention of Section 5.10, where \(f\) was positive for a convex lens and negative for a concave one.

For example, the power of a convex lens of focal length 20 cm is \(+5.0\ \text{D}\), while that of a concave lens of the same focal length is \(-5.0\ \text{D}\).

Power \(P\) is measured in dioptre (D), where \(1\ \text{D} = 1\ \text{m}^{-1}\). Use \(f\) in metre, never centimetre. \(P\) is positive for a convex lens and negative for a concave one, and a larger numerical value always means a shorter focal length.

Example 5.9 — Power From A Focal Length

  1. Given: a convex lens of focal length \(f = 25\ \text{cm}\), which is positive because the lens is convex. To find: the power \(P\) in dioptre.

  2. Relation used: \(P = \dfrac{1}{f}\), which requires \(f\) in metre, so first \(f = +25\ \text{cm} = +0.25\ \text{m}\).

  3. Working: \(P = \dfrac{1}{+0.25\ \text{m}} = +4.0\ \text{D}\).

  4. Answer: the power of the lens is \(+4.0\ \text{D}\). The positive sign is consistent with a convex lens, and a focal length shorter than a metre must give a power greater than 1 D.

Example 5.10 — Focal Length And Kind Of Lens From A Power

  1. Given: a lens of power \(P = -2.5\ \text{D}\). To find: its focal length \(f\), and the kind of lens.

  2. Relation used: \(P = \dfrac{1}{f}\), rearranged to make \(f\) the subject as \(f = \dfrac{1}{P}\), with \(f\) coming out in metre.

  3. Working: \(f = \dfrac{1}{-2.5\ \text{D}} = -0.40\ \text{m} = -40\ \text{cm}\).

  4. Answer: the focal length is \(40\ \text{cm}\), and both the negative power and the negative focal length show that the lens is concave. Check: a power smaller in size than 5 D should give a focal length longer than 20 cm, and 40 cm is.

5.13.3 Two Thin Lenses In Contact

If two thin lenses are placed in contact, the combination has a power equal to the algebraic sum of the powers of the individual lenses. The word algebraic matters: the signs are added along with the numbers.

The neatest case is a pair that cancels. If a convex lens of power \(+2.0\ \text{D}\) is kept in contact with a concave lens of power \(-2.0\ \text{D}\), the combination will have zero power, and it will behave like a glass plate — light passes through it undeviated, as in Fig. 5.13-2.

Diagram

Fig. 5.13-2 — A Convex And A Concave Lens Of Equal And Opposite Power In Contact A thin bi-convex lens and a thin bi-concave lens drawn touching each other, their two facing surfaces matching so that the pair together forms a shape of uniform thickness, standing on a horizontal principal axis. Three horizontal parallel rays with arrowheads enter from the left, pass through the pair, and leave on the right still horizontal and parallel, at the same heights at which they entered, with no convergence or divergence. Label the two lenses ‘convex, +2·0 D’ and ‘concave, −2·0 D’, and label the emergent beam ‘parallel — the pair behaves as a glass plate’.

Example 5.11 — Power Of A Combination

  1. Given: a convex lens of power \(P_1 = +5.0\ \text{D}\) placed in contact with a concave lens of power \(P_2 = -3.0\ \text{D}\). To find: the power \(P\) of the combination in dioptre, and its focal length in cm.

  2. Relations used: for two thin lenses in contact, \(P = P_1 + P_2\), the powers being added with their signs; then \(f = \dfrac{1}{P}\) to obtain the focal length in metre.

  3. Working: \(P = (+5.0\ \text{D}) + (-3.0\ \text{D}) = +2.0\ \text{D}\). Then \(f = \dfrac{1}{+2.0\ \text{D}} = +0.50\ \text{m} = +50\ \text{cm}\).

  4. Answer: the combination has a power of \(+2.0\ \text{D}\) and a focal length of \(50\ \text{cm}\), and behaves as a convex lens. Check: the converging lens was the stronger of the two, so the pair had to come out converging but weaker than either the \(+5.0\ \text{D}\) lens alone.

A concave lens of focal length \(50\ \text{cm}\) is placed in contact with a convex lens of focal length \(25\ \text{cm}\). Find the power of each lens, the power of the combination, and the focal length of the combination.

5.14 Magnifying Glass Or Simple Microscope

5.14.1 Principle Of The Magnifying Glass

To observe a tiny object distinctly, we prefer to place it as near to our eye as possible. But the human eye has a limitation. In order to see an object by the naked eye, it is necessary to place it at least at a distance of 25 cm from the eye, which is the least distance of distinct vision, denoted by D, for a normal eye.

An object placed at a distance of less than 25 cm is not seen distinctly. So an eye sees an object distinctly, and at its maximum size, when it is placed at the distance D from it. Bringing a printed letter closer than that only makes it blurred, never bigger.

Why should nearness make an object look bigger at all? The size of an image on the retina is actually determined by the angle subtended by the object at the eye. The smaller the angle subtended by the object at the eye, the smaller is the size of the image formed on the retina. And the eye is not able to see an object distinctly if it subtends an angle less than \(1'\), that is one minute of arc, which is \(\left(\dfrac{1}{60}\right)^\circ\).

Some objects are so small that even at 25 cm they subtend an angle smaller than \(1'\) at the eye. To observe such an object, we take the help of a convex lens of short focal length.

We hold the convex lens in front of our eye at such a distance that the object lies within its focal length. From Section 5.7.6 you already know what a convex lens does with an object placed between the optical centre and the focus: it forms an erect, virtual and magnified image on the same side of the lens and behind the object. The lens is positioned so that this image is at the distance D from the lens.

The lens chosen has so much power that this magnified image now subtends on the eye an angle greater than \(1'\), so it is distinctly seen by the eye. This is the principle of a simple microscope, also called a magnifying glass or a reading glass.

One minute of arc is a sixtieth of a degree. Held at arm’s length, a full stop on this page subtends roughly that angle — which is why a full stop is about the smallest mark the unaided eye can pick out at all.

5.14.2 Construction And Way Of Using It

A simple microscope is a convex lens of short focal length mounted in a lens holder, as shown in Fig. 5.14-1.

Diagram

Fig. 5.14-1 — A Magnifying Glass, Or Simple Microscope A hand magnifier drawn in side-on outline: a broad oval frame enclosing a circular convex lens shown lightly shaded, with a short neck joining the frame to a straight cylindrical handle. Label the LENS, the FRAME that holds it, and the HOLDER, which is the handle.

The lens is placed near the eye, and its distance from the object is adjusted such that the image is formed behind the object at a distance D from the lens, where D is the least distance of distinct vision, equal to 25 cm for a normal eye. In practice you do this by sliding the lens gently to and fro until the print snaps into focus.

Hold a magnifier close to a printed word and slowly lift it away. The word grows, then blurs, then suddenly reappears upside down. The flip marks the moment the object passed the focus of the lens.

5.14.3 Ray Diagram And Magnifying Power

In Fig. 5.14-2, AB is a tiny object — a word of a print, or a part of a watch — placed in front of a convex lens. The position of the lens is so adjusted that the object AB lies in between its optical centre O and its focus \(\text{F}_1\), and the image is formed at the least distance of distinct vision D. Thus, for a lens of a given focal length, \(u\) is adjusted to make \(v = \text{D} = 25\ \text{cm}\).

Diagram

Fig. 5.14-2 — Ray Diagram For Location Of The Image In A Magnifying Glass A thin convex lens outline with a vertical line ruled through its optical centre O, standing on a horizontal principal axis, with F₁ marked on the axis to the left of O and F₂ to the right. A short upright object arrow AB stands on the axis between F₁ and O, close to the lens, foot B on the axis and tip A above it. From A draw two rays: a horizontal ray AD meeting the lens line at D and bending to pass through F₂, drawn onward as DF₂; and a ray AO through the optical centre continuing straight on to the right as OO′. Produce both refracted rays backwards as dotted lines to meet at a point A′ far to the left, well above the axis; drop a perpendicular from A′ to the axis at B′, giving the large upright virtual image arrow A′B′. Draw an eye symbol just beyond the lens on the right at O′, looking back through the lens. Mark the distance from B′ to the lens with a long double-headed arrow labelled D. Label A, B, D, O, O′, F₁, F₂, A′ and B′.

Follow the two rays. The ray of light AD from the point A of the object, incident parallel to the principal axis, after refraction from the lens passes through its focus \(\text{F}_2\) as \(\text{DF}_2\). The other incident ray AO, passing through the optical centre O, gets refracted as \(\text{OO}'\) without deviation.

The two refracted rays \(\text{DF}_2\) and \(\text{OO}'\) do not actually meet, but they appear to come from a point \(\text{A}'\) when the rays are produced backwards. Thus \(\text{A}'\) is the virtual image of the point A of the object. Similarly, for the point B of the object, \(\text{B}'\) is the virtual image. Hence \(\text{A}'\text{B}'\) is the virtual, magnified and erect image of the object AB. To observe the image, the eye is kept very close to the lens, on the other side of the object, at the position shown in Fig. 5.14-2.

Q2. He had to hold the lens almost touching his eye, and the water drop very close to the lens. Why would moving the drop further away have spoiled the view?

A. Because the object must stay inside the focal length of the lens. His bead lenses had a focal length of only a few millimetres; move the drop past the focus and the image stops being the erect magnified virtual one and turns real and inverted. Keeping the eye at the lens simply lets it catch the whole spread of the emerging rays.

How Much It Magnifies

The magnifying power of a simple microscope is given as

\[M = 1 + \frac{\text{D}}{f}\]

where \(f\) is the focal length of the lens and D is the least distance of distinct vision, equal to 25 cm for a normal eye. Both lengths must be in the same unit, so that \(M\) itself is a pure number with no unit. Note that this D, a distance in centimetre, has nothing to do with the symbol D for the dioptre used in Section 5.13.

Read the relation as a statement about the lens. The shorter the focal length, the larger the fraction \(\text{D}/f\), and so the greater the magnifying power. For example, the magnifying power of a convex lens of focal length 100 cm is only 1·25, while that of a convex lens of focal length 5 cm is 6.

The magnifying power of the microscope can therefore be increased by using a lens of shorter focal length. But it cannot be increased indefinitely, because a lens of very short focal length must be very small and very steeply curved, and the image it gives soon becomes too faint and too distorted to be of use.

Magnifying power and linear magnification. Linear magnification \(m = v/u\) compares image height with object height. Magnifying power \(M = 1 + \text{D}/f\) compares the angle the image subtends at the eye with the angle the object itself would subtend at 25 cm.

5.14.4 Uses Of A Magnifying Glass

A simple microscope is used to see and read small letters and figures. It is used by watch-makers to see the small parts and screws of a watch.

In optical instruments such as a travelling microscope or a spectrometer, a magnifying glass is provided above the vernier scale, so as to read the scale accurately. Because of this last use it is also called the reading lens.

5.15 Application Of Lenses

Lenses are used in a large number of optical instruments. Everything in the list below is an application of one of the cases you have already worked through, so as you read each one, try to name the case it uses.

Lenses That Form Real Images
  1. The objective lens of a telescope, a camera or a slide projector is a convex lens, which forms a real and inverted image of the object.
  2. Our eye lens is also a convex lens. The eye lens forms the inverted image of the object on our retina. The brain enables us to perceive it as an upright image.
Lenses That Correct Eyesight
  1. When our eyesight weakens, we use spectacles having either a convex or a concave lens. A person suffering from long sightedness, or hypermetropia — that is, unable to see nearby objects distinctly — wears spectacles having a convex lens. A person suffering from short sightedness, or myopia — unable to see faraway objects distinctly — wears spectacles having a concave lens.

Some people are both short sighted and long sighted. They use either separate spectacles for seeing near and far objects, or spectacles with a bifocal lens. In spectacles having a bifocal lens, the lower part is a convex lens, which is used for reading and to see nearby objects, while the upper part is a concave lens, which is used to see distant objects. Fig. 5.15-1 shows how the two parts are arranged, and FC 5.1 sets out the choice of lens as a whole.

Diagram

Fig. 5.15-1 — A Bifocal Spectacle Lens Front view of a single spectacle lens in its rim, drawn as a rounded rectangle. A horizontal line low across the lens divides it into two regions. Label the large upper region ‘upper part — concave, for distant objects’ and the small lower segment ‘lower part — convex, for reading and nearby objects’. Beside the front view, draw a small side-on cross-section of the same lens showing the upper portion thin in the middle and thick at the rim, and the lower segment thicker in the middle.

Flowchart

FC 5.1 — Choosing The Corrective Lens Eyesight is weak → is the difficulty with distant objects, with nearby objects, or with both? → distant objects blurred (myopia): concave lens → nearby objects blurred (hypermetropia): convex lens → both blurred: separate spectacles, or one bifocal lens with a concave upper part and a convex lower part

Myopia and hypermetropia. Myopia is short sightedness — far things blur, and a concave lens corrects it. Hypermetropia is long sightedness — near things blur, and a convex lens corrects it. The name tells you what the person can see, not what they cannot.

Lenses In Instruments
  1. A magnifying glass is a convex lens of short focal length, fitted in a steel or plastic frame provided with a handle, as in Fig. 5.14-1.
  2. In the collimator of a spectroscope, a convex lens is used for obtaining a parallel beam of light.
  3. A concave lens is used as the eye lens in a Galilean telescope, to obtain the final erect image of the object.
A Defect, And Its Cure
  1. A single convex lens usually forms a coloured or blurred image, a defect known as chromatic aberration. Therefore a combination of concave and convex lenses is used to overcome such a defect.

The cause is worth a sentence, because it follows from Section 5.3.8. The focal length of a lens depends on the refractive index of its material, and the refractive index is not quite the same for every colour of light. Violet light is bent a little more than red, so a single lens brings violet to a focus slightly nearer itself than red, and the edges of the image are fringed with colour. A concave lens of a different kind of glass, placed with the convex one, bends the colours back together again.

Such a pair is called an achromatic doublet: a converging lens of crown glass cemented to a diverging lens of flint glass. The two glasses spread the colours by different amounts, so the pair can converge light while cancelling most of the colour fringing.

For each of the following, name the kind of lens used and state the one property of that lens which makes it suitable: (i) the objective of a camera, (ii) the eye lens of a Galilean telescope, (iii) the spectacle lens of a person who cannot read small print held at 25 cm, (iv) the lens of a collimator.

5.16 Experimental Determination Of Focal Length Of A Convex Lens

5.16.1 Estimation By The Distant Object Method

This is the quickest way to find roughly how strong a lens is, and it needs nothing but a metre rule and a wall.

Principle: A beam of parallel rays from a distant object incident on a convex lens gets converged in the focal plane of the lens.

Method: In an open space, against a white or light-coloured wall, place a metre rule horizontally with its 0 cm end touching the wall and its other end pointing towards the illuminated object, which must be at a very large distance — a tree, or an electric pole. Hold the given lens vertically on the metre rule and focus the object on the wall by moving the lens to and fro along the length of the metre rule, until the picture on the wall is at its sharpest. The set-up is shown in Fig. 5.16-1.

Diagram

Fig. 5.16-1 — Determination Of Focal Length By The Distant Object Method A metre rule lying horizontally on a table with its 0 cm end touching an upright white wall on the right and its far end to the left, its scale marked 0, 10, 20, 30, 40, 50, 60 cm. A hand holds a convex lens vertically, standing on the rule at about the 48 cm mark. On the wall, in line with the lens, a small sharp inverted image of a leafy tree is drawn. Label WHITE WALL, the lens, and mark the distance from the lens to the wall with a double-headed arrow along the rule labelled FOCAL LENGTH.

Since the light rays incident from a distant object are nearly parallel, the image formed on the wall is almost in the focal plane of the lens. The distance of the lens from the wall, read directly on the metre rule, therefore gives the approximate focal length of the lens. In Fig. 5.16-1 that reading is 48 cm, so the approximate focal length of that lens is 48 cm. The ray diagram for it is drawn in Fig. 5.16-2.

Diagram

Fig. 5.16-2 — Ray Diagram For The Distant Object Method A leafy tree drawn at the far left, with three horizontal parallel rays with arrowheads travelling from it towards a thin convex lens drawn upright on a horizontal principal axis, with its optical centre marked O. After the lens the three rays converge and meet at a point F on the axis to the right, where a small inverted image is formed. Mark the distance from O to F with a long double-headed arrow labelled f. Label the tree, O and F.

Try this against a wall of your room with a hand lens, using a lamp-post or a distant window as the object. Never point the lens at the sun and never look through it at the sun — the concentrated beam can burn skin, paper and the retina within seconds.

5.16.2 Exact Focal Length By An Auxiliary Plane Mirror Method

If we are given a vertical stand, a plane mirror, a lens and a pin, we can find the focal length exactly rather than approximately. Proceed as follows.

  1. Place the lens L on a plane mirror \(\text{MM}'\) kept on the horizontal surface of the vertical stand. Arrange the pin P horizontally in the clamp so that its tip is vertically above the centre O of the lens L, as shown in Fig. 5.16-3.
  2. Adjust the height of the pin on the stand till it has no parallax with its inverted image, as seen from vertically above the pin. Parallax simply means a difference in position between two things being compared: when the pin and its image are at different heights, they appear to shift relative to each other as you move your head. To check for parallax, keep the eye vertically above the tip of the pin P at a distance of nearly 25 cm from it and move it sideways. If the tip of the pin and its image shift together, then there is no parallax, which implies that the image is in line with the object.
  3. Measure the distance \(x\) of the pin P from the lens and the distance \(y\) of the pin from the mirror, using a metre rule and a plumb line. Calculate the average of the two distances. This gives the focal length of the lens, that is

\[f = \frac{x + y}{2}\]

where \(x\) and \(y\) are the two measured distances in cm and \(f\) is the focal length in cm. The two differ only by the small thickness of the lens, and averaging them removes it.

  1. Take three different observations, removing the parallax each time, and find the mean value of the focal length \(f\) of the lens.

Diagram

Fig. 5.16-3 — Determination Of Focal Length Of A Convex Lens By An Auxiliary Plane Mirror A heavy vertical stand rising from a flat horizontal base. On the base lies a plane mirror MM′, face upwards, with a convex lens L resting flat on the mirror; mark the centre of the lens O. A horizontal pin P is held in a clamp fixed to the vertical stand, its tip directly above O, pointing towards the stand. Draw the inverted image of the pin just above the pin itself and label it IMAGE. An eye symbol is drawn at the top, looking vertically down on the pin. Mark two vertical double-headed measuring arrows at the left: one from the pin down to the lens labelled x, and one from the pin down to the mirror labelled y. Label EYE, CLAMP, PIN, IMAGE, VERTICAL STAND, L, M, M′ and O.

Why The Method Works

When the object pin O lies in the focal plane of the lens L, the rays of light starting from O, after refraction from the lens, become parallel to its principal axis. Travelling parallel to the axis, they strike the plane mirror M squarely — the angle of incidence is \(i = 0^\circ\), so the angle of reflection is \(r = 0^\circ\).

Rays reflected straight back retrace their own path. They re-enter the lens as a parallel beam, and the lens converges them in its focal plane, forming the image I just above the object pin O. The image I is inverted, real, and of the same size as that of the object O. Fig. 5.16-4 shows the ray diagram. The distance OL, which equals IL, is therefore the focal length of the lens.

Diagram

Fig. 5.16-4 — Ray Diagram For The Pin And Plane Mirror Method Drawn on its side for clarity. A thin convex lens L stands upright at the centre, with a plane mirror M drawn as a vertical hatched line close behind it on the right. The object point O is marked at the left on the principal axis, with the image point I marked just beside it. Two rays leave O, strike the lens and emerge parallel to the principal axis, travel to the mirror, are reflected straight back along the same lines, pass through the lens again and converge to the point I next to O. Mark the distance from O to the lens with a double-headed arrow labelled f, and mark the two positions x₁ and x₂. Label O, I, L, M and f.

Note that the position of the plane mirror relative to the lens L — that is, the distance LM — does not affect the position of the image, as long as the rays from the lens fall normally on the plane mirror. The mirror M is nevertheless kept close to the convex lens L, so that the image I can be distinctly seen.

The two methods. The distant object method gives an approximate focal length, because a real object is never truly at infinity. The plane mirror method gives an exact value, because the parallel beam is produced by the lens itself.

Example 5.12 — Finding The Mean Focal Length From Three Observations

  1. Given: three parallax-free settings of a pin above a convex lens resting on a plane mirror. Observation 1: \(x = 15.2\ \text{cm}\), \(y = 15.6\ \text{cm}\). Observation 2: \(x = 15.0\ \text{cm}\), \(y = 15.4\ \text{cm}\). Observation 3: \(x = 15.4\ \text{cm}\), \(y = 15.6\ \text{cm}\). To find: the mean focal length of the lens in cm.

  2. Relation used: \(f = \dfrac{x + y}{2}\) for each observation, since the pin sits in the focal plane and \(x\) and \(y\) differ only by the thickness of the lens; the three values are then averaged.

  3. Working: \(f_1 = \dfrac{15.2 + 15.6}{2}\ \text{cm} = 15.4\ \text{cm}\); \(f_2 = \dfrac{15.0 + 15.4}{2}\ \text{cm} = 15.2\ \text{cm}\); \(f_3 = \dfrac{15.4 + 15.6}{2}\ \text{cm} = 15.5\ \text{cm}\). Mean \(f = \dfrac{15.4 + 15.2 + 15.5}{3}\ \text{cm} = \dfrac{46.1}{3}\ \text{cm} = 15.4\ \text{cm}\).

  4. Answer: the focal length of the lens is \(15.4\ \text{cm}\). Check: the three separate values differ by no more than \(0.3\ \text{cm}\), so a mean quoted to the nearest \(0.1\ \text{cm}\) is as much precision as the readings allow.

5.16.3 Alternative Method Using An Optical Bench

The above experiment can also be performed by using a pin, a convex lens and a plane mirror fixed on separate vertical stands of an optical bench, as shown in Fig. 5.16-5. We proceed as follows.

  1. Keep the lens L close to the plane mirror M. Adjust the height of the object pin O such that its tip lies at the centre of the convex lens L.
  2. Move the object pin O away from the lens L along the optical bench till its inverted image is seen, viewed by keeping the eye behind the object pin O at a distance of about 25 cm from it.
  3. Now adjust the position of the object pin O by gradual to-and-fro movement of it, so that the tip of the inverted image I coincides with the tip of the object pin O. In this situation, on moving the eye sideways, both the tip of the object pin and the tip of its image appear to move together, that is, there is no parallax.
  4. Note the position of the object pin, say \(x_1\), and the position of the lens, say \(x_2\), on the optical bench. The difference \(x_2 - x_1\) gives the focal length of the convex lens, that is

\[f = x_2 - x_1\]

where \(x_1\) and \(x_2\) are the two positions read off the bench scale in cm, and \(f\) is the focal length in cm.

Diagram

Fig. 5.16-5 — Determination Of Focal Length By The One Pin Method On An Optical Bench A horizontal optical bench drawn as a long graduated scale marked 0, 10, 20, 30, 40, 50, 60, 70, 80 cm, with three uprights standing on it. At the 30 cm mark, an upright carrying a vertical object pin O, with its small inverted image I drawn just above it. At the 60 cm mark, an upright carrying a convex lens L. Immediately behind the lens, an upright carrying a plane mirror M drawn as a hatched vertical line. An eye symbol at the far left looks along the bench towards the pin. Mark the position of the pin as x₁ and the position of the lens as x₂, and mark the distance between them with a double-headed arrow labelled f. Label O, I, L, M, x₁, x₂ and f.

In Fig. 5.16-5, \(x_1 = 30.0\ \text{cm}\) and \(x_2 = 60.0\ \text{cm}\), so

\[f = 60.0\ \text{cm} - 30.0\ \text{cm} = 30.0\ \text{cm}\]

The explanation is exactly the one given in Section 5.16.2. The pin is again sitting in the focal plane of the lens, the mirror is again returning a parallel beam along its own path, and the image again forms at the object itself.

In a one pin experiment on an optical bench, a student records the position of the lens as 72.5 cm and, after removing the parallax, the position of the object pin as 54.0 cm. Find the focal length of the lens, and state its power in dioptre.

5.17 To Differentiate Between A Convex And A Concave Lens

Handed an unmarked lens, you can name it in a few seconds. There are two simple tests, and neither needs any apparatus.

By Touching

If the lens is thick in the middle and thin at the edges, the lens is convex. If the lens is thin in the middle and thick at the edges, the lens is concave. Run a fingertip from the rim to the centre and the answer is in your hand.

By Seeing The Image

The second test uses the image, and can be done in two ways.

On keeping the lens near a printed page, if the letters appear magnified, the lens is convex, and if the letters appear diminished, the lens is concave. This follows from the cases of Sections 5.7.6 and 5.8.2: held close to the print, a convex lens gives an erect magnified virtual image, while a concave lens always gives an erect diminished one.

On seeing a distant object through the lens, if its inverted image is seen, the lens is convex, and if an upright image is seen, the lens is concave. A distant object is far beyond the focus of the convex lens, so its image is real and inverted; a concave lens can never invert anything.

Diagram

Fig. 5.17-1 — Telling The Two Lenses Apart By The Image Two drawings side by side, each showing a lens held a short distance above a printed line of text reading ‘lens’. (a) A convex lens, thick in the middle: through it the word ‘lens’ is seen clearly larger than the surrounding print, upright. Caption: convex — letters magnified. (b) A concave lens, thin in the middle: through it the word ‘lens’ is seen clearly smaller than the surrounding print, upright. Caption: concave — letters diminished.

Table 5.6 puts the tests side by side.

Table 5.6 — Tests To Identify A Convex And A Concave Lens | Test | Convex lens | Concave lens | |—|—|—| | By touching | Thick in the middle, thin at the edges | Thin in the middle, thick at the edges | | Letters of a printed page held near the lens | Appear magnified | Appear diminished | | A distant object seen through the lens | Its image is inverted | Its image is upright |

Q. Looking through an unmarked lens at a lamp-post across the road, you see it standing on its head. Which lens is it?

A. Convex. Only a convex lens can form an inverted image, and a distant object lies well beyond its focus.

Borrow two spectacle lenses from different people in your family and try both tests on each. Note that a lens for reading behaves one way and a lens for distance the other, and check that touching and looking give you the same answer every time.

A student holds a lens close to a page and finds the letters diminished, then turns towards a window and sees the trees outside upright. Name the lens, and state which one of the two observations alone would have been enough to identify it.

5.18 Summary

  1. A lens is a transparent refracting medium bounded either by two spherical surfaces, or by one spherical surface and one plane surface. A plane surface is treated as a spherical surface of infinite radius of curvature.
  2. A convex, or converging, lens is thick in the middle and thin at its periphery. A concave, or diverging, lens is thin in the middle and thick at its periphery.
  3. Convex lenses are of three kinds — bi-convex, plano-convex and concavo-convex. Concave lenses are of three kinds — bi-concave, plano-concave and convexo-concave. A concavo-convex lens is thicker in the middle and converges light; a convexo-concave lens is thinner in the middle and diverges light.
  4. A lens may be treated as a set of prisms, or more simply as a rectangular slab at the centre with one prism on either side of it. Since a prism bends light towards its base, a convex lens converges parallel rays to a point F, and a concave lens diverges them so that they appear to come from a point F on the side of the incident rays.
  5. The centre of curvature of a surface is the centre of the sphere of which that surface is a part; its radius is the radius of curvature. A lens has two of each. A lens is equi-convex or equi-concave when the two radii of curvature are equal.
  6. The principal axis is the line joining the two centres of curvature. The optical centre O is the point on it such that a ray directed towards it passes undeviated through a thin lens.
  7. A lens has two principal foci. Rays coming from the first focus \(\text{F}_1\), or appearing to meet at it, emerge parallel to the principal axis. Rays incident parallel to the principal axis pass through the second focus \(\text{F}_2\), or appear to diverge from it.
  8. The focal length is the distance of a focus from the optical centre, measured in m or cm. If the medium on both sides of the lens is the same, \(f_1 = f_2\) numerically, and the word focus by itself means the second focus.
  9. A focal plane is a plane through a focus, normal to the principal axis. An oblique parallel beam converges at, or appears to diverge from, a point in the second focal plane where the ray through the optical centre meets it.
  10. A convex lens has a real focus; a concave lens has a virtual focus.
  11. The focal length of a lens depends only on the refractive index of its material relative to the surrounding medium, and on the radii of curvature of its two surfaces. Placed in water, a lens has a longer focal length; a thick lens has a shorter focal length than a thin lens of the same material.
  12. If a part of a lens is covered, the focal length, the position, the size and the nature of the image do not change. Only the amount of light entering the lens falls, so the image becomes less intense.
  13. Light is refracted at both surfaces of a lens and bends the same way at each, so the total deviation is the sum of the two. In every lens the deviation is towards the thicker portion.
  14. Three principal rays are used to construct ray diagrams: the ray through the optical centre, which is undeviated; the ray incident parallel to the principal axis, which goes through \(\text{F}_2\) or appears to come from it; and the ray through \(\text{F}_1\), or directed towards it, which emerges parallel to the principal axis. Any two of them locate an image point.
  15. A real image is formed where refracted rays actually meet; it can be caught on a screen and is inverted. A virtual image is formed where they only appear to meet; it cannot be caught on a screen and is erect.
  16. For a convex lens, as the object moves in from infinity to the optical centre, the image moves out from \(\text{F}_2\) to infinity and then becomes virtual: at infinity the image is at \(\text{F}_2\), real, inverted and highly diminished; beyond \(2\text{F}_1\) it is between \(\text{F}_2\) and \(2\text{F}_2\), real, inverted and diminished; at \(2\text{F}_1\) it is at \(2\text{F}_2\), real, inverted and of the same size; between \(\text{F}_1\) and \(2\text{F}_1\) it is beyond \(2\text{F}_2\), real, inverted and magnified; at \(\text{F}_1\) it is at infinity, real, inverted and highly magnified; and between the lens and \(\text{F}_1\) it is on the same side, virtual, erect and magnified.
  17. For a concave lens the image is always virtual, erect and diminished, and lies between the optical centre and the focus, on the same side as the object, whatever the position of the object.
  18. The image formed by a convex lens may be real or virtual, magnified, diminished or of the same size, and inverted or erect. The image formed by a concave lens is always virtual, always diminished and always erect.
  19. By the Cartesian sign convention, all distances are measured from the optical centre along the principal axis. Distances measured in the direction of the incident light are positive and those opposite to it are negative; heights above the principal axis are positive and those below it are negative. Hence \(u\) is always negative, \(f\) is positive for a convex lens and negative for a concave lens, and \(v\) is positive for a real image and negative for a virtual image.
  20. The lens formula, the same for a convex and a concave lens, is \[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\] where \(u\) is the object distance, \(v\) the image distance and \(f\) the focal length, all measured from the optical centre in the same unit, m or cm, and all carrying their proper signs. It holds for a thin lens with the same medium on both sides.
  21. The linear magnification is \[m = \frac{\text{I}}{\text{O}} = \frac{v}{u}\] where I is the height of the image and O the height of the object in the same unit, and \(u\) and \(v\) are in the same unit. It has no unit. A negative \(m\) means a real, inverted image and a positive \(m\) a virtual, erect one; a numerical value above 1 means magnified, 1 means the same size, and below 1 means diminished. For a concave lens \(m\) is always positive and always less than 1.
  22. The power of a lens is the measure of the deviation it produces, and is the reciprocal of its focal length: \[P = \frac{1}{f}\] where \(f\) is the focal length in metre (m) and \(P\) is the power in dioptre (D). A lens of focal length 1 m has a power of 1 D. The power is positive for a convex lens and negative for a concave lens.
  23. Two thin lenses placed in contact have a power equal to the algebraic sum of their individual powers. A \(+2.0\ \text{D}\) convex lens with a \(-2.0\ \text{D}\) concave lens gives zero power and behaves like a glass plate.
  24. The least distance of distinct vision D for a normal eye is 25 cm, and the eye cannot see distinctly an object subtending an angle less than \(1'\). A magnifying glass, or simple microscope, is a convex lens of short focal length that is held with the object within its focal length, giving a virtual, erect and magnified image at the distance D.
  25. The magnifying power of a simple microscope is \[M = 1 + \frac{\text{D}}{f}\] where D is the least distance of distinct vision and \(f\) the focal length of the lens, both in the same unit; \(M\) has no unit. A shorter focal length gives a greater magnifying power, but it cannot be increased indefinitely.
  26. Convex lenses are used as the objective of a telescope, camera and slide projector, as the eye lens, as a magnifying glass, in the collimator of a spectroscope, and in spectacles for hypermetropia. Concave lenses are used in spectacles for myopia and as the eye lens of a Galilean telescope. A bifocal lens has a concave upper part and a convex lower part.
  27. A single convex lens forms a coloured or blurred image because the focal length differs slightly for different colours. This defect is called chromatic aberration, and a combination of convex and concave lenses is used to overcome it.
  28. The focal length of a convex lens is estimated by the distant object method, in which the sharp image of a far object is thrown on a wall and the lens-to-wall distance read off a metre rule. It is found exactly by the plane mirror method, in which a pin is adjusted above a lens resting on a plane mirror until there is no parallax between the pin and its inverted image, giving \[f = \frac{x + y}{2}\] where \(x\) is the distance of the pin from the lens and \(y\) its distance from the mirror, both in cm. On an optical bench the same experiment gives \(f = x_2 - x_1\), where \(x_1\) is the position of the pin and \(x_2\) that of the lens in cm.
  29. An unmarked lens is identified by touch, by whether print held near it looks magnified or diminished, and by whether a distant object seen through it looks inverted or upright.

Flowchart

FC 5.2 — What A Convex Lens Does As The Object Comes In Object at infinity: image at F₂, real, inverted, highly diminished → object beyond 2F₁: image between F₂ and 2F₂, real, inverted, diminished → object at 2F₁: image at 2F₂, real, inverted, same size → object between F₁ and 2F₁: image beyond 2F₂, real, inverted, magnified → object at F₁: image at infinity, real, inverted, highly magnified → object between the lens and F₁: image on the same side, virtual, erect, magnified

5.19 Key Terms

  1. Centre of curvature — the centre of the sphere of which a surface of the lens is a part; a lens has two of them, one for each surface.

  2. Concave lens — a lens that is thin in the middle and thick at its periphery, and spreads out the light falling on it.

  3. Construction rays — the three rays whose path after refraction is already known, used to locate an image in a ray diagram.

  4. Converging lens — another name for a convex lens, from its action of bringing a beam of light closer together.

  5. Convex lens — a lens that is thick in the middle and thin at its periphery, and brings the light falling on it closer together.

  6. Diverging lens — another name for a concave lens, from its action of spreading a beam of light apart.

  7. Equi-concave — describing a concave lens whose two surfaces have equal radii of curvature.

  8. Equi-convex — describing a convex lens whose two surfaces have equal radii of curvature.

  9. First focal length — the distance of the first focal point from the optical centre, symbol \(f_1\); SI unit metre (m), commonly measured in cm.

  10. First focal point — the point \(\text{F}_1\) on the principal axis from which rays come, or towards which they are directed, so as to emerge parallel to the principal axis.

  11. Focal length — the distance of a focus from the optical centre, symbol \(f\); SI unit metre (m), commonly measured in cm. Unqualified, it means the second focal length.

  12. Focal plane — a plane passing through a focus and normal to the principal axis; a lens has two of them.

  13. Least distance of distinct vision — the nearest distance at which a normal eye sees an object distinctly, symbol D, equal to 25 cm; SI unit metre (m).

  14. Lens — a transparent refracting medium bounded either by two spherical surfaces, or by one spherical surface and one plane surface.

  15. Lens formula — the relation \(\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\), connecting object distance, image distance and focal length for a thin lens.

  16. Linear magnification — the ratio of the height of the image to the height of the object, symbol \(m\), equal to \(v/u\); it is a ratio and has no unit.

  17. Magnifying power — the number of times a simple microscope increases the angle an object subtends at the eye, symbol \(M\), equal to \(1 + \text{D}/f\); it has no unit.

  18. Optical centre — the point O on the principal axis such that a ray of light directed towards it passes undeviated through a thin lens.

  19. Power of a lens — a measure of the deviation a lens produces, equal to the reciprocal of the focal length in metre, symbol \(P\); SI unit dioptre (D), where \(1\ \text{D} = 1\ \text{m}^{-1}\).

  20. Principal axis — the straight line joining the centres of curvature of the two surfaces of a lens.

  21. Principal foci — the two focal points of a lens, one on either side of it.

  22. Radius of curvature — the radius of the sphere of which a surface of the lens is a part; SI unit metre (m), commonly measured in cm.

  23. Real image — an image formed where the refracted rays actually meet; it can be obtained on a screen and is inverted.

  24. Second focal length — the distance of the second focal point from the optical centre, symbol \(f_2\); SI unit metre (m), commonly measured in cm.

  25. Second focal point — the point \(\text{F}_2\) through which rays incident parallel to the principal axis pass, or from which they appear to diverge, after refraction.

  26. Virtual image — an image formed where the refracted rays only appear to meet; it cannot be obtained on a screen and is erect.

5.20 Points To Ponder

  1. When an object is placed exactly at the first focus of a convex lens, we say the image is formed at infinity. No screen anywhere can ever catch it. In what sense, then, does such an image exist at all — and what would an eye placed in the emerging beam actually see?

  2. A convex lens made of glass converges light in air. Suppose it could be surrounded by a medium of refractive index greater than that of its own glass. What would the same lens then do to a parallel beam, and would it still be right to call it a converging lens?

  3. Cover the lower half of a convex lens and the image stays in the same place, at the same size, only fainter. Cut the same lens in half instead, along its principal axis, and each half still has the original focal length. What does this tell you about where in a lens the focusing is done?

  4. When two thin lenses are placed in contact their powers add, but their focal lengths do not. Which of the two quantities, focal length or power, describes what a lens does to light more directly, and why does the more useful description turn out to be the reciprocal of the more easily measured one?

  5. The magnifying power of a simple microscope rises as its focal length falls, yet no one grinds a lens of focal length one millimetre for classroom use. What physical limits stop us from shrinking the focal length as far as we please, and which of them are limits of the lens rather than of the eye?

  6. Every ray diagram in this chapter used two rays out of an infinite number leaving each point of the object. The other rays were never drawn. What would happen to the image if they were, and what does their omission assume about the lens?

5.21 Objective Questions

Multiple Choice
  1. A lens is thin at its middle and thick at its edges. The lens is: (1 mark)
    1. bi-convex
    2. concavo-convex
    3. bi-concave
    4. plano-convex
  2. A ray of light directed towards the optical centre of a thin lens: (1 mark)
    1. emerges parallel to the principal axis
    2. passes undeviated through the lens
    3. passes through the second focus
    4. is totally reflected at the second surface
  3. The focal length of a thin lens does NOT depend upon: (1 mark)
    1. the radii of curvature of its two surfaces
    2. the refractive index of the material of the lens
    3. the medium surrounding the lens
    4. the area of the lens left uncovered
  4. An object is placed between \(\text{F}_1\) and \(2\text{F}_1\) of a convex lens. The image formed is: (1 mark)
    1. real, inverted and diminished
    2. real, inverted and of the same size
    3. real, inverted and magnified
    4. virtual, erect and magnified
  5. The power of a concave lens of focal length 50 cm is: (1 mark)
    1. \(+0.5\ \text{D}\)
    2. \(-0.5\ \text{D}\)
    3. \(+2.0\ \text{D}\)
    4. \(-2.0\ \text{D}\)
  6. A lens gives a magnification \(m = +0.4\). The lens and the image are: (1 mark)
    1. convex, image real
    2. convex, image magnified
    3. concave, image virtual
    4. concave, image inverted
Assertion–Reason

For each of the following, choose: a. both A and R are true and R is the correct explanation of A b. both A and R are true but R is not the correct explanation of A c. A is false but R is true d. A is true but R is false

  1. Assertion (A): A virtual image cannot be obtained on a screen. Reason (R): The refracted rays forming a virtual image meet only when they are produced backwards. (1 mark)

  2. Assertion (A): Covering the upper half of a convex lens removes the upper half of the image. Reason (R): Covering a part of a lens does not change its focal length. (1 mark)

Match The Following
  1. Match the position of the object in front of a convex lens in Column A with the image in Column B. (4 marks)
Column A Column B
a. At infinity i. Image at infinity, highly magnified
b. At \(2\text{F}_1\) ii. Image at \(\text{F}_2\), highly diminished
c. At \(\text{F}_1\) iii. Image on the same side, virtual and magnified
d. Between the lens and \(\text{F}_1\) iv. Image at \(2\text{F}_2\), of the same size
Fill In The Blanks
  1. Fill in each blank. (3 marks)
    1. The image formed by a concave lens is always virtual, erect and ……….. .
    2. A ray incident parallel to the principal axis of a concave lens, after refraction, appears to come from its ……….. .
    3. For two thin lenses in contact, the power of the combination is the ……….. sum of their individual powers.
True Or False
  1. Write T for true and F for false. (5 marks)
    1. A concavo-convex lens has a diverging action on a beam of light.
    2. For a convex lens, the object distance \(u\) is always negative.
    3. A concave lens can form an image twice the size of the object.
    4. A lens of shorter focal length has a greater power.
    5. The first and second focal lengths of a lens are numerically equal when the medium on both sides of it is the same.
Name The Following
  1. Name the following. (4 marks)
    1. The point on the principal axis through which a ray of light passes undeviated.
    2. The lens used as the eye lens of a Galilean telescope.
    3. The plane passing through the second focus and normal to the principal axis.
    4. The defect because of which a single convex lens forms a coloured image.
Give One Word Or Phrase For
  1. Give one word or phrase for each. (2 marks)
    1. The nearest distance at which a normal eye can see an object distinctly.
    2. The ratio of the height of the image to the height of the object.
State The SI Unit Of
  1. State the SI unit of each. (3 marks)
    1. focal length of a lens
    2. power of a lens
    3. linear magnification
Odd One Out
  1. Pick the odd one out and give a reason: bi-convex, plano-convex, concavo-convex, convexo-concave. (1 mark)

5.22 Check Your Concepts

  1. Define the power of a lens and state its unit. (2 marks)

  2. Give a reason: a glass convex lens has a longer focal length in water than it has in air. (2 marks)

  3. Why is the focus of a concave lens said to be virtual, while that of a convex lens is real? Answer with reference to a beam incident parallel to the principal axis. (3 marks)

  4. State the three principal rays used to construct the ray diagram for a lens, giving the path of each after refraction through a convex lens. (3 marks)

  5. Differentiate between a convex and a concave lens, stating the basis of difference in each case: (a) shape, (b) action on a parallel beam of light, (c) nature of the focus. (3 marks)

  6. The upper half of a convex lens forming a sharp image on a screen is covered with a card. State what happens to the position, the size, the nature and the brightness of the image, and give a reason for each answer. (4 marks)

  7. State any four rules of the sign convention used in the measurement of distances for a lens, and hence state the sign of \(u\), of \(f\) and of \(v\) for a concave lens. (4 marks)

  8. Explain how a convex lens is used as a magnifying glass. State where the object must be placed with respect to the focus, where the image is formed, and the three characteristics of the image. (4 marks)

  9. Describe how the position, the size and the nature of the image formed by a convex lens change as an object is brought steadily from infinity up to the lens. (5 marks)

5.23 Diagrams, Graphs & Constructions

  1. Draw a ray diagram to show the formation of the image when an object is placed at \(2\text{F}_1\) of a convex lens. Mark \(\text{F}_1\), \(\text{F}_2\), \(2\text{F}_1\) and \(2\text{F}_2\), and state three characteristics of the image. (4 marks)

  2. Fig. 5.6-6 shows the three construction rays for a concave lens. Redraw the diagram for an object placed at a distance equal to twice the focal length of the lens, locate the image, and state its nature and size. (4 marks)

  3. In Fig. 5.3-6, an oblique parallel beam falls on a convex lens and is brought together at the point B. Explain how the position of B is found, and give a reason why the beam does not converge at \(\text{F}_2\). (3 marks)

  4. Draw a labelled diagram of the apparatus used to determine the focal length of a convex lens by the auxiliary plane mirror method. Mark on it the two distances that are measured, and write the relation used. (4 marks)

  5. Draw a ray diagram to show how a converging lens is used as a magnifying glass. Mark the two foci and the position of the eye, and label the image. (4 marks)

  6. Fig. 5.17-1 shows the same printed word seen through two different lenses. Identify each lens and give one reason for each identification. (2 marks)

  7. Using a ray diagram, show that a diverging lens cannot form a real image of an object placed anywhere on its principal axis. (3 marks)

5.24 Practice With Numericals

  1. Find the power of (a) a convex lens of focal length 40 cm, and (b) a lens of focal length 20 cm which forms only virtual erect diminished images. (2 marks)

  2. An object is placed 30 cm in front of a convex lens of focal length 20 cm. Find the position of the image and state its nature. (3 marks)

  3. An object is placed 10 cm in front of a concave lens of focal length 15 cm. Find the position of the image and the magnification. (3 marks)

  4. A convex lens forms a sharp image on a screen placed 45 cm behind it, when the object is 90 cm in front of it. Find the focal length of the lens and the magnification of the image. (3 marks)

  5. An object 5.0 cm tall is placed 20 cm in front of a convex lens of focal length 12 cm. Find the position of the image, its magnification and its height, and state whether it is erect or inverted. (4 marks)

  6. A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 25 cm. Find the power of each lens, the power of the combination, and the focal length of the combination in cm. (3 marks)

  7. A lens forms an image one-quarter the size of an object placed 60 cm in front of it, and the image can be caught on a screen. Find the image distance and the focal length of the lens, and name the kind of lens. (4 marks)

  8. A lens of power \(+4.0\ \text{D}\) forms a real image twice the size of the object. Find the distance of the object from the lens and the distance of the image from the lens. (4 marks)