Chapter 4: Refraction of Light at Plane Surfaces
About a thousand years ago, a mathematician named Ibn Sahl was working in Baghdad on a problem that sounds very practical: how to shape a piece of glass so that it gathers sunlight into a single hot point. Burning instruments — lenses and curved mirrors used to start a fire from sunlight — were his subject. To design one, he had to know exactly where a ray of light goes after it enters the glass.
He already knew that a ray entering glass does not carry straight on. It turns at the surface. What nobody had set down was by how much. So Ibn Sahl drew the incoming ray, the surface it struck, and the turned ray inside the glass, and completed a right-angled triangle on each of the two rays. When he compared one particular side of the first triangle with the matching side of the second, he found that their lengths always stood in the same ratio — no matter how slanted the incoming ray was.
That single unchanging ratio is what this chapter will call the refractive index. His manuscript ended up scattered between two libraries, and what he had actually done was recognised only in recent decades. But the physics in his hand is the physics you are about to meet: light travels at different speeds in different transparent materials, and that change of speed is what makes a ray bend.
Diagram
Fig. 4.0-1 — Ibn Sahl’s Construction For A Bending Ray A horizontal line marks the flat surface of a glass block, air above and glass below, labelled ‘air’ and ‘glass’. A slanting ray comes down from the upper left and meets the surface at a point O; below the surface it continues into the glass along a different, steeper line. Draw a dashed vertical normal through O. Take equal lengths along the incident ray above O and along the refracted ray below O, and from each end point drop a dashed perpendicular onto the normal, forming two right-angled triangles that share the normal. Label the two perpendiculars clearly as the two lengths whose ratio stays the same. Caption the figure so the reader sees that only the ratio of these two dashed lengths matters.
4.1 Refraction Of Light
In Class IX you studied the reflection of light — the return of light into the same medium after it strikes a polished surface. You met its two laws there: the angle of reflection equals the angle of incidence, and the incident ray, the normal at the point of incidence and the reflected ray all lie in one plane.
This chapter takes up what happens to the rest of the light — the part that does not come back, but goes on into the second medium. That part changes direction, and in this chapter you will follow it through plane surfaces: a water surface, a glass block, a prism.
4.1.1 The Speed Of Light In Different Media
Light travels fastest in vacuum. Its speed there is \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), and this is the largest speed anything can have. In every transparent material substance light travels more slowly than this.
The speed in air is very nearly the same as in vacuum, so for our purposes both are taken as \(3 \\times 10^{8}\\ \\text{m s}^{-1}\). In water and in glass light is distinctly slower, as Table 4.1 shows.
| Medium | Speed of light |
|---|---|
| Vacuum | \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) |
| Air | \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) (very nearly) |
| Water | \(2.25 \\times 10^{8}\\ \\text{m s}^{-1}\) |
| Glass | \(2 \\times 10^{8}\\ \\text{m s}^{-1}\) |
Table 4.1 Speed of light in some common media
Within any one transparent homogeneous medium — a medium that is the same all through, with no change in its material from place to place — light travels in a straight line at a constant speed. A change of speed therefore happens only where one medium ends and another begins.
4.1.2 Optically Denser And Optically Rarer Media
Because light has a different speed in each medium, we can compare any two media by the speed light has in them.
While passing from one medium to another, if light slows down, the second medium is said to be optically denser than the first. If light speeds up, the second medium is said to be optically rarer than the first.
So water and glass are optically denser than air, which is another way of saying air is optically rarer than water and glass. Glass is optically denser than water; water is optically rarer than glass.
Now a warning that is easy to miss. Optical density has nothing to do with the ordinary density of a medium. Ordinary density depends on how closely the molecules of a substance are packed together; optical density depends only on the speed of light in the substance. Kerosene is less dense than water — that is why it floats on water — and yet it is optically denser than water, because light travels more slowly in kerosene. Table 4.2 sets the two ideas side by side.
4.1.3 Partial Reflection And Partial Refraction At A Boundary
Consider a ray of light travelling in one transparent medium that strikes the surface of another transparent medium obliquely — that is, at any angle other than straight on. The light does not all do the same thing at that surface. Part of it goes back into the first medium, obeying the laws of reflection; this part is the reflected light. The remaining part enters the second medium and travels on in a straight line, but in a direction different from its original one; this part is the refracted light.
Thus, at the boundary separating two media, light suffers partial reflection and partial refraction. This is shown in Fig. 4.1-1, where SS′ is the surface separating medium 1 (air) from medium 2 (glass).
Diagram
Fig. 4.1-1 — Refraction From A Rarer Medium To A Denser Medium A horizontal line SS′ separates two regions: above it ‘medium 1, air (rarer)’, below it ‘medium 2, glass (denser)’. A dashed normal NOM is drawn perpendicular to SS′ through the point O on the surface. An incident ray AO comes down from the upper left to O, labelled ‘incident ray’, with the angle AON between it and ON marked as the angle of incidence i. A weaker reflected ray OC goes up to the upper right, labelled ‘reflected ray’, with its angle to ON marked. A refracted ray OB continues below the surface, bent closer to the normal OM than the dashed straight-on continuation of AO, labelled ‘refracted ray’; the angle BOM is marked as the angle of refraction r. Draw the straight-on continuation of AO as a faint dashed line and mark the small angle between it and OB as the deviation δ. Show r less than i.
The refracted light is always less bright than the incident light, because some of the incident light’s intensity has been carried away by the reflected part.
Fig. 4.1-1 also fixes the names we shall use throughout the chapter. For the ray AO, called the incident ray, OB is the refracted ray and OC the reflected ray, and NOM is the normal — the line drawn perpendicular to the surface at the point of incidence O. The angle the incident ray makes with the normal, \(\\angle\)AON, is the angle of incidence \(i\); the angle the refracted ray makes with the normal, \(\\angle\)BOM, is the angle of refraction \(r\). Note carefully that \(r\) is not equal to \(i\) — the ray OB does not lie along the straight-on direction of AO.
Refraction is essentially a surface phenomenon: the ray changes direction only at the boundary, and travels straight both before it and after it.
4.1.4 How A Ray Bends At A Surface
Experiment shows that a ray bends in one of three ways at a boundary, and which one it is depends only on the two media and on how the ray arrives.
Rarer To Denser
When a ray of light travels from a rarer medium to a denser medium — say from air to glass — it bends towards the normal, so that \(\\angle r < \\angle i\). This is the case already drawn in Fig. 4.1-1.
Before we go further we need one more quantity. The ray has been turned away from the path it was following, and the angle through which it has been turned is called the angle of deviation, denoted by \(\\delta\). It is the angle between the direction of the refracted ray and the direction of the incident ray, and like every other angle here it is measured in degrees.
For refraction from a rarer to a denser medium,
\[\\delta = i - r\]
where \(i\) is the angle of incidence in degrees, \(r\) the angle of refraction in degrees, and \(\\delta\) the angle of deviation in degrees. Since the ray has bent towards the normal, \(r\) is the smaller angle, and the difference is positive.
Denser To Rarer
When a ray of light travels from a denser medium to a rarer medium — say from glass to air — it bends away from the normal, so that \(\\angle r > \\angle i\). This case is drawn in Fig. 4.1-2. Now the refracted ray is the one further from the normal, so the deviation is
\[\\delta = r - i\]
with the same meanings and the same unit as before.
Diagram
Fig. 4.1-2 — Refraction From A Denser Medium To A Rarer Medium The same layout as Fig. 4.1-1 but with the media interchanged: above the surface SS′ write ‘medium 1, glass (denser)’, below it ‘medium 2, air (rarer)’. Dashed normal NOM through O. Incident ray AO from the upper left with angle of incidence i marked between AO and ON; weak reflected ray OC to the upper right; refracted ray OB below the surface, bent away from the normal so that the angle of refraction r between OB and OM is clearly larger than i. Show the faint dashed straight-on continuation of AO and mark the deviation δ between it and OB.
Normal Incidence
A ray that strikes the surface normally — that is, along the normal itself, so that \(\\angle i = 0^\\circ\) — passes into the second medium undeviated. It suffers no bending at all: \(\\angle r = 0^\\circ\) and \(\\delta = 0^\\circ\). Fig. 4.1-3 shows this.
Diagram
Fig. 4.1-3 — Refraction At Normal Incidence A horizontal surface SS′ with ‘medium 1, air (rarer)’ above and ‘medium 2, glass or water (denser)’ below. A ray AO comes vertically down onto the point O, making an angle of \(90^\\circ\) with the surface; mark the angle of incidence as i = 0° and mark the right angle at the surface. The ray continues vertically downwards as OB below the surface, labelled ‘refracted ray’, with the angle of refraction marked r = 0°. No bending is shown.
From this point onward, the reflected ray at a boundary will not be drawn in our diagrams, so that the refracted path stays clear. Remember that it is always there.
4.1.5 The Cause Of Refraction
Why should a ray bend at all? The answer is the one thing we established at the start of this section: light travels at different speeds in different media.
When a ray passes from one medium to another, its direction changes because its speed changes. If light slows down on entering the second medium, the ray bends towards the normal. If it speeds up, the ray bends away from the normal. And if the speed of light happens to be the same in both media, the ray does not bend at all — it passes undeviated, however slanted it may be.
Normal incidence is the one case where speed and direction come apart. A ray arriving along the normal does change its speed as it crosses the boundary, but it has no direction to change: it carries straight on. Bending needs the ray to strike the surface obliquely.
4.2 Laws Of Refraction And Refractive Index
Section 4.1 told us which way a ray bends. It did not tell us by how much. That is what the laws of refraction do, and they turn refraction from a description into something you can calculate.
4.2.1 The Two Laws Of Refraction
Refraction of light obeys two laws, given by the Dutch scientist Willebrord Snell, and known after his name as Snell’s laws.
The first law is about where the rays lie. Look again at any of the diagrams in Section 4.1: the incident ray, the normal and the refracted ray have all been drawn on the flat page. That is not an accident of drawing. The three always lie in one plane, so a refracted ray never swings sideways out of the plane in which the light arrived.
The second law is about the size of the angles. If you send rays into the same block of glass at many different angles of incidence, both \(i\) and \(r\) change from trial to trial — but the ratio of their sines does not. It stays fixed for that pair of media.
The second law is written
\[\\frac{\\sin i}{\\sin r} = {}*{1}\\mu*{2}\]
where \(i\) is the angle of incidence in medium 1 in degrees, \(r\) the angle of refraction in medium 2 in degrees, and \(*{1}\\mu*{2}\) a constant for that pair of media. Because it is a ratio of two sines, and a sine is a ratio of two lengths, the constant carries no unit at all.
4.2.2 Refractive Index Of One Medium With Respect To Another
The constant \(*{1}\\mu*{2}\) has a name of its own. It is called the refractive index of the second medium with respect to the first, and it is written with the Greek letter \(\\mu\) (read as ‘mew’). The small figures tell you the order: the light starts in medium 1 and enters medium 2.
So a single number, measured once, settles every angle for that pair of media. Fig. 4.2-1 shows what the first law means in three dimensions, which a flat diagram can easily hide.
Diagram
Fig. 4.2-1 — The Incident Ray, The Normal And The Refracted Ray Lie In One Plane Show a horizontal boundary surface drawn in perspective as a shaded rectangle, air above and glass below. A vertical dashed normal rises from a point O on the surface. An incident ray comes down to O from the upper left and a refracted ray continues below the surface. Draw a semi-transparent tinted vertical plane that contains all three lines — incident ray, normal and refracted ray — and let it cut the surface along a straight line, showing that the three lines are flat within this one plane. Label the plane ‘plane of incidence’. Add one faint dotted ray going sideways out of this plane, with a cross through it, labelled ‘never happens’.
4.2.3 Effect Of Refraction On Speed, Frequency And Wavelength
A light wave crossing a boundary has three things that could change: its speed, its frequency and its wavelength. Exactly two of them do.
The speed changes. When light is refracted from a rarer to a denser medium, its speed decreases; when it is refracted from a denser to a rarer medium, its speed increases.
The frequency does not change. The frequency \(f\) is the number of waves sent out each second, and that is settled by the source of the light, not by the medium the light happens to be passing through. A source that emits \(5 \\times 10^{14}\) waves each second goes on delivering that many each second into glass, water or air alike.
The wavelength therefore changes. The speed \(V\) of light in a medium, its frequency \(f\) and its wavelength \(\\lambda\) in that medium are connected by
\[V = f\\lambda \\qquad \\text{or} \\qquad \\lambda = \\frac{V}{f}\]
where \(V\) is the speed in that medium in \(\\text{m s}^{-1}\), \(f\) the frequency in hertz (Hz), and \(\\lambda\) the wavelength in that medium in m.
Read the second form as a statement rather than a rule: with \(f\) pinned down by the source, the wavelength has no choice but to follow the speed. When light passes from a rarer to a denser medium the speed falls, so the wavelength decreases. When it passes from a denser to a rarer medium the speed rises, so the wavelength increases. Fig. 4.2-2 shows this, and FC 4.1 sets out the whole chain in order.
Diagram
Fig. 4.2-2 — Wavelength Shortens In A Denser Medium Show a horizontal boundary line with ‘air (rarer)’ above and ‘glass (denser)’ below. Above the line draw a row of evenly spaced parallel wave crests as short parallel lines perpendicular to the direction of travel, travelling downwards; below the line draw the same wave continuing with the crests clearly closer together. Mark the spacing above as λ and the spacing below as λ′, with λ′ smaller. Label ‘frequency f is the same on both sides’ with an arrow spanning the boundary, and label the two regions ‘speed c’ and ‘speed V, smaller’.
Flowchart
FC 4.1 — What Changes When Light Enters An Optically Denser Medium Light crosses the boundary → speed of light decreases → frequency stays unchanged (it is fixed by the source) → wavelength decreases → ray bends towards the normal
4.3 Refractive Index And The Speed Of Light
Section 4.2 defined refractive index from angles you can measure with a protractor. Section 4.1 said that bending is caused by a change of speed. These two accounts must fit together, and this section shows how: the refractive index is nothing but a comparison of two speeds.
4.3.1 Absolute Refractive Index
The speed of light is greatest in vacuum, where it equals \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), and it is denoted by the symbol \(c\). In air the speed is so nearly the same that we treat the two as equal. In every other transparent medium the speed is less than this.
For that reason the refractive index of a medium is usually quoted not against some other medium but against vacuum (or air). Quoted this way it is called the absolute refractive index of the medium, or simply its refractive index, and it is written \(\\mu\).
\[\\mu = \\frac{\\text{speed of light in vacuum or air } (c)}{\\text{speed of light in that medium } (V)}\]
where \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\) and \(V\) is the speed of light in the medium in \(\\text{m s}^{-1}\).
Two consequences follow at once from this one line. First, the refractive index of a transparent medium is always greater than 1 and can never be less, because \(V\) is always less than \(c\). Second, for air or vacuum itself \(\\mu = 1\), since there \(V\) and \(c\) are the same.
So \(\\mu\) is a statement about slowing down. Saying that the refractive index of diamond is \(2.41\) says that light travels \(2.41\) times faster in air than in diamond, or that its speed in diamond is \(\\frac{1}{2.41}\) of its speed in air. Since the speed in air is \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), the speed of light in diamond is \(1.245 \\times 10^{8}\\ \\text{m s}^{-1}\).
4.3.2 Refractive Indices Of Some Common Substances
Table 4.3 lists the refractive indices of some common transparent substances. Reading down it is a way of reading how much each substance slows light down.
| Substance | \(\\mu\) | Substance | \(\\mu\) |
|---|---|---|---|
| Vacuum | 1.00 | Paraffin oil | 1.44 |
| Air | 1.00 (1.0003) | Glycerine | 1.47 |
| Ice | 1.31 | Turpentine oil | 1.47 |
| Water | 1.33 | Ordinary glass | 1.5 |
| Methylated spirit | 1.36 | Crown glass | 1.53 |
| Ether | 1.36 | Quartz | 1.54 |
| Alcohol | 1.37 | Rock salt | 1.56 |
| Kerosene | 1.41 | Carbon disulphide | 1.63 |
| Sulphuric acid | 1.43 | Flint glass | 1.65 |
| Ruby | 1.76 | ||
| Diamond | 2.41 |
Table 4.3 Refractive indices of some common substances, measured with respect to air
Notice the entry for kerosene. Its refractive index, \(1.41\), is larger than that of water, \(1.33\) — so kerosene is optically denser than water, even though it is light enough to float on it. This is the point made in Section 4.1.2, now in numbers.
4.3.3 Relative Refractive Index In Terms Of Speeds
The refractive index of one medium with respect to another can be written in speeds too. In general,
\[*{1}\\mu*{2} = \\frac{\\text{speed of light in medium 1}}{\\text{speed of light in medium 2}} = \\frac{V\_1}{V\_2}\]
where \(V\_1\) and \(V\_2\) are the speeds of light in medium 1 and medium 2 in \(\\text{m s}^{-1}\), and \(*{1}\\mu*{2}\) has no unit.
This can be turned into something more useful. From the definition of absolute refractive index, \(V\_1 = c/\\mu\_1\) and \(V\_2 = c/\\mu\_2\), where \(\\mu\_1\) and \(\\mu\_2\) are the absolute refractive indices of the two media. Substituting these,
\[*{1}\\mu*{2} = \\frac{V\_1}{V\_2} = \\frac{c/\\mu\_1}{c/\\mu\_2} = \\frac{\\mu\_2}{\\mu\_1}\]
The \(c\) in the numerator and denominator cancels, leaving a result worth remembering as a sentence: the refractive index of medium 2 with respect to medium 1 is the absolute refractive index of medium 2 divided by that of medium 1. Note the order carefully — the medium the light is entering goes on top.
One thing follows that surprises students. A relative refractive index can be less than 1. If light goes from an optically denser medium into a rarer one, \(\\mu\_2\) is smaller than \(\\mu\_1\) and the ratio comes out below 1. Only the absolute refractive index is barred from being less than 1.
4.3.4 Conditions For A Light Ray To Pass Undeviated
A ray passes from medium 1 into medium 2 without any bending in either of two cases.
When the angle of incidence at the boundary is zero, that is \(\\angle i = 0^\\circ\), so that \(\\angle r = 0^\\circ\). This is the normal incidence already drawn in Fig. 4.1-3.
When the refractive index of medium 2 is the same as that of medium 1, that is \(\\mu\_1 = \\mu\_2\), so that \(i = r\). This is drawn in Fig. 4.3-1.
The second case is worth pausing on. If the two refractive indices are equal, the speed of light is the same in both media, and Section 4.1.5 told us that a ray with no change of speed has no reason to bend. The ray goes straight on even when it strikes the surface at a large angle — and since nothing marks where one medium ends and the other begins, the second medium becomes invisible in the first.
Diagram
Fig. 4.3-1 — No Deviation When The Two Media Have The Same Refractive Index A horizontal surface SS′ with a dashed normal through the point O. An incident ray AO comes down obliquely from the upper left, making a marked angle i with the normal. Label the region above the surface μ and the region below the surface μ as well, to show the two are equal. The ray continues below the surface as OB along the same straight line as AO, with the angle r marked equal to i. Show clearly that AOB is one unbroken straight line.
4.3.5 Relationship Between The Wavelengths In Two Media
Section 4.2.3 established that the wavelength shortens when light enters a denser medium. We can now say by how much.
Take a ray of light of frequency \(f\) and wavelength \(\\lambda\) in air, where its speed is \(c\), entering a medium in which its speed is \(V\) and its wavelength becomes \(\\lambda'\). The frequency is unchanged, so it may be written from either side:
\[f = \\frac{c}{\\lambda} \\qquad \\text{and} \\qquad f = \\frac{V}{\\lambda'}\]
Since both expressions equal the same \(f\),
\[\\frac{c}{\\lambda} = \\frac{V}{\\lambda'} \\qquad \\text{so} \\qquad \\lambda' = \\frac{V}{c},\\lambda\]
But \(c/V\) is just the refractive index \(\\mu\) of the medium, so \(V/c\) is \(1/\\mu\). Hence
\[\\lambda' = \\frac{\\lambda}{\\mu}\]
where \(\\lambda\) is the wavelength in air in m, \(\\lambda'\) the wavelength in the medium in m, and \(\\mu\) the absolute refractive index of the medium.
The result reads exactly as it should. Going from a rarer to a denser medium, \(\\mu > 1\), so \(\\lambda'\) is less than \(\\lambda\) and the wavelength decreases. Going from a denser to a rarer medium the relative refractive index is less than 1, and the wavelength increases.
4.3.6 Factors Affecting The Refractive Index Of A Medium
The refractive index of a medium is not a fixed label attached to the substance for all time. It depends on three things.
Nature of the medium. Since \(\\mu = c/V\), the slower light travels in a medium compared with air, the greater is the refractive index of that medium. In glass \(V\_{\\text{glass}} = 2 \\times 10^{8}\\ \\text{m s}^{-1}\) and \(\\mu\_{\\text{glass}} = 1.5\); in water \(V\_{\\text{water}} = 2.25 \\times 10^{8}\\ \\text{m s}^{-1}\) and \(\\mu\_{\\text{water}} = 1.33\).
Physical condition, such as temperature. With an increase in temperature the refractive index of a medium decreases. By \(\\mu = c/V\) this means the speed of light in the medium increases as it is warmed.
Colour, or wavelength, of the light. In air or vacuum all colours travel at the same speed, but in any other transparent medium the speed of light is different for different colours. In a given medium red light travels fastest and violet light slowest. The refractive index of the medium is therefore greatest for violet light and least for red light, that is \(\\mu\_{\\text{violet}} > \\mu\_{\\text{red}}\).
Since the wavelength of visible light increases as we go from the violet end to the red end of the spectrum, the third statement can be put in one line: the refractive index of a medium decreases with an increase in the wavelength of light. This single fact will return again and again in the rest of the chapter, so it is worth fixing now.
4.4 Principle Of Reversibility Of The Path Of Light
There is a small and rather satisfying rule about light paths that we shall use repeatedly for the rest of the chapter. It says that light does not care which way round you send it.
Fig. 4.4-1 shows what this means at a single surface. A ray AO in medium 1 strikes the surface SS′ at an angle of incidence \(i\) and is refracted along OB in medium 2 at an angle of refraction \(r\). For this passage,
\[*{1}\\mu*{2} = \\frac{\\sin i}{\\sin r} \\qquad \\dots \\text{(i)}\]
Diagram
Fig. 4.4-1 — The Reversible Path Of A Refracted Ray A horizontal surface SS′ with a dashed normal through the point O; label the region above ‘medium 1’ and the region below ‘medium 2’. A ray AO comes down from the upper left to O, with the angle i marked between AO and the normal above the surface. It continues below the surface as OB, bent towards the normal, with the angle r marked between OB and the normal below the surface. Put an arrowhead on AO pointing towards O and on OB pointing away from O. Alongside, draw a second identical figure with both arrowheads reversed: light now travels from B to O and out along OA, with the same angles r and i marked. Caption the second one ‘the same path, travelled the other way’.
Now suppose the light is sent the other way, from medium 2 into medium 1. The principle of reversibility requires that a ray travelling along BO, striking O at an angle of incidence \(r\) in medium 2, must be refracted along OA at an angle of refraction \(i\) in medium 1 — and along no other direction whatever. For this passage the medium the light starts in is medium 2, so
\[*{2}\\mu*{1} = \\frac{\\sin r}{\\sin i} \\qquad \\dots \\text{(ii)}\]
Multiplying (i) and (ii),
\[*{1}\\mu*{2} \\times {}*{2}\\mu*{1} = \\frac{\\sin i}{\\sin r} \\times \\frac{\\sin r}{\\sin i}\]
The sines cancel in pairs, leaving
\[*{1}\\mu*{2} \\times {}*{2}\\mu*{1} = 1\]
or, rearranged,
\[*{2}\\mu*{1} = \\frac{1}{*{1}\\mu*{2}} \\qquad \\text{and} \\qquad *{1}\\mu*{2} = \\frac{1}{*{2}\\mu*{1}}\]
Neither quantity has a unit. In words: reversing the direction of travel replaces the refractive index by its reciprocal. So if the refractive index of glass with respect to air is \(\\dfrac{3}{2}\), then the refractive index of air with respect to glass is \(\\dfrac{1}{3/2} = \\dfrac{2}{3}\).
This also explains, without any new work, why one of the two numbers in such a pair is bound to be less than 1. Passing into the denser medium gives a value above 1; the reversed passage must give its reciprocal, which lies below 1.
4.5 Experimental Verification Of The Laws Of Refraction And Determination Of The Refractive Index Of Glass
Everything in the last three sections rests on a claim that can be tested on a drawing board with four pins. The experiment below both checks the laws of refraction and measures the refractive index of the glass used.
4.5.1 Procedure With Pins
The difficulty of tracing a ray through glass is that you cannot see the ray, only where it enters and where it leaves. The pins get round this: two pins fix the incoming direction, two more fix the outgoing direction, and the rest is geometry. The arrangement is shown in Fig. 4.5-1. FC 4.2 sets out the same steps in order.
Place a rectangular glass block on a white sheet of paper fixed to a drawing board, and draw its boundary line PQRS with a pencil.
Remove the block. On the boundary line PQ take a point O near its middle, and draw the normal NOM to PQ at O.
Draw a line AO inclined to the normal NOM at an angle \(i\), say \(40^\\circ\). This will be the incident ray.
Replace the glass block exactly on its boundary line.
Fix two pins \(a\) and \(b\) vertically on the board, about 5 cm apart, on the line AO. Two pins are needed because two points are what fix a line, and the gap between them keeps that line well defined.
Now look from the other side RS of the block, keeping the eye close to the plane of the board, and fix two more pins \(c\) and \(d\) such that the bases of all four pins \(a\), \(b\), \(c\) and \(d\) appear to lie in one straight line as seen through the block.
Remove the pins one by one, marking the position of each with a fine pencil dot. Remove the block. Join \(c\) and \(d\) by a line BC, produced back to meet the boundary line RS at the point B. Join O and B by a straight line. OB is the path of the light ray inside the glass block.
Measure the angles \(i\) and \(r\). Read the values of \(\\sin i\) and \(\\sin r\) from a sine table and calculate the ratio \(\\dfrac{\\sin i}{\\sin r}\).
Repeat steps 3 to 8 for angles of incidence of \(50^\\circ\), \(60^\\circ\), \(70^\\circ\) and \(80^\\circ\), finding the ratio in each case.
Record the observations in a table of the form of Table 4.4.
In this arrangement AO is the incident ray, OB is the refracted ray through the glass block, and BC is the ray that finally comes out, called the emergent ray. NOM is the normal to the glass surface at the point of incidence O, \(\\angle\)AON is the angle of incidence \(i\), and \(\\angle\)BOM is the angle of refraction \(r\).
Diagram
Fig. 4.5-1 — Verifying The Laws Of Refraction With A Glass Block A white sheet on a drawing board. Draw the pencilled rectangle PQRS marking the outline of a rectangular glass block, with PQ the upper long side and RS the lower long side; label the block ‘glass block’ inside the rectangle. Mark a point O near the middle of PQ and draw a long dashed normal NOM perpendicular to PQ through O, N above and M below. Draw the incident ray AO coming down to O from the upper left, with the angle AON marked i. Show two dots labelled a and b on AO above O, about 5 cm apart, marked ‘pins’. Inside the rectangle draw the refracted ray OB going down to a point B on RS, bent towards the normal, with the angle BOM marked r. Below RS continue the emergent ray BC to the lower right, with two dots labelled c and d on it, marked ‘pins’, and an eye symbol beyond looking back along BC. Draw a faint dashed circle centred at O for the alternative method, cutting AO at D and OB at E, with dashed perpendiculars DF and EG dropped onto the normal NOM, and label the points D, E, F and G.
Flowchart
FC 4.2 — Sequence Of Steps In The Experiment Draw the outline PQRS of the block → mark O on PQ and draw the normal NOM → draw the incident ray AO at a chosen angle i → replace the block on its outline → fix pins a and b on AO → sight through the block and fix pins c and d in line with a and b → remove the pins and the block and mark the dots → join c and d to meet RS at B → join O to B to get the ray inside the glass → measure i and r and find sin i / sin r → repeat for a new angle of incidence
4.5.2 Alternative Method Without Measuring Angles
The angles can be avoided altogether, and two lengths measured with a scale used instead. The construction is included in Fig. 4.5-1.
With the point O as centre, draw a circle of any convenient radius. Let it cut the incident ray AO at D and the refracted ray OB at E. From D and E drop perpendiculars DF and EG onto the normal NOM. Measure the lengths DF and EG with a scale and find the ratio DF/EG. This ratio is constant, and it is the refractive index of the glass.
Here is why it works. In the right-angled triangle OFD, the angle at O is the angle of incidence, so
\[\\sin i = \\frac{\\text{DF}}{\\text{OD}}\]
and in the right-angled triangle OGE, the angle at O is the angle of refraction, so
\[\\sin r = \\frac{\\text{EG}}{\\text{OE}}\]
Therefore
\[\\mu = \\frac{\\sin i}{\\sin r} = \\frac{\\text{DF}/\\text{OD}}{\\text{EG}/\\text{OE}}\]
Now OD and OE are both radii of the same circle, so OD \(=\) OE and they cancel, leaving
\[\\mu = \\frac{\\text{DF}}{\\text{EG}}\]
where DF and EG are lengths in cm (or in any one unit, since only their ratio is used), and \(\\mu\) has no unit. Drawing the circle is what makes the two denominators equal; without it the cancellation would not be available.
4.5.3 Observations And Conclusion
The readings are entered in a table of this form.
| S.No. | \(i\) | \(r\) | \(\\sin i\) or DF | \(\\sin r\) or EG | \(\\dfrac{\\sin i}{\\sin r}\) or \(\\dfrac{\\text{DF}}{\\text{EG}} = \\mu\) |
|---|---|---|---|---|---|
| 1 | \(40^\\circ\) | ||||
| 2 | \(50^\\circ\) | ||||
| 3 | \(60^\\circ\) | ||||
| 4 | \(70^\\circ\) | ||||
| 5 | \(80^\\circ\) | ||||
| Average \(\\mu =\) |
Table 4.4 Form of the observation table for the experiment
Two conclusions come out of it.
From the last column, the ratio \(\\dfrac{\\sin i}{\\sin r}\) — equivalently \(\\dfrac{\\text{DF}}{\\text{EG}}\) — comes out to be the same number for every angle of incidence, even though \(i\) and \(r\) themselves change from row to row. This verifies the second law of refraction. The constant so obtained is the refractive index \(\\mu\) of the material of the block, and the same experiment with a block of any other transparent material will give the refractive index of that material.
The first law is verified at the same time, and rather more quietly: the incident ray AO, the normal NOM at the point of incidence, and the refracted ray OB all lie flat on the sheet of paper, that is, in the same plane.
4.6 Refraction Of Light Through A Rectangular Glass Block
A single surface bends a ray once. A glass block has two surfaces, and what happens at the second one undoes, in one respect, what happened at the first.
4.6.1 Path Of The Ray Through The Block
Fig. 4.6-1 shows a rectangular glass block PQRS with a light ray AO falling on the face PQ. NOM is the normal to PQ at the point of incidence O.
At the surface PQ the ray AO passes from air, the rarer medium, into glass, the denser medium. It therefore slows down and bends towards the normal NOM, and travels in a straight path inside the glass along OB.
At the surface RS the ray OB meets a second boundary and suffers a second refraction. Here \(\\text{N}\_1\\text{BM}\_1\) is the normal at the point of incidence B. The ray now passes from glass into air, that is from the denser medium to the rarer one, so it speeds up and bends away from the normal \(\\text{N}\_1\\text{BM}\_1\). It travels along BC in air.
The three parts have names. AO is the incident ray, OB the refracted ray, and BC the emergent ray — the ray that finally leaves the block. The angle \(\\angle\)AON is the angle of incidence \(i\), the angle \(\\angle\)BOM is the angle of refraction \(r\), and the angle \(\\angle\\text{CBM}\_1\) that the emergent ray makes with the normal at B is called the angle of emergence \(e\). All three are measured in degrees.
Diagram
Fig. 4.6-1 — Refraction Of Light Through A Rectangular Glass Block Draw a rectangular glass block PQRS lying horizontally, P upper left, Q upper right, R lower right, S lower left; label it ‘glass block’, with ‘air’ above PQ and ‘air’ below SR. A ray AO comes down from the upper left onto a point O on the top face PQ. Draw the dashed normal NOM at O, perpendicular to PQ, and mark the angle AON as i. Inside the block draw the refracted ray OB going down to a point B on the bottom face SR, bent towards the normal, with the angle BOM marked r. At B draw the dashed normal N₁BM₁ perpendicular to SR, and draw the emergent ray BC leaving downwards to the right, bent away from that normal, with the angle CBM₁ marked e. Draw the original direction of AO produced as a long faint dotted line OD passing through the block and beyond. Mark the perpendicular distance between the emergent ray BC and this dotted line OD as XY, labelled ‘lateral displacement x’. Show BC parallel to AO.
Now compare the two refractions. The faces PQ and RS are parallel, so their normals are parallel too, and the angle \(\\angle\)MOB inside the glass equals the angle \(\\angle\\text{N}\_1\\text{BO}\). The passage from air into glass at O and the passage from glass into air at B are therefore the same refraction taken in opposite directions, and by the principle of reversibility of Section 4.4,
\[\\angle e = \\angle i\]
The angle of emergence equals the angle of incidence. Since BC makes the same angle with its normal as AO makes with its own parallel normal, the emergent ray BC is parallel to the incident ray AO.
This result holds only for a block with parallel faces, with the same medium — here air — on both sides of it. Neither condition is idle: Section 4.9 deals with a prism, whose faces are not parallel, and there the emergent ray is not parallel to the incident ray at all.
4.6.2 Lateral Displacement
Parallel is not the same as identical. Look again at Fig. 4.6-1: the emergent ray BC travels in the same direction as the incident ray AO, but not along the same line. It has been shifted sideways. Had the glass block not been there, the ray would have carried straight on along the dotted line OD.
The lateral displacement \(x\) depends on three things.
The thickness of the block. The thicker the medium, the greater the lateral displacement, because the ray travels further along its slanted path inside the glass before it is turned back parallel.
The angle of incidence. The greater the angle of incidence, the greater the lateral displacement. At normal incidence the displacement is zero.
The refractive index of the medium and the wavelength of the light. The greater the refractive index, the greater the lateral displacement, since a larger \(\\mu\) means a sharper bend at the first face. And because the refractive index of a medium increases as the wavelength decreases (Section 4.3.6), the lateral displacement increases as the wavelength decreases — it is greater for violet light than for red light through the same block.
These three are gathered up in a single relation, which is given here only for completeness and is not needed for the work of this chapter:
\[x = \\frac{t \\sin (i - r)}{\\cos r}\]
where \(t\) is the thickness of the block in cm, \(i\) the angle of incidence in degrees, \(r\) the angle of refraction in degrees, and \(x\) the lateral displacement in cm. You can read the three dependences straight out of it: \(x\) grows with \(t\), and grows as \(i\) pulls further away from \(r\).
4.7 Multiple Images In A Thick Plane Glass Plate Or Thick Mirror
Stand square in front of a mirror and you see one image. Stand well to one side and look along the glass at a slant, and you may notice something else: a faint image, a bright one, and a short trail of fainter ones behind it. A thick mirror gives not one image but many.
If a pin or an illuminated object is placed in front of a thick plane glass plate, or a thick mirror, and is viewed obliquely, a number of images are seen. Out of these images, the second image is the brightest, while the others are of decreasing brightness.
To see why, follow the light. In Fig. 4.7-1, LMNP represents a thick plane mirror. The front face LM is plain glass; the back face NP is the silvered surface, the coating that does the real reflecting. An illuminated object A is kept in front of the mirror. Of the rays leaving A, consider one that falls normally on the mirror and another, AB, that falls obliquely on it.
Diagram
Fig. 4.7-1 — Multiple Reflections In A Thick Mirror Draw a thick horizontal glass slab LMNP: LM is the upper front surface, NP the lower surface, which is drawn as a heavily hatched line labelled ‘silvered surface’. Place the object A at the upper left, above LM. Draw the ray AB striking LM obliquely at B. From B draw a weak ray BP going up and to the right, labelled ‘about 4% reflected’. From B draw a stronger ray BB′ refracted down through the glass to the point B′ on the silvered surface, labelled ‘about 96% refracted’. From B′ draw the strongly reflected ray B′C back up to the point C on LM. At C, draw a strong ray CQ refracted out into the air above, and a weaker ray CC′ reflected back down inside the glass to C′ on the silvered surface. Continue this zig-zag through D, D′, E, F, showing rays R, S, T emerging above LM, each drawn progressively fainter. Below the slab, on the far side of the silvered surface, mark the images A₁, A₂, A₃, A₄, A₅ in a vertical line, with dashed construction lines running back from the emerging rays to them; label A₁ ‘faint first image’ and A₂ ‘brightest second image’.
When the ray AB falls on the front surface LM, a small fraction of the light — nearly 4 per cent — is reflected in the direction BP. This weak reflected light forms a faint virtual image at \(\\text{A}\_1\); a virtual image is one from which the light only appears to come, so that it cannot be caught on a screen.
The larger fraction, nearly 96 per cent, is refracted into the glass along BB′. The ray BB′ strikes the silvered surface at B′ and is now strongly reflected back inside the glass as B′C. At C this ray meets the front surface again, and there it divides: it is partly refracted out into the air along CQ, and partly reflected back into the glass along CC′.
The refracted ray CQ forms the virtual image \(\\text{A}\_2\). This is the brightest image of all, because the light forming it has been strongly reflected once at the silvered surface and has lost only a little at each of its two passages through the front face.
The reflected ray CC′ is not finished. It suffers further reflections at C′, D, D′, and so on, and further refractions out of the front surface at D, E, F, giving rise to more virtual images \(\\text{A}\_3\), \(\\text{A}\_4\), \(\\text{A}\_5\) and beyond. Each round trip inside the glass loses a little more light, so these images are of gradually decreasing brightness. FC 4.3 sets out the whole chain.
Flowchart
FC 4.3 — Why A Thick Mirror Gives Many Images Ray AB strikes the front glass surface → about 4% reflected, forming the faint image A₁ → about 96% refracted into the glass → strong reflection at the silvered back surface → at the front surface, partly refracted out, forming the brightest image A₂ → the rest reflected back into the glass → reflected again at the silvered surface → partly refracted out, forming A₃ → the cycle repeats, giving A₄, A₅ … each fainter than the one before
Two points about the drawing are worth noting. In Fig. 4.7-1 the emerging rays BP, CQ and the rest have had to be drawn well separated so that they can be told apart, but in reality they are very close together and all enter the eye at the same moment — which is why the images appear together as a small trail. And a thick plane glass plate, with no silvering at all, behaves in the same way, since its rear surface also reflects part of the light back.
4.8 The Prism
So far every piece of glass we have used has had parallel faces, and a ray has always come out of it travelling in its original direction. The next few sections take that condition away.
Two opposite parallel surfaces of a prism are identical triangles, while the other three surfaces are rectangular and inclined to each other.
Fig. 4.8-1 shows a prism in which the opposite parallel surfaces ABC and DEF are the triangles. The triangle ABC is called the principal section of the prism — it is the shape you see when you look at the prism end on, and it is the shape every ray diagram in this chapter will use.
Diagram
Fig. 4.8-1 — The Faces Of A Prism Draw a triangular prism in perspective, lying with its length going back and to the right. The near triangular face has vertices A at the top, B at the lower left and C at the lower right; the far triangular face has vertices D at the top, E at the lower left and F at the lower right, with AD, BE and CF as the long edges. Shade the two rectangular faces ABED and ACFD lightly and label them ‘refracting surfaces’. Mark the angle BAC on the near triangle as A and label it ‘angle of prism’. Label the top edge AD ‘refracting edge’. Label the lower rectangular face BCFE ‘base’ and draw it with a rough, granular texture. Label the near triangle ABC ‘principal section’.
The angle \(\\angle\)BAC of the principal section is the angle of prism, denoted by the letter \(A\) and measured in degrees. It is the angle between the two refracting surfaces.
The two rectangular surfaces ABED and ACFD, shown shaded in Fig. 4.8-1, are polished, and it is through these that light enters and leaves. They are called the refracting surfaces. The line AD along which they meet is the refracting edge. The remaining rectangular surface BCFE is the base of the prism, and it is usually ground or made rough.
Two things about this arrangement decide everything that follows. First, the two refracting surfaces are not parallel to each other, unlike the two faces of the glass block in Section 4.6; they are inclined at the angle \(A\). Second, in ray diagrams a prism is drawn only by its principal section, the triangle ABC, as in Fig. 4.8-2 — so that a triangle in a diagram from here on means a solid prism seen end on.
Diagram
Fig. 4.8-2 — How A Prism Is Drawn In Ray Diagrams Draw the principal section only: a triangle with apex A at the top, B at the lower left and C at the lower right. Mark the angle at A as A, labelled ‘angle of prism’. Label the side AB and the side AC each as ‘refracting surface’. Label the side BC as ‘base’, drawn with a rough, hatched texture. Keep the triangle clean, with no rays drawn on it.
4.9 Refraction Of Light Through A Glass Prism
In a glass block the two refractions cancelled each other’s turning. In a prism the faces are inclined, and the two refractions add instead. That is the whole story of this section, told carefully.
4.9.1 Path Of A Ray Through A Prism
Fig. 4.9-1 shows the principal section ABC of a glass prism. The angle of the prism is \(\\angle\)BAC \(= A\).
Let a monochromatic ray of light — that is, a ray of a single colour — strike the face AB at a point P, at an angle of incidence \(i\_1\). Only one colour is used so that a single refractive index applies throughout; Section 4.3.6 warned that \(\\mu\) is different for different colours.
At the face AB the ray passes from air, the rarer medium, into glass, the denser medium. So it bends towards the normal PN, making an angle of refraction \(r\_1\), and travels along PQ inside the prism. PQ is the refracted ray.
The refracted ray PQ now strikes the second face AC at the point Q, at an angle of incidence \(r\_2\). Here the light passes from glass into air, that is from the denser to the rarer medium, so it bends away from the normal NQ and emerges out of the prism along QR, at an angle of emergence \(i\_2\). QR is the emergent ray.
Diagram
Fig. 4.9-1 — Deviation Of A Ray By A Prism Draw the principal section of a prism as a triangle with apex A at the top, B at the lower left and C at the lower right; mark the angle at A as A. An incident ray OP comes from the left and strikes the face AB at the point P; mark the angle of incidence i₁ between OP and a dashed normal PN drawn perpendicular to AB at P. Inside the prism draw the refracted ray PQ running to the point Q on the face AC, with the angle of refraction r₁ marked between PQ and the normal at P, and the angle of incidence r₂ marked between PQ and a dashed normal NQ drawn perpendicular to AC at Q. From Q draw the emergent ray QR leaving to the right, bent away from the normal at Q, with the angle of emergence i₂ marked. Produce the incident ray OP forward as a dotted line beyond P; produce the emergent ray QR backwards as a dotted line; let these two dotted lines meet at a point M above the prism, and mark the angle between them there as δ, labelled ‘angle of deviation’. Mark the small deviation at P between PQ produced and the dotted line as δ₁, and the deviation at Q as δ₂. Label OP ‘incident ray’, PQ ‘refracted ray’ and QR ‘emergent ray’.
Because of the principle of reversibility, a ray sent in along RQ would follow the path QPO exactly, emerging along PO. So the diagram may be read either way round.
4.9.2 The Angle Of Deviation
The ray has been turned twice, and both turns are in the same sense — towards the base of the prism. This is the crucial difference from the glass block.
At the first face AB, the ray OP, instead of carrying straight on, has bent at P along PQ. It has therefore suffered a deviation \(\\delta\_1\) at this face. At the second face AC, the ray PQ has bent at Q along QR, suffering a further deviation \(\\delta\_2\).
In the absence of the prism the incident ray OP would simply have travelled straight on. The prism has sent it away along QR instead, and the emergent ray, produced backwards, meets the original direction at M. The angle between the direction of the incident ray produced forward and the emergent ray produced backwards is the total turning the prism has produced. It is called the angle of deviation of the prism, and denoted by \(\\delta\).
From Fig. 4.9-1, the total deviation is the sum of the two,
\[\\delta = \\delta\_1 + \\delta\_2 \\qquad \\dots \\text{(i)}\]
At the first face the ray turned from the direction of incidence, making an angle \(i\_1\) with the normal, to the direction of refraction, making \(r\_1\) with the same normal, so
\[\\delta\_1 = i\_1 - r\_1\]
and in the same way at the second face,
\[\\delta\_2 = i\_2 - r\_2\]
Here \(i\_1\) is the angle of incidence, \(r\_1\) the angle of refraction at the first face, \(r\_2\) the angle of incidence at the second face and \(i\_2\) the angle of emergence, all in degrees; \(\\delta\), \(\\delta\_1\) and \(\\delta\_2\) are angles of deviation in degrees. Substituting both into (i),
\[\\delta = (i\_1 - r\_1) + (i\_2 - r\_2)\]
and gathering the terms,
\[\\delta = (i\_1 + i\_2) - (r\_1 + r\_2)\]
In a prism, then, the emergent ray is not parallel to the incident ray. It has been deviated, and always towards the base of the prism. Table 4.5 sets the prism and the glass block side by side.
4.9.3 Relation Between The Angle Of Prism And The Angles Of Refraction
The expression for \(\\delta\) still contains the two angles inside the glass, \(r\_1\) and \(r\_2\), which cannot be measured from outside. A piece of geometry removes them.
Look at the quadrilateral APNQ in Fig. 4.9-1, formed by the two faces and the two normals. The normals are perpendicular to the faces, so
\[\\angle \\text{APN} = \\angle \\text{AQN} = 90^\\circ\]
The four angles of a quadrilateral add up to \(360^\\circ\). Taking away the two right angles leaves
\[\\angle \\text{PNQ} + \\angle \\text{PAQ} = 180^\\circ\]
But \(\\angle\)PAQ is the angle of the prism, \(A\). Therefore
\[\\angle \\text{PNQ} = 180^\\circ - A \\qquad \\dots \\text{(ii)}\]
Now turn to the triangle PNQ, whose angles at P and Q are \(r\_1\) and \(r\_2\). Since the three angles of a triangle add up to \(180^\\circ\),
\[\\angle \\text{PNQ} = 180^\\circ - (r\_1 + r\_2) \\qquad \\dots \\text{(iii)}\]
Comparing (ii) and (iii), the two expressions for the same angle must be equal:
\[180^\\circ - (r\_1 + r\_2) = 180^\\circ - A\]
so that
\[r\_1 + r\_2 = A\]
where \(r\_1\) and \(r\_2\) are the two angles inside the glass and \(A\) is the angle of the prism, all in degrees. This is a useful result in itself: the two angles inside the prism always add up to the angle of the prism.
Substituting it into the expression for \(\\delta\) obtained in Section 4.9.2,
\[\\delta = (i\_1 + i\_2) - A\]
or, written the way it is most often used,
\[i\_1 + i\_2 = A + \\delta\]
where \(i\_1\) is the angle of incidence, \(i\_2\) the angle of emergence, \(A\) the angle of the prism and \(\\delta\) the angle of deviation, all in degrees. Every quantity in it can be measured outside the prism. Where the angle of emergence is written \(e\) instead of \(i\_2\), and the angle of incidence simply \(i\), the same relation reads \(i + e = A + \\delta\).
4.9.4 Factors Affecting The Angle Of Deviation
The angle of deviation produced by a prism depends on four things:
- the angle of incidence \(i\),
- the material of the prism, that is, its refractive index \(\\mu\),
- the angle of prism \(A\), and
- the colour, or wavelength \(\\lambda\), of the light used.
Each is taken in turn.
4.9.4.1 Dependence On The Angle Of Incidence; The i–δ Curve
It is found by experiment that as the angle of incidence increases, the angle of deviation first decreases, reaches a minimum value at a certain angle of incidence, and then, on increasing the angle of incidence further, begins to increase again.
To obtain this graph in the laboratory, the direction of the incident ray is kept unchanged and the prism itself is rotated. On rotating the prism, the direction of the normal at the face changes, so the angle which the incident ray makes with the normal — the angle of incidence — changes. The deviation \(\\delta\) is noted for each angle of incidence \(i\), and a graph is drawn of \(\\delta\) against \(i\). Such a graph is shown in Fig. 4.9-2, and is called the i–δ curve.
Diagram
Fig. 4.9-2 — The i–δ Curve For A Prism Draw a graph with the horizontal axis labelled ‘angle of incidence i (degrees)’ and the vertical axis labelled ‘angle of deviation δ (degrees)’. Plot a smooth U-shaped curve that comes down steeply from a high value on the left, flattens through a single lowest point, and rises again to the right — the right-hand branch rising more steeply than the left falls. Mark the lowest point of the curve and draw a horizontal dashed line from it to the vertical axis, labelled δ_min. Drop a vertical dashed line from the same lowest point to the horizontal axis. Draw a second horizontal dashed line higher up, at some value δ₁, cutting the curve at two points, and drop vertical dashed lines from these two intersections to the horizontal axis, labelled i₁ and i₂, to show that one value of δ above the minimum corresponds to two angles of incidence.
4.9.4.2 The Position Of Minimum Deviation
It is observed that the angle of deviation is minimum when the angle of incidence is equal to the angle of emergence, that is when \(i\_1 = i\_2\), or equivalently when \(r\_1 = r\_2\).
This has a visible consequence. If the principal section ABC is equilateral, then when \(r\_1 = r\_2\) the refracted ray PQ inside the prism runs parallel to the base BC, as Fig. 4.9-3 shows. So the ray inside the glass tells you at a glance whether the prism is set for minimum deviation.
Diagram
Fig. 4.9-3 — A Prism In The Position Of Minimum Deviation Draw the principal section of an equilateral prism, apex A at the top, B at the lower left, C at the lower right, with the angle at A marked 60°. An incident ray strikes the face AB at P and is refracted along PQ inside the prism to the point Q on AC; draw PQ clearly parallel to the base BC, and mark this with a pair of matching arrowheads on PQ and BC. Draw the dashed normals at P and at Q, and mark the two angles inside the glass as r₁ and r₂, labelled equal. Mark the angle of incidence at P as i and the angle of emergence at Q as i, labelled equal. Produce the incident ray forward and the emergent ray backwards as dotted lines meeting above the prism, and mark the angle between them as δ_min.
In this position, writing \(i\_1 = i\_2 = i\), the relation of Section 4.9.3 becomes
\[2i = A + \\delta\_{\\text{min}}\]
or
\[\\delta\_{\\text{min}} = 2i - A\]
where \(i\) is the common value of the angle of incidence and the angle of emergence in degrees, \(A\) the angle of the prism in degrees, and \(\\delta\_{\\text{min}}\) the angle of minimum deviation in degrees. Note the condition: this relation holds only in the position of minimum deviation, and nowhere else on the curve.
For a given prism and a given colour of light, \(\\delta\_{\\text{min}}\) is unique. The reason can be read straight off Fig. 4.9-2: only one horizontal line can be drawn touching the lowest point of the i–δ curve, so there is only one value of the angle of incidence for which the deviation is least.
4.9.4.3 Deviation At Different Angles Of Incidence
Fig. 4.9-4 shows the deviation of a light ray by an equilateral glass prism of \(\\mu = 1.5\) at three different angles of incidence.
In Fig. 4.9-4(a), the angle of incidence \(i\_1 = 70^\\circ\), the angle of emergence \(i\_2 = 33^\\circ\) and the angle of deviation \(\\delta = 43^\\circ\). In Fig. 4.9-4(b), \(i\_1 = 48^\\circ\), \(i\_2 = 48^\\circ\) and \(\\delta = 36^\\circ\). In Fig. 4.9-4(c), \(i\_1 = 30^\\circ\), \(i\_2 = 77^\\circ\) and \(\\delta = 47^\\circ\).
Diagram
Fig. 4.9-4 — Deviation By An Equilateral Prism At Different Angles Of Incidence Three separate triangles side by side, each an equilateral prism ABC with apex A at the top, angle 60°, labelled (a), (b) and (c). In each, draw the incident ray striking face AB at P, the refracted ray PQ inside to Q on AC, and the emergent ray from Q, with dashed normals at P and Q, and dotted produced lines meeting to show the deviation δ. (a) Mark i₁ = 70°, r₁ = 38.5°, r₂ = 21.5°, i₂ = 33°, δ = 43°; the ray PQ slopes downwards towards the base. (b) Mark i₁ = 48°, r₁ = 30°, r₂ = 30°, i₂ = 48°, δ = 36°; draw PQ parallel to the base BC and mark it so. (c) Mark i₁ = 30°, r₁ = 19.5°, r₂ = 40.5°, i₂ = 77°, δ = 47°; the ray PQ slopes upwards away from the base. Caption each with its angle of incidence.
These angles are worked out in three steps, and it is worth seeing the route.
For refraction at the first face, \(\\mu = \\dfrac{\\sin i\_1}{\\sin r\_1}\), from which \(r\_1\) is obtained. Then \(r\_1 + r\_2 = A\), from which \(r\_2\), the angle of incidence at the second face, is calculated. For refraction at the second face, where the light passes from glass to air, \(\\dfrac{1}{\\mu} = \\dfrac{\\sin r\_2}{\\sin i\_2}\), from which \(i\_2\), the angle of emergence, is obtained. Finally \(\\delta = (i\_1 + i\_2) - A\) is calculated.
Compare the three. The deviation has its least value, \(36^\\circ\), in Fig. 4.9-4(b), where the angle of incidence is \(48^\\circ\) and the refracted ray PQ is parallel to the base BC. There the angle of emergence is also \(48^\\circ\), equal to the angle of incidence. The minimum deviation for this prism can be checked from the relation:
\[\\delta\_{\\text{min}} = 2i - A = 2 \\times 48^\\circ - 60^\\circ = 36^\\circ\]
Notice too that the deviation in (a), where \(i\_1 = 70^\\circ\), and in (c), where \(i\_1 = 30^\\circ\), are both larger than this — one on each side of the minimum, exactly as the i–δ curve of Fig. 4.9-2 predicts.
4.9.4.4 Dependence On The Material Of The Prism
It is found that for a given angle of incidence, a prism with a higher refractive index produces a greater deviation than a prism which has a lower refractive index.
A flint glass prism therefore produces more deviation than a crown glass prism of the same refracting angle, since \(\\mu\_{\\text{flint}} > \\mu\_{\\text{crown}}\). Table 4.3 bears this out: crown glass stands at \(1.53\), flint glass higher up the column.
4.9.4.5 Dependence On The Angle Of Prism
It is found that the angle of deviation \(\\delta\) increases with an increase in the angle of prism \(A\). A thin prism, with a small angle between its faces, turns light only slightly; a fat one turns it a great deal. For a prism of small angle \(A\) the two are simply proportional, \(\\delta = (\\mu - 1)A\), with \(\\delta\) and \(A\) in degrees and \(\\mu\) the refractive index of the material.
4.9.4.6 Dependence On The Colour Of Light
The refractive index of a given transparent medium is different for light of different colours, and it decreases as the wavelength of the light increases. So for visible light the refractive index of the material of a prism is maximum for violet colour and minimum for red colour.
Since a higher refractive index means a greater deviation, it follows that a given prism deviates violet light the most and red light the least, that is \(\\delta\_{\\text{violet}} > \\delta\_{\\text{red}}\), because \(\\mu\_{\\text{violet}} > \\mu\_{\\text{red}}\).
4.10 Real Depth And Apparent Depth
The sections so far have followed single rays. This one follows two of them at once, because that is what an eye does, and it explains something you have almost certainly noticed in a bucket of water.
An object placed in a denser medium, when viewed from a rarer medium, appears to be at a depth lesser than its real depth. This is because of refraction of light.
4.10.1 Why An Object Under Water Appears Raised
In Fig. 4.10-1, consider a point object O kept at the bottom of a transparent medium such as water or glass, separated from the air above by the surface PQ.
A ray of light OA, starting from the object O, is incident on the surface PQ normally, so it passes undeviated along the path AA′. Another ray OB, also starting from O, strikes the boundary surface PQ at B and suffers refraction there. Since this ray travels from a denser medium into a rarer one, it bends away from the normal N′BN drawn at the point of incidence B, and travels along BC in air.
Diagram
Fig. 4.10-1 — Real Depth And Apparent Depth Draw a horizontal line PQ as the surface of water, with ‘air (rarer)’ marked above it and ‘water or glass (denser)’ marked below. Place a point object O well below the surface. From O draw a vertical ray up to the point A on the surface, continuing straight up as AA′, marked ‘normal incidence, no bending’. From O draw a second ray OB to a point B on the surface, slightly to the right of A, and from B draw the refracted ray BC going up and to the right, bent away from the dashed normal N′BN at B; mark the angle of incidence i between OB and BN′ below the surface, and the angle of refraction r between BC and BN above the surface, with r larger than i. Produce A′A downwards and CB backwards as dotted lines, meeting at a point I between O and A on the vertical line; label I ‘image’ and mark it clearly nearer the surface than O. Mark the distance AO as ‘real depth’ and the distance AI as ‘apparent depth’ with arrowed lines beside the vertical. Draw an eye above and to the right, receiving both AA′ and BC.
When these rays are received by the eye, the ray BC appears to be coming from a point I, obtained by producing A′A and CB backwards. The point I is the virtual image of O. So any object placed at O — a coin lying on the bed of a stream, say — when seen from above in air, appears to be at I, which is at a lesser depth AI than its actual depth AO.
In the figure the two rays OA′ and BC have been drawn well apart for clarity. In reality the point B lies very close to the point A, and both rays enter the eye at the same moment; that is what allows the eye to place the image at I.
4.10.2 Relation Between Refractive Index, Real Depth And Apparent Depth
Fig. 4.10-1 contains everything needed to turn this into a formula.
For the incident ray OB, the angle of incidence is \(i = \\angle\)OBN′ and the angle of refraction is \(r = \\angle\)CBN. Two pairs of parallel lines let these angles be moved to more convenient places. Since AO and BN′ are parallel and OB is a transversal line cutting them,
\[\\angle \\text{AOB} = \\angle \\text{OBN}' = i\]
Similarly, IA′ and BN are parallel and IC is the transversal line, so
\[\\angle \\text{BIA}' = \\angle \\text{CBN} = r\]
Now use these two angles inside two right-angled triangles.
In the right-angled triangle BAO, the angle at O is \(i\), BA is the side opposite to it and OB is the hypotenuse, so
\[\\sin i = \\frac{\\text{BA}}{\\text{OB}}\]
In the right-angled triangle IAB, the angle at I is \(r\), BA is the side opposite to it and IB is the hypotenuse, so
\[\\sin r = \\frac{\\text{BA}}{\\text{IB}}\]
For refraction from the medium into air, Snell’s law gives the refractive index of air with respect to the medium as
\[*{m}\\mu*{a} = \\frac{\\sin i}{\\sin r} = \\frac{\\text{BA}/\\text{OB}}{\\text{BA}/\\text{IB}} = \\frac{\\text{IB}}{\\text{OB}}\]
the length BA cancelling from numerator and denominator. By the principle of reversibility, the refractive index of the medium with respect to air is the reciprocal of this:
\[*{a}\\mu*{m} = \\frac{1}{{}*{m}\\mu*{a}} = \\frac{\\text{OB}}{\\text{IB}}\]
Finally, since the point B is very close to the point A — which is to say, since the object is viewed from a point vertically above it — we may take \(\\text{IB} = \\text{IA}\) and \(\\text{OB} = \\text{OA}\). Hence
\[*{a}\\mu*{m} = \\frac{\\text{OA}}{\\text{IA}} = \\frac{\\text{real depth}}{\\text{apparent depth}}\]
or, rearranged,
\[\\text{apparent depth} = \\frac{\\text{real depth}}{{}*{a}\\mu*{m}}\]
where the real depth and the apparent depth are both distances, measured in the same unit — both in cm, or both in m — and \(*{a}\\mu*{m}\), being a ratio of two lengths, has no unit. Note the condition carefully: this relation holds only when the object is viewed from vertically above.
Two examples of what it means. For glass, \(*{a}\\mu*{g} = \\dfrac{3}{2}\), so a glass slab appears only two-thirds of its real thickness when it is viewed from air with the eye vertically above it. For water, \(*{a}\\mu*{w} = \\dfrac{4}{3}\), so a pond appears three-fourths of its real depth when seen from air in a nearly vertical direction — it appears shallow. This is also why a fish, when seen from air, appears to be nearer to the surface of the water than it actually is.
4.10.3 The Shift And The Factors It Depends On
The image is raised, and the amount by which it is raised has a name of its own: the shift.
\[\\text{shift OI} = \\text{real depth} - \\text{apparent depth}\]
Substituting for the apparent depth from Section 4.10.2 and taking out the common factor,
\[\\text{shift} = \\text{real depth} \\times \\left(1 - \\frac{1}{{}*{a}\\mu*{m}}\\right)\]
where the shift and the real depth are distances in the same unit, and \(*{a}\\mu*{m}\) has no unit. Since \(*{a}\\mu*{m}\) is greater than 1, the bracket is always positive — the image is always raised, never lowered.
The shift by which the object appears to be raised depends on three things.
The refractive index of the medium. The higher the refractive index of the medium, the more is the shift. A slab of diamond would raise an image far more than a slab of water of the same thickness.
The thickness of the denser medium. For a given medium, the shift is directly proportional to the thickness of the medium. The thicker the medium, the more is the shift.
The colour, or wavelength, of the light. The shift decreases with an increase in the wavelength of the light used. Since \(\\mu\_{\\text{violet}} > \\mu\_{\\text{red}}\), the shift is more for violet light than for red light in a given medium.
One caution completes the section. The apparent depth of an object lying in a denser medium is always less than its real depth, from whatever direction in the rarer medium it is viewed. But the three relations of this section — for the refractive index, the apparent depth and the shift — are valid only when the object is seen from vertically above.
4.11 Apparent Bending Of A Stick Under Water
Section 4.10 raised a single point object. A stick is a whole line of point objects, one behind the other, and when each of them is raised the result is something stranger than a shallow pond: the stick looks broken at the water line.
Why The Stick Looks Broken
Fig. 4.11-1 shows a straight stick, or a pencil, XOP placed obliquely in water. The point O is where it crosses the surface, so the part XO is in air and the part OP is under water. Seen from above, the immersed part OP appears shortened and raised up, along OP′.
Follow the tip P first. Rays of light coming from P travel up through the water and cross the water–air surface. There they pass from a denser medium into a rarer one, so each of them bends away from the normal at the surface. To an eye receiving them, they appear to be coming from a point P′ lying above P. That point P′ is the virtual image of the point P.
Now the same argument holds for every point of the stick between P and O, each one being raised by an amount that depends on how deep it lies. Since O itself lies exactly at the surface, it is not raised at all. So the immersed part PO appears as P′O — shorter than it really is, and tilted.
The part XO in air is not affected, because no refraction happens to its light. The eye therefore sees XO in its true position joined to a raised P′O, and the whole stick XOP appears as the bent line XOP′. The stick appears to be raised and therefore bent at the point O, at the surface of the water.
Diagram
Fig. 4.11-1 — Bending Of A Stick Due To Refraction Draw a horizontal line for the water surface with ‘air’ above and ‘water’ below. A straight stick XOP runs from X above the surface, down through the point O on the surface, to the tip P deep under water on the right. From P draw two rays going up to two nearby points on the surface; at each point draw the refracted ray bending away from the dashed normal, both refracted rays travelling up to an eye drawn at the top right. Produce the two refracted rays backwards as dotted lines to meet at a point P′ above P and label P′ ‘virtual image of P’. Draw the apparent stick OP′ as a thick dashed line from O to P′, clearly shorter and less steep than OP. Label the stick, the points X, O, P and P′, the eye, and mark ‘apparent bend here’ at O.
Looking The Other Way Round
Everything so far has assumed an eye in air looking into water. Turn the situation round and the effect reverses.
An object placed in a rarer medium, when viewed from a denser medium, appears to be at a greater distance than its real distance. In Fig. 4.11-2, an object O placed in air is viewed by an eye inside a body of water. The rays from O bend towards the normal as they enter the water, and traced back they meet at I, which is higher above the surface than O itself. So a fish looking up at a bird sees it farther away, and higher, than it really is.
Diagram
Fig. 4.11-2 — An Object In A Rarer Medium Viewed From A Denser Medium Draw a horizontal water surface with ‘air (rarer)’ above and ‘water (denser)’ below. Place a point object O in the air a short distance above the surface. From O draw a vertical ray straight down to a point A on the surface and continuing straight down to S below; from O draw a second ray to a nearby point B on the surface, and from B draw the refracted ray inside the water bent towards the dashed normal, travelling down to an eye drawn inside the water at the lower right. Produce this refracted ray backwards as a dotted line to meet the produced vertical line at a point I above O; label I ‘image, farther than the object’. Label O, A, B, I, S, the eye, and the normal at B.
4.12 Some Consequences Of Refraction Of Light
Refraction is not a laboratory effect that needs a glass block and a set of pins to show itself. In daily life we come across many everyday things that are caused by refraction of light. Some of them are given below, and each one is only the last few sections applied outside the laboratory.
In The Sky
A star appears twinkling in the sky. This happens due to the fluctuation in the refractive index of air with temperature. Air at different heights is at different temperatures, so its refractive index keeps changing slightly and unevenly, and the starlight reaching the eye is bent by a slightly different amount from moment to moment.
The sun is seen a few minutes before it rises above the horizon in the morning, and in the evening it is seen for a few minutes longer after it sets. Sunlight entering the atmosphere is refracted as it passes through air of increasing density, so it reaches the eye along a bent path from a sun that is still below the horizon.
Diagram
Fig. 4.12-1 — Why The Sun Is Seen Before It Rises Draw the curved edge of the earth with an observer standing on it, and the horizon line drawn as a tangent at the observer’s position. Show a band of atmosphere above the earth’s surface as two or three faint concentric arcs, labelled ‘atmosphere, denser near the earth’. Place the actual sun below the horizon line and draw a ray from it entering the atmosphere and curving gently downwards towards the observer, bending a little more at each arc. Produce the final part of this ray backwards as a dotted straight line to a position above the horizon and draw a faint sun there labelled ‘apparent position of the sun’. Label the actual sun ‘actual position, still below the horizon’, the observer, and the horizon.
In Everyday Objects
- A coin kept in a vessel and not visible when seen from just below the edge of the vessel can be viewed from the same position when water is poured into the vessel. The rim blocks the straight line to the coin, but light from the coin bends at the water surface and can then reach the eye, as shown in Fig. 4.12-2.
Diagram
Fig. 4.12-2 — A Hidden Coin Made Visible By Water Draw the same vessel twice, side by side, in section. In the left figure the empty vessel holds a coin on its base at the far side, and an eye is placed low, just above the near rim; draw a straight dotted line of sight from the eye grazing the near rim and striking the base short of the coin, labelled ‘coin not visible’. In the right figure the vessel is filled with water up to a marked level; draw a ray from the coin travelling up to the water surface, bending away from the dashed normal at the surface, and then travelling to the eye in the same low position; label ‘coin now visible’, and mark the angles of incidence and refraction at the surface with the angle of refraction the larger of the two.
The print on paper appears to be raised when a glass slab is placed over it.
A piece of paper stuck at the bottom of a glass slab appears to be raised when seen from above.
A water tank appears shallower than its actual depth.
A person’s legs, as seen from outside, appear to be short when the person is standing in a pool or a water tank.
The last four are all the same effect wearing four costumes. In each of them light travels from an object in a denser medium out into air, bends away from the normal as it leaves, and so reaches the eye along steeper lines than it set out on. Traced back, those lines place the image nearer the surface than the object — which is exactly the raising established in Section 4.10.
4.13 Transmission Of Light From A Denser Medium To A Rarer Medium At Different Angles Of Incidence
Everything so far in this chapter has treated the two directions of refraction as a matched pair: light entering a denser medium bends towards the normal, light leaving it bends away. That symmetry is real, but it does not last. Going from denser to rarer, something happens that has no counterpart in the other direction, and this section watches it happen.
Consider a light ray travelling in a denser medium, such as glass, which falls on the surface separating it from a rarer medium, such as air. Part of the light is reflected back into the denser medium and part of it is refracted into the rarer medium. The refracted ray bends away from the normal at the point of incidence, obeying the laws of refraction. Now let the angle of incidence be increased step by step, and watch both rays.
Case (i): A Small Angle Of Incidence
In Fig. 4.13-1, a light ray AO travelling in glass is incident at the glass–air interface at a small angle of incidence \(i\). It is partly reflected and partly refracted. We get a weak reflected ray OC inside the glass and a strong refracted ray OB in the air. Since the incident ray bends away from the normal when it suffers refraction from glass to air, the angle of refraction \(r\) is greater than the angle of incidence \(i\).
Diagram
Fig. 4.13-1 — Refraction From Glass To Air When The Angle Of Incidence Is Small Draw a horizontal boundary line with ‘air’ marked above and ‘glass’ marked below, and a dashed normal NON′ through the point of incidence O, perpendicular to the boundary. Draw the incident ray AO travelling up through the glass from the lower left to O, making a small angle \(i\) with the normal below the boundary. Draw the strong refracted ray OB going up into the air, bent away from the normal, making a clearly larger angle \(r\) with the normal above the boundary. Draw the weak reflected ray OC going back down into the glass to the lower right, making an angle equal to \(i\) with the normal. Draw OB as a thick line and OC as a thin line, and label them ‘strong refracted ray’ and ‘weak reflected ray’. Mark the angles \(i\), \(r\) and the reflection angle, and label the rays AO, OB, OC and the normal.
Now suppose the angle of incidence \(i\) is gradually increased. Two things happen together. The angle of refraction \(r\) also increases, since it is always the larger of the two. And the intensity of the refracted ray keeps on decreasing, while the reflected ray inside the glass grows correspondingly stronger — the light that fails to get out has to go somewhere, and it goes back into the glass.
The angle of refraction cannot grow without limit. A refracted ray lies in the air, so the largest angle it can possibly make with the normal is \(90^\\circ\), at which it would be grazing along the surface itself. Finally, therefore, at one particular angle of incidence the angle of refraction reaches its maximum possible value of \(90^\\circ\). That angle of incidence in the denser medium is called the critical angle, written \(C\) — the angle beyond which the light can no longer find a way out into the rarer medium. It is measured in degrees, and it has a different value for every pair of media.
Case (ii): The Angle Of Incidence Equal To The Critical Angle
At an angle of incidence equal to the critical angle, that is when \(i = C\), the angle of refraction becomes \(90^\\circ\), as shown in Fig. 4.13-2. The refracted ray OB now travels along the glass–air interface itself, and it is very weak. The reflected ray OC inside the glass has by this stage taken almost all of the light.
Diagram
Fig. 4.13-2 — Refraction From Glass To Air When The Angle Of Incidence Equals The Critical Angle Draw the same horizontal boundary with ‘air’ above and ‘glass’ below, and a dashed normal NON′ at the point of incidence O. Draw the incident ray AO in the glass making an angle marked \(i = C\) with the normal, noticeably larger than in the previous figure. Draw the refracted ray OB running along the boundary line to the right, marked \(r = 90^\\circ\), drawn as a very thin faint line labelled ‘very weak refracted ray, grazing the surface’. Draw the reflected ray OC in the glass to the lower right as a thick line, at an angle equal to \(C\) with the normal, labelled ‘reflected ray’. Mark the angles \(C\), \(90^\\circ\) and the angle of reflection.
Case (iii): An Angle Of Incidence Greater Than The Critical Angle
Fig. 4.13-3 shows the incident ray AO arriving at an angle of incidence \(i\) greater than the critical angle \(C\). No refracted ray is obtained at all. The whole of the incident light is reflected back into the glass as OC, with \(\\angle \\text{AON} = \\angle \\text{CON}\), that is, obeying the ordinary law of reflection. The boundary surface separating the two media now behaves as a reflecting surface — a mirror made of nothing but a change of medium.
Diagram
Fig. 4.13-3 — Total Reflection When The Angle Of Incidence Is Greater Than The Critical Angle Draw the same horizontal boundary with ‘air’ above and ‘glass’ below and a dashed normal NON′ at O. Draw the incident ray AO in the glass making an angle \(i\) with the normal, clearly larger than the critical angle, and mark it \(i > C\). Draw no ray at all in the air above the boundary. Draw a single thick reflected ray OC going back down into the glass to the lower right, making an angle with the normal equal to \(i\), and label it ‘the entire light is reflected back’. Mark the two equal angles at O.
The three cases are worth holding side by side, since a question will usually give you one of them and ask for the other two.
4.14 Critical Angle
Section 4.13 met the critical angle as the angle at which the escaping ray runs out of room. This section pins that idea down, connects it to the refractive index by a short piece of algebra, and then asks what the critical angle for a given pair of media depends on.
4.14.1 Meaning Of The Critical Angle
We have read that when a ray of light passes from a denser medium to a rarer medium, at a certain angle of incidence \(i = C\) the angle of refraction becomes \(90^\\circ\). The angle \(C\) is called the critical angle.
Two words in that definition carry the whole meaning, and both are easy to lose. The angle is measured in the denser medium, because that is where the light is travelling; and the definition belongs to a pair of media, not to a single substance, since the value depends on both sides of the boundary. When a critical angle is quoted for water or for glass without any further mention, the second medium is understood to be air.
4.14.2 Relationship Between Critical Angle And Refractive Index
Fig. 4.13-2 has already been drawn for exactly this situation: a ray AO incident at the glass–air interface at an angle of incidence \(i = C\), for which the angle of refraction \(r\) is \(90^\\circ\). Apply Snell’s law to it.
The refractive index of air with respect to glass is the ratio of the sine of the angle of incidence in glass to the sine of the angle of refraction in air, so
\[*{g}\\mu*{a} = \\frac{\\sin C}{\\sin 90^\\circ}\]
But \(\\sin 90^\\circ = 1\), and therefore
\[*{g}\\mu*{a} = \\sin C\]
This is the refractive index of the rarer medium with respect to the denser one, which is not the quantity usually quoted. By the principle of reversibility of Section 4.4, the refractive index of glass with respect to air is its reciprocal:
\[*{a}\\mu*{g} = \\frac{1}{{}*{g}\\mu*{a}} = \\frac{1}{\\sin C} = \\operatorname{cosec} C\]
or, turned round to give the critical angle directly,
\[\\sin C = \\frac{1}{{}*{a}\\mu*{g}}\]
where \(C\) is the critical angle for the pair of media in degrees, and \(*{a}\\mu*{g}\) is the refractive index of the denser medium with respect to the rarer one, which has no unit. The relation holds only for light travelling from the denser medium towards the rarer one, since only then can a critical angle exist at all.
So knowing the refractive index of the denser medium with respect to the rarer medium, we can calculate the critical angle \(C\) for that pair of media. Two standard cases follow at once.
For water, the refractive index with respect to air is \(*{a}\\mu*{w} = \\dfrac{4}{3}\), so
\[\\sin C = \\frac{1}{4/3} = \\frac{3}{4}\]
and since \(\\sin 49^\\circ = \\dfrac{3}{4}\), the critical angle for the water–air pair is \(C = 49^\\circ\).
4.14.3 Critical Angles For Some Substances
Table 4.7 gives the critical angles of some substances with respect to air, each one calculated from the refractive index beside it.
Table 4.7 — Critical angles for some substances with respect to air
| Substance | Refractive index \(\\mu\) | Critical angle \(C\) |
|---|---|---|
| Water | 1.33 | \(48^\\circ 45' \\approx 49^\\circ\) |
| Turpentine | 1.47 | \(42^\\circ 53' \\approx 43^\\circ\) |
| Glass | 1.5 | \(41^\\circ 49' \\approx 42^\\circ\) |
| Flint glass | 1.58 | \(39^\\circ 16' \\approx 39^\\circ\) |
| Diamond | 2.41 | \(24^\\circ 31' \\approx 25^\\circ\) |
Read the table down the columns rather than across the rows, and its pattern is plain: as the refractive index rises, the critical angle falls. Diamond, at the bottom, has the highest refractive index of the five and a critical angle of only about \(25^\\circ\), which means light inside a diamond escapes only through a narrow cone of directions and is turned back over a very wide range of others.
4.14.4 Factors Affecting The Critical Angle
The critical angle for a given pair of media depends on their refractive indices — that is what Section 4.14.2 established. But the refractive index of a medium is itself affected by two factors, so the critical angle is affected by the same two.
Dependence on the colour of light. The refractive index of a transparent medium decreases with an increase in the wavelength of light; it is the most for violet light and the least for red light. Therefore the critical angle for a pair of media is the least for violet light and the most for red light. In short, the critical angle increases with an increase in the wavelength of light.
Dependence on temperature. On increasing the temperature of a medium, its refractive index decreases. Therefore the critical angle increases with an increase in temperature.
Both statements follow from the same single fact, so neither needs to be remembered separately: anything that lowers the refractive index raises the critical angle, because \(\\sin C\) is \(1\) divided by that refractive index.
4.15 Total Internal Reflection
The third case of Section 4.13 was left standing without a name. It has one, and it is one of the most useful pieces of physics in this chapter.
4.15.1 The Phenomenon
When light travels from a rarer medium to a denser medium, at all angles of incidence a part of it is reflected and the rest of it is refracted at the boundary surface. Reflection and refraction occur together, and no angle of incidence can prevent the refraction: there is always a ray that gets through.
When light travels from a denser medium to a rarer medium, this is no longer true. Under a certain condition — when the angle of incidence is greater than the critical angle — no part of the light is refracted, and the entire light is reflected back into the same medium. This is the case drawn in Fig. 4.13-3, where the ray AO gets entirely reflected as OC, obeying the laws of reflection, and suffers no refraction at all. This phenomenon is called total internal reflection: total because all of the light is turned back, internal because the light stays inside the denser medium, and reflection because the boundary is doing exactly what a mirror does.
In the process of total internal reflection, \(100%\) of the energy, or intensity, of the light is reflected back. No other device, such as a plane mirror, produces \(100%\) reflection, because at an ordinary mirror some part of the light is absorbed and some part is refracted. This single property is why total internal reflection is worth engineering with, and Section 4.17 returns to it.
4.15.2 Essential Conditions For Total Internal Reflection
There are two necessary conditions for total internal reflection. Both must hold; either one alone is not enough.
The light must travel from a denser medium to a rarer medium.
The angle of incidence must be greater than the critical angle for the pair of media.
The second condition is the reason the first one cannot be dropped. A critical angle exists only for light going from denser to rarer, since only in that direction does the angle of refraction reach \(90^\\circ\) before the angle of incidence does. Light going the other way has no critical angle to exceed.
4.15.3 Refraction And Total Internal Reflection At Increasing Angles Of Incidence
It helps to see all the cases at once, coming from a single source. Fig. 4.15-1 shows light from a point source A kept in a denser medium, striking the boundary with a rarer medium at four increasing angles of incidence.
Diagram
Fig. 4.15-1 — Refraction And Total Internal Reflection At Increasing Angles Of Incidence Draw a horizontal boundary line across the figure with ‘rarer medium’ marked above and ‘denser medium’ marked below. Place a point source A low down in the denser medium, and from it draw four rays striking the boundary at four points spaced from left to right: O, P, Q and R, with the angle of incidence increasing from O to R. At O the ray is vertical: draw it continuing straight up into the rarer medium as OA′, marked ‘i = 0, r = 0, no bending’. At P, mark the angle of incidence i less than C: draw a thick refracted ray PC going up into the rarer medium bent away from the dashed normal at P with r greater than i, and a thin reflected ray PB going back down into the denser medium. At Q, mark the angle of incidence equal to C: draw a very faint refracted ray QC′ grazing along the boundary at 90 degrees to the normal, and a thicker reflected ray QB′ going back down. At R, mark the angle of incidence greater than C: draw no ray above the boundary at all, and one thick totally reflected ray RB″ going back down into the denser medium at an angle of reflection equal to the angle of incidence. Show a dashed normal at each of the four points, label the angles i and r at P, C and 90 degrees at Q, and the two equal angles at R.
Read Fig. 4.15-1 from left to right and the whole of Sections 4.13 to 4.15 is laid out in one picture.
For the ray AO, the angle of incidence is zero, and it is refracted as OA′ at an angle of refraction of zero — that is, it passes straight through, undeviated.
For the ray AP, the angle of incidence \(i\) is less than the critical angle \(C\), so it is partly reflected as PB and partly refracted as PC, at an angle of refraction \(r\) greater than \(i\).
For the ray AQ, the angle of incidence is equal to the critical angle for the pair of media, so it is partly reflected as QB′ and partly refracted as QC′ at an angle of refraction equal to \(90^\\circ\).
For the ray AR, the angle of incidence is greater than the critical angle, so it is totally reflected as RB″ at an angle of reflection \(r = i\), and no refraction occurs.
Notice what happens to the thickness of the lines across the four rays: the light leaving the medium fades away as the angle grows, and the light staying inside strengthens to take all of it. Nothing is lost at any stage; it is only shared out differently.
4.16 Total Internal Reflection In A Prism
A prism has flat faces set at known angles, and glass has a critical angle of about \(42^\\circ\) with air. Put those two facts together and a prism becomes a device for steering light exactly where you want it, using total internal reflection rather than any coating. We shall consider the phenomenon in three different prisms: a \(45^\\circ\), \(90^\\circ\), \(45^\\circ\) prism, a prism with each angle \(60^\\circ\), and a \(30^\\circ\), \(90^\\circ\), \(60^\\circ\) prism.
4.16.1 The Total Reflecting Prism
A prism having an angle of \(90^\\circ\) between its two refracting surfaces, and the other two angles each equal to \(45^\\circ\), is called a total reflecting prism, because light incident normally on any of its faces suffers total internal reflection inside the prism. It is also described as a right-angled isosceles prism, from the shape of its principal section.
The reason lies in one comparison. A ray entering normally through one face travels on undeviated and meets a second face at \(45^\\circ\). For a glass–air interface the critical angle is about \(42^\\circ\), and \(45^\\circ\) is greater than \(42^\\circ\), so the ray is totally internally reflected there. Nothing needs to be silvered, and no light is lost.
Because of this behaviour, a total reflecting prism is used for the following three purposes:
to deviate a ray of light through \(90^\\circ\),
to deviate a ray of light through \(180^\\circ\), and
to erect an inverted image without producing deviation in its path.
4.16.1.1 To Deviate A Ray Of Light Through 90°
In Fig. 4.16-1, ABC is a total reflecting prism with the right angle at B. A beam of light is incident normally at the face AB, so it passes undeviated into the prism and strikes the face AC at an angle of incidence equal to \(45^\\circ\).
For the glass–air interface the critical angle is about \(42^\\circ\). Therefore the beam of light suffers total internal reflection at the face AC, because the angle of incidence there is greater than the critical angle. The reflected beam inside the prism then strikes the face BC, where it is incident normally and therefore passes out undeviated.
As a result, the incident beam gets deviated through \(90^\\circ\) while emerging out of the prism. Fig. 4.16-1 shows the path of two such rays, PQ and P′Q′, through the prism and then out of it.
Diagram
Fig. 4.16-1 — Deviation Through 90° By A Total Reflecting Prism Draw the principal section of a right-angled isosceles prism ABC with the right angle at B at the lower left, the \(45^\\circ\) angle at A at the top, and the \(45^\\circ\) angle at C at the right, so that AB is the vertical left face, BC is the horizontal bottom face and AC is the sloping hypotenuse. Mark the angles \(90^\\circ\) at B, \(45^\\circ\) at A and \(45^\\circ\) at C. Draw two parallel horizontal rays PQ and P′Q′ arriving from the left and striking the face AB normally at Q and Q′; continue each straight through the glass to the hypotenuse AC, marking the angle of incidence there as \(45^\\circ\) against a dashed normal. From each point on AC draw the totally reflected ray turning downwards through a right angle, striking the bottom face BC normally, and continuing straight down out of the prism as two parallel vertical rays. Label the incident rays, the points of total internal reflection on AC, the emergent rays, and mark ‘deviation = 90°’ with a curved arrow from the incident to the emergent direction.
By the principle of reversibility of the path of light, if light is incident normally on the face BC of the prism instead, it will emerge out from the face AB after suffering total internal reflection at the face AC. In either case the face AC behaves as a reflecting surface.
Use: this action of a prism is used in a periscope, where a total reflecting prism is preferred over a plane mirror.
4.16.1.2 To Deviate A Ray Of Light Through 180°
In Fig. 4.16-2, a beam of light from the object PQ falls normally on the face AC, that is on the hypotenuse face of the prism, so it enters undeviated inside the prism and strikes the glass–air interface AB.
Since the beam of light is travelling from glass to air at an angle of incidence equal to \(45^\\circ\), which is greater than the critical angle of about \(42^\\circ\) for glass and air, the beam of light suffers total internal reflection at AB. The reflected beam inside the prism then strikes the face BC at an angle of incidence equal to \(45^\\circ\), so it again suffers total internal reflection there.
The beam of light inside the prism now falls normally on the face AC and therefore passes undeviated out of the prism, forming the image P′Q′ of the object PQ. The beam thus gets deviated by \(90^\\circ\) at each reflection, and the total deviation due to the two reflections becomes \(180^\\circ\).
Diagram
Fig. 4.16-2 — Deviation Through 180° By A Total Reflecting Prism Draw the principal section of a right-angled isosceles prism with the right angle at B at the top, and the two \(45^\\circ\) angles at A on the left and C on the right, so that the hypotenuse AC is the horizontal bottom face and AB and BC are the two sloping faces meeting at the top. Mark \(90^\\circ\) at B and \(45^\\circ\) at A and at C. Below the prism draw an upright object PQ on the left with an arrow, and from it draw two horizontal rays travelling to the right that turn up to strike the hypotenuse AC normally. Inside the prism, take each ray up to the face AB, mark the angle of incidence \(45^\\circ\) against a dashed normal, reflect it horizontally across to the face BC, mark \(45^\\circ\) there too, and reflect it downwards to strike AC normally and pass straight out. Below the prism on the right draw the inverted image P′Q′ with its arrow reversed. Mark ‘total internal reflection’ at both points on AB and BC, and mark ‘total deviation = 180°’.
Use: this action of a prism is used in a binocular and in a camera, to invert the image without any loss of intensity.
4.16.1.3 To Erect An Inverted Image Without Deviation
In Fig. 4.16-3, a beam of light from the object PQ is incident parallel to the face AC, the hypotenuse, and strikes the face AB of the prism. It suffers refraction from air to glass at AB and then strikes the face AC of the prism.
These rays, travelling through the glass, strike the glass–air interface at AC at an angle of incidence greater than the critical angle of about \(42^\\circ\), so the beam suffers total internal reflection there. The beam inside the prism now strikes the face BC at an angle of incidence less than the critical angle, so at BC it suffers refraction from glass to air and bends away from the normal. The beam emerges parallel to the face AC.
As a result of refraction, on emergence the rays are inverted, and so for the inverted object PQ the erect image P′Q′ is obtained. A prism used in this manner is called an erecting prism.
Diagram
Fig. 4.16-3 — An Erecting Prism Draw the principal section of a right-angled isosceles prism with the right angle at B at the top, and the \(45^\\circ\) angles at A on the lower left and C on the lower right, so that AC is the horizontal bottom face. Mark \(90^\\circ\) at B and \(45^\\circ\) at A and C. On the left draw the object PQ as a vertical arrow pointing downwards, labelled ‘object, inverted’. From the top and bottom of the object draw two horizontal rays travelling to the right, parallel to the face AC, entering the prism through the sloping face AB where each bends towards the dashed normal. Inside the prism carry each ray to the hypotenuse AC, mark the point ‘total internal reflection’, and reflect it up to the face BC. At BC show each ray refracting out into the air, bending away from the dashed normal, and emerging to the right parallel to AC. On the right draw the image P′Q′ as a vertical arrow pointing upwards, labelled ‘image, erect’. Show the two rays crossing between AC and BC so that the top of the object arrives at the bottom of the image.
Use: this action of a prism is used in a slide projector.
4.16.2 Total Internal Reflection In An Equilateral Prism
A prism of each angle \(60^\\circ\) can be used to deviate a light ray through \(60^\\circ\) by total internal reflection.
Fig. 4.16-4 shows a prism ABC of each angle \(60^\\circ\). A light ray PQ, incident normally on the face AB of the prism, passes undeviated as QR into the prism and strikes the face AC at an angle of incidence \(60^\\circ\) at the point R. Now the light ray QR suffers total internal reflection at the glass–air interface, since the angle of incidence, \(60^\\circ\), is greater than the critical angle, which is \(42^\\circ\); the geometry is such that \(\\angle \\text{QRS} = 120^\\circ\). The ray RS obtained after total internal reflection falls normally on the face BC of the prism, and so it passes out undeviated through the face BC. Thus ST is the emergent ray.
From Fig. 4.16-4 it is clear that the incident ray PQ has turned through an angle of \(60^\\circ\) towards the base from its initial direction, and it emerges out as ST.
Further, in a similar manner, rays incident on the lower half portion of the face AB, such as P′Q′, get internally reflected as R′T′ by the face BC, which is also a glass–air interface. Thus such a ray also turns through \(60^\\circ\), to emerge perpendicular to the face AC.
Thus we find that a parallel beam of light incident normally on the face AB, after refraction, splits into two equal beams travelling in different directions, making an angle of \(120^\\circ\) with each other.
Diagram
Fig. 4.16-4 — Deviation Through 60° By An Equilateral Prism Draw the principal section of an equilateral prism ABC with the apex A at the top, B at the lower left and C at the lower right, marking \(60^\\circ\) at each vertex. Draw a horizontal ray PQ from the left striking the upper half of the face AB normally at Q, continuing straight into the prism as QR to a point R on the face AC; at R draw a dashed normal, mark the angle of incidence \(60^\\circ\), mark ‘total internal reflection’, and mark \(\\angle \\text{QRS} = 120^\\circ\). From R draw the reflected ray RS striking the base BC normally at S, and continuing straight down out of the prism as the emergent ray ST. Draw a second horizontal ray P′Q′ striking the lower half of the face AB normally at Q′, continuing to a point R′ on the base BC, where it is totally internally reflected and travels as R′T′ to emerge perpendicular to the face AC on the right. Mark the deviation of each ray as \(60^\\circ\) with a curved arrow, and mark the angle between the two emergent beams as \(120^\\circ\).
4.16.3 Total Internal Reflection And Refraction In A 30°, 90°, 60° Prism
A \(30^\\circ\), \(90^\\circ\), \(60^\\circ\) prism can be used to deviate a light ray through an angle less than \(60^\\circ\) by total internal reflection.
Fig. 4.16-5 shows a prism ABC of angles \(30^\\circ\), \(90^\\circ\) and \(60^\\circ\). A light ray PQ, incident normally on the face BC of the prism — the face opposite to the refracting angle of \(30^\\circ\) — passes undeviated as QR into the prism, and it strikes the face AC at an angle of incidence \(60^\\circ\) at the point R. The ray QR therefore suffers total internal reflection at the glass–air interface, since the angle of incidence is greater than the critical angle of \(42^\\circ\).
The ray RS obtained after total internal reflection is incident at the face AB of the prism at an angle of incidence equal to \(30^\\circ\). This angle of incidence at the glass–air interface is less than the critical angle, so the ray RS is not reflected there. It is refracted from glass to air as ST, with an angle of refraction greater than \(30^\\circ\). Thus the emergent ray ST has turned through an angle \(\\delta\) which is less than \(60^\\circ\) from the direction of the incident ray PQ.
Diagram
Fig. 4.16-5 — Deviation Less Than 60° By A 30°, 90°, 60° Prism Draw the principal section of a prism ABC with the apex A at the top marked \(30^\\circ\), the right angle at B at the lower left marked \(90^\\circ\), and C at the lower right marked \(60^\\circ\); AB is the vertical left face, BC the horizontal base and AC the sloping hypotenuse. Draw a vertical ray PQ arriving from below and striking the base BC normally at Q, continuing straight up inside the prism as QR to a point R on AC; at R draw a dashed normal, mark the angle of incidence \(60^\\circ\) and mark ‘total internal reflection’. From R draw the reflected ray RS across to a point S on the vertical face AB; at S draw a dashed normal, mark the angle of incidence \(30^\\circ\), and show the ray refracting out into the air as ST, bending away from the normal with an angle of refraction clearly greater than \(30^\\circ\). Produce PQ forward as a dotted line and mark the angle \(\\delta\) between it and the emergent ray ST, labelled ‘deviation less than 60°’.
When Total Internal Reflection Does Not Occur
The same prism does not always produce total internal reflection; it depends on which face the light enters. Fig. 4.16-6 shows two such cases.
In the first, a light ray PQ is incident normally on the face AB, the face opposite to the refracting angle of \(60^\\circ\). In the second, a light ray PQ is incident normally on the upper part of the hypotenuse face AC of the prism ABC. In both the cases, the light ray does not suffer total internal reflection inside the prism, and gets refracted out as RS, obeying the laws of refraction. The reason is the same in each case: the angle of incidence at the other face inside the prism is \(30^\\circ\), which is less than the critical angle of \(42^\\circ\).
Diagram
Fig. 4.16-6 — No Total Internal Reflection In A 30°, 90°, 60° Prism Draw two separate principal sections of the same \(30^\\circ\), \(90^\\circ\), \(60^\\circ\) prism side by side, each with A at the top marked \(30^\\circ\), B at the lower left marked \(90^\\circ\) and C at the lower right marked \(60^\\circ\). In figure (a) draw a horizontal ray PQ from the left striking the vertical face AB normally at Q, continuing straight to a point R on the hypotenuse AC; at R draw a dashed normal, mark the angle of incidence \(30^\\circ\), and show the ray refracting out as RS into the air, bending away from the normal. In figure (b) draw a ray PQ arriving perpendicular to the hypotenuse and striking the upper part of AC normally at Q, continuing straight inside the prism to a point R on the vertical face AB; at R draw a dashed normal, mark the angle of incidence \(30^\\circ\), and show the ray refracting out as RS, bending away from the normal. Label the two figures (a) and (b) and mark on each ‘angle of incidence 30°, less than the critical angle, so the light gets out’.
There is one exception worth knowing, because it looks at first like the second case above. A ray of light incident normally on the hypotenuse face of the prism, but below the foot of the perpendicular drawn on it from the opposite corner, suffers total internal reflection and gets deviated by an angle greater than \(60^\\circ\).
In Fig. 4.16-7, a light ray PQ is incident normally on the face AC, the hypotenuse, in the portion DC, where D is the foot of the perpendicular from B on AC. It passes undeviated as QR inside the prism and strikes the face BC of the prism at an angle of incidence \(60^\\circ\), which is greater than the critical angle of \(42^\\circ\). The ray suffers total internal reflection inside the prism as RS. The reflected ray RS then strikes the face AB at an angle of incidence \(30^\\circ\), so it suffers refraction from glass to air and emerges out of the prism as ST, bending away from the normal at the face AB. Thus the incident ray PQ gets deviated through an angle greater than \(60^\\circ\).
Diagram
Fig. 4.16-7 — Total Internal Reflection Through The Lower Part Of The Hypotenuse Draw the principal section of the same \(30^\\circ\), \(90^\\circ\), \(60^\\circ\) prism with A at the top marked \(30^\\circ\), B at the lower left marked \(90^\\circ\) and C at the lower right marked \(60^\\circ\). Drop a dotted perpendicular from B to the hypotenuse AC and mark its foot D. Draw the ray PQ arriving perpendicular to AC and striking it normally at a point Q lying in the portion DC, below D. Continue the ray straight inside the prism as QR to a point R on the base BC; at R draw a dashed normal, mark the angle of incidence \(60^\\circ\) and mark ‘total internal reflection’. From R draw the reflected ray RS to a point S on the vertical face AB; at S draw a dashed normal, mark the angle of incidence \(30^\\circ\), and show the ray emerging into the air as ST, bending away from the normal. Mark the deviation between PQ produced and ST as ‘greater than 60°’, and label the points D, Q, R, S and T.
4.17 Use Of A Total Internal Reflecting Prism In Place Of A Plane Mirror
Both a plane mirror and a total reflecting prism can turn a beam of light through a right angle. Instrument makers nevertheless pay for the prism. This short section says why.
A total internal reflecting prism is used in place of a plane mirror to deviate a light ray by \(90^\\circ\) in a periscope, and by \(180^\\circ\) in a binocular as well as in a camera. The reason is that due to total internal reflection in the prism, the entire incident light, that is \(100%\) of it, is reflected back into the denser medium. In ordinary reflection from a plane mirror, some light is refracted and absorbed, so the reflection is not \(100%\).
Thus the image obtained by the use of a total reflecting prism is much brighter than that obtained by using a plane mirror. Further, the brightness of the image formed by a total reflecting prism always remains unchanged, while in a plane mirror the silvering deteriorates after a long use, and the image formed by it becomes faint.
Diagram
Fig. 4.17-1 — Turning Light Through 90°: Prism And Mirror Compared Two side-by-side sketches, each showing a horizontal ray arriving from the left and leaving vertically downwards. In the left sketch draw a plane mirror set at \(45^\\circ\) to the incoming ray, with its silvered back marked by hatching; show the incoming ray as a thick line, the reflected ray leaving downwards as a slightly thinner line, and two faint stray lines passing into and through the glass of the mirror, labelled ‘some light absorbed and refracted’. Label the sketch ‘plane mirror: reflection is not complete’. In the right sketch draw a right-angled isosceles prism with the light entering normally through a \(45^\\circ\) face, striking the hypotenuse at \(45^\\circ\), and leaving normally through the other face downwards; draw the incoming and outgoing rays with the same thickness, and mark the hypotenuse ‘total internal reflection, 100% of the light’. Label the sketch ‘total reflecting prism: nothing is lost’.
4.17.1 Total Internal Reflection Compared With Reflection From A Plane Mirror
The two processes look alike on a ray diagram and are quite different in physics. Table 4.8 sets them side by side.
4.18 Some Consequences Of Total Internal Reflection
Total internal reflection needs no apparatus and no silvering — only a denser medium, a rarer one, and a steep enough angle. Those three things turn up often, and in daily life we observe many consequences of them. Some of these are given below.
On a hot sunny day, a driver may see a pool of water, or a wet road, in front of him at some distance. It is the phenomenon of mirage, which is often observed in a desert.
An empty test tube placed in water in a beaker, with its mouth outside the water surface, shines like a mirror when seen at certain angles.
A crack in a glass vessel often shines like a mirror.
A piece of diamond sparkles when viewed from certain directions.
An optical fibre is used to transmit a light signal over a long distance without any loss of energy.
Why The Road Looks Wet
The mirage deserves a closer look, because it is the one item on the list where the two media are both air. On a hot day the layer of air close to the road surface is heated strongly, and Section 4.14.4 established that a rise in temperature lowers the refractive index of a medium. So the air near the ground is optically rarer than the cooler air above it.
Light from the sky travelling down towards the road therefore passes from optically denser air into optically rarer air, bending away from the normal a little at each stage, until it meets a layer at an angle of incidence greater than the critical angle for those two layers. There it suffers total internal reflection and curves back upwards to the driver’s eye. The driver, tracing the light back in a straight line, sees a patch of sky on the road ahead — and a patch of sky on the ground looks exactly like a sheet of standing water.
Diagram
Fig. 4.18-1 — How A Mirage Is Formed On A Hot Road Draw a long horizontal road surface with a car and driver on the right, seen from the side. Above the road draw four or five faint horizontal lines dividing the air into layers, labelled from the ground upwards ‘hottest air, optically rarest’ up to ‘cooler air, optically denser’. From a point in the sky on the upper left draw a ray of light travelling down and to the right, bending away from the dashed normal a little more at each layer boundary so that its path curves and becomes more nearly horizontal as it nears the ground; at the lowest boundary it reaches an angle of incidence greater than the critical angle, so mark that point ‘total internal reflection’ and curve the ray back upwards to the driver’s eye. Produce the final part of the ray backwards as a dotted straight line down to the road surface, and mark that point ‘apparent pool of water’. Label the sky, the layers, the driver and the road.
Why A Diamond Sparkles
Diamond has a refractive index of \(2.41\) with respect to air, the highest in Table 4.7, and therefore a critical angle of only about \(25^\\circ\). Light that has entered a diamond can escape only if it meets a face within \(25^\\circ\) of the normal; at every steeper angle it is totally internally reflected back inside. A cut diamond has many faces set at many angles, so light entering it bounces from face to face several times before finally leaving through one of the few faces it can escape from — and it leaves concentrated, in flashes, as the stone is turned. Ordinary glass, with a critical angle of about \(42^\\circ\), lets light out far more easily and so never sparkles in the same way, whatever shape it is cut into.
Light Sent Down A Fibre
An optical fibre is a very thin thread of glass, coated with a second layer of glass of lower refractive index. Light entering one end strikes the boundary between the two glasses at a large angle of incidence — larger than the critical angle for that pair — and so is totally internally reflected. It meets the boundary again a short distance along, and again, and so on, travelling down the fibre in a long series of total internal reflections. Since each reflection is total, no energy is lost at any of them, and the signal arrives at the far end even if the fibre is many kilometres long or is bent round corners.
Flowchart
FC 4.4 — Path Of Light Along An Optical Fibre Light signal enters one end of the fibre → travels through the inner glass thread → strikes the boundary with the outer layer at an angle greater than the critical angle → suffers total internal reflection with no loss of energy → travels on to the next point on the boundary → total internal reflection repeats many thousands of times along the fibre → light signal emerges at the far end
Diagram
Fig. 4.18-2 — Total Internal Reflection Inside An Optical Fibre Draw a long horizontal fibre in section, gently curved along its length, showing two layers: a narrow inner thread labelled ‘inner glass, higher refractive index’ and a thin outer coating on both sides labelled ‘outer glass, lower refractive index’. Show a ray of light entering the left end at a small angle to the axis, striking the inner boundary at a large angle of incidence, and zig-zagging down the fibre by repeated reflections off the upper and lower boundaries, following the curve of the fibre and emerging at the right end. At one reflection point draw a dashed normal, mark the angle of incidence as greater than the critical angle, and label ‘total internal reflection, no light escapes’. Mark the entering and emerging rays.
4.19 Summary
Light travels fastest in vacuum, at \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\), and at a smaller speed in every other transparent medium: \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) in air, \(2.25 \\times 10^{8}\\ \\text{m s}^{-1}\) in water and \(2 \\times 10^{8}\\ \\text{m s}^{-1}\) in glass. Within a homogeneous medium the speed is constant and the path is straight.
If light slows down on entering the second medium, that medium is optically denser than the first; if it speeds up, the second medium is optically rarer. Optical density depends on the speed of light in the medium, and is unrelated to ordinary density, which depends on molecular packing.
At the boundary separating two media, light suffers partial reflection and partial refraction. The refracted light is less intense than the incident light, since part of the intensity has been lost by reflection.
Refraction is the change in the direction of the path of light when it passes from one transparent medium to another. It is a surface phenomenon.
Going from a rarer to a denser medium, the ray bends towards the normal, so \(r < i\), and the deviation is \(\\delta = i - r\). Going from a denser to a rarer medium, the ray bends away from the normal, so \(r > i\), and the deviation is \(\\delta = r - i\). Here \(i\) is the angle of incidence, \(r\) the angle of refraction and \(\\delta\) the angle of deviation, all in degrees.
A ray incident normally on the surface, with \(i = 0^\\circ\), passes undeviated, so \(r = 0^\\circ\) and \(\\delta = 0^\\circ\). Its speed still changes, but its direction does not.
Refraction is caused by the change in the speed of light on passing from one medium to another. If the speed does not change, the ray does not bend.
The laws of refraction, given by Willebrord Snell, are: (i) the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane; and (ii) for a given pair of media,
\[\\frac{\\sin i}{\\sin r} = {}*{1}\\mu*{2}\]
where \(*{1}\\mu*{2}\) is the refractive index of medium 2 with respect to medium 1, a ratio of two similar quantities and therefore having no unit.
- On refraction the speed changes; the frequency \(f\) does not change, since it is fixed by the source; and the wavelength therefore changes, the three being connected in any medium by
\[V = f\\lambda\]
where \(V\) is the speed of light in the medium in \(\\text{m s}^{-1}\), \(f\) is the frequency in \(\\text{Hz}\) and \(\\lambda\) is the wavelength in that medium in \(\\text{m}\).
Going from a rarer to a denser medium the wavelength decreases; going from a denser to a rarer medium it increases.
The absolute refractive index of a medium is
\[\\mu = \\frac{c}{V}\]
where \(c\) is the speed of light in vacuum or air in \(\\text{m s}^{-1}\) and \(V\) is the speed of light in that medium in \(\\text{m s}^{-1}\); \(\\mu\) has no unit. Since \(V < c\) always, \(\\mu\) is always greater than 1, and \(\\mu = 1\) for air or vacuum.
- For two media,
\[*{1}\\mu*{2} = \\frac{V\_1}{V\_2} = \\frac{\\mu\_2}{\\mu\_1}\]
where \(V\_1\) and \(V\_2\) are the speeds of light in \(\\text{m s}^{-1}\) in media 1 and 2, and \(\\mu\_1\) and \(\\mu\_2\) are their absolute refractive indices.
A light ray passes undeviated in either of two cases: when the angle of incidence is zero, or when the two media have the same refractive index, so that \(\\mu\_1 = \\mu\_2\) and \(i = r\).
If a ray of wavelength \(\\lambda\) in air enters a medium of refractive index \(\\mu\), its wavelength in that medium becomes
\[\\lambda' = \\frac{\\lambda}{\\mu}\]
with \(\\lambda\) and \(\\lambda'\) both in \(\\text{m}\) and \(\\mu\) having no unit.
The refractive index of a medium depends on three factors: the nature of the medium, since \(\\mu = c/V\); its temperature, a rise in temperature lowering \(\\mu\); and the colour of the light, \(\\mu\) being greatest for violet and least for red, so that \(\\mu\) decreases as the wavelength increases.
By the principle of reversibility of the path of light, a ray retraces its own path when its direction is reversed. Hence
\[*{1}\\mu*{2} \\times {}*{2}\\mu*{1} = 1, \\qquad {}*{2}\\mu*{1} = \\frac{1}{{}*{1}\\mu*{2}}\]
The laws of refraction can be verified, and the refractive index of glass found, by tracing rays through a rectangular glass block with pins and measuring \(i\) and \(r\) for several values of \(i\). The ratio \(\\dfrac{\\sin i}{\\sin r}\) comes out constant, and equals \(\\mu\) for the glass. In the alternative method the same ratio is obtained as \(\\dfrac{\\text{DF}}{\\text{EG}}\), the lengths of two normals drawn to the same normal line from a circle of any radius centred at the point of incidence.
In a rectangular glass block the two refracting surfaces are parallel, so the angle of emergence \(e\) equals the angle of incidence \(i\), and the emergent ray is parallel to the incident ray but not along the same line.
The perpendicular distance between the path of the emergent ray and the direction of the incident ray is the lateral displacement,
\[x = \\frac{t \\sin (i - r)}{\\cos r}\]
where \(x\) is the lateral displacement and \(t\) the thickness of the block, both in the same unit of length, and \(i\) and \(r\) are in degrees. It increases with the thickness of the block, with the angle of incidence, and with the refractive index of the medium; since \(\\mu\) is greater for violet light than for red, the lateral displacement is greater for violet light.
A pin viewed obliquely in a thick glass plate or thick mirror gives a number of images of gradually decreasing brightness. The second image is the brightest, because it is formed by light that has suffered a strong first reflection at the silvered surface.
A prism is a transparent medium bounded by five plane surfaces with a triangular cross section. The angle between its two refracting surfaces is the angle of prism \(A\), measured in degrees.
In passing through a prism, a ray is refracted at two inclined faces and bends towards the base at each refraction, so the emergent ray is not parallel to the incident ray. For the prism,
\[r\_1 + r\_2 = A, \\qquad \\delta = (i\_1 + i\_2) - A, \\qquad i\_1 + i\_2 = A + \\delta\]
where \(i\_1\) is the angle of incidence, \(i\_2\) the angle of emergence, \(r\_1\) and \(r\_2\) the angles of refraction at the two faces, \(A\) the angle of prism and \(\\delta\) the angle of deviation, all in degrees.
- As the angle of incidence is increased, the angle of deviation first decreases to a minimum value and then increases again. At minimum deviation \(i\_1 = i\_2 = i\) and \(r\_1 = r\_2\), the refracted ray inside the prism is parallel to the base for an equilateral prism, and
\[\\delta\_{\\min} = 2i - A\]
with \(\\delta\_{\\min}\), \(i\) and \(A\) in degrees. For a given prism and a given colour of light, \(\\delta\_{\\min}\) is unique.
The angle of deviation produced by a prism depends on the angle of incidence, the material of the prism, the angle of prism, and the colour of the light. A prism deviates violet light the most and red light the least, since \(\\mu\_{\\text{violet}} > \\mu\_{\\text{red}}\).
An object placed in a denser medium and viewed from a rarer medium appears raised. Viewed from vertically above,
\[\\mu = \\frac{\\text{real depth}}{\\text{apparent depth}}, \\qquad \\text{apparent depth} = \\frac{\\text{real depth}}{\\mu}\]
with both depths in the same unit and \(\\mu\) having no unit.
- The amount by which the object appears raised is the shift,
\[\\text{shift} = \\text{real depth} \\times \\left(1 - \\frac{1}{\\mu}\\right)\]
the shift and the real depth being in the same unit. The shift increases with the refractive index of the medium and with its thickness, and decreases as the wavelength of the light increases.
A stick placed obliquely in water appears bent at the surface, because every immersed point of it is seen raised while the part in air is not. An object in a rarer medium, viewed from a denser medium, appears farther away than it really is.
Refraction explains the twinkling of stars, the Sun being seen before it rises and after it sets, a hidden coin becoming visible when water is poured into a vessel, print appearing raised under a glass slab, a tank appearing shallow, and a person’s legs appearing short in a pool.
For light going from a denser to a rarer medium: when \(i < C\) it is partly reflected and partly refracted with \(r > i\); when \(i = C\) the angle of refraction becomes \(90^\\circ\) and the refracted ray grazes the surface; and when \(i > C\) there is no refracted ray at all.
The critical angle \(C\) is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \(90^\\circ\). It is measured in degrees, and
\[\\sin C = \\frac{1}{\\mu}, \\qquad \\mu = \\frac{1}{\\sin C} = \\operatorname{cosec} C\]
where \(\\mu\) is the refractive index of the denser medium with respect to the rarer one. For glass and air \(C \\approx 42^\\circ\); for water and air \(C \\approx 49^\\circ\); for diamond and air \(C \\approx 25^\\circ\).
The critical angle for a pair of media increases with the wavelength of the light — it is least for violet and most for red — and increases with a rise in temperature, because both changes lower the refractive index.
Total internal reflection is the return of the whole of the light into the denser medium when a ray travelling in it strikes the boundary with a rarer medium at an angle of incidence greater than the critical angle. Its two essential conditions are that the light travels from a denser to a rarer medium, and that \(i > C\).
In total internal reflection \(100%\) of the light is reflected, which no plane mirror can achieve. A total reflecting prism, of angles \(45^\\circ\), \(90^\\circ\), \(45^\\circ\), deviates light through \(90^\\circ\) (used in a periscope), deviates it through \(180^\\circ\) by two reflections (used in a binocular and a camera), and erects an inverted image without deviation (used in a slide projector).
A prism of each angle \(60^\\circ\) deviates a normally incident ray through \(60^\\circ\) by total internal reflection. In a \(30^\\circ\), \(90^\\circ\), \(60^\\circ\) prism, a ray entering normally through the face opposite the \(30^\\circ\) angle is totally internally reflected and deviated by less than \(60^\\circ\), while a ray entering normally through the face opposite the \(60^\\circ\) angle, or through the upper part of the hypotenuse, escapes by refraction, since it meets the second face at \(30^\\circ\), which is less than the critical angle.
Total internal reflection explains a mirage on a hot road or in a desert, the silvery shine of an empty test tube in water and of a crack in glass, the sparkle of diamond, and the transmission of a signal along an optical fibre.
Flowchart
FC 4.5 — What Happens To Light Meeting A Rarer Medium, As The Angle Of Incidence Grows Ray travelling in the denser medium reaches the boundary → at \(i = 0^\\circ\), it passes straight through undeviated → at \(0^\\circ < i < C\), it is partly reflected and partly refracted, with \(r > i\) and the refracted ray weakening as \(i\) grows → at \(i = C\), the angle of refraction reaches \(90^\\circ\) and the refracted ray grazes the surface, very weak → at \(i > C\), no refracted ray remains and the whole of the light is reflected back into the denser medium as total internal reflection
4.20 Key Terms
Absolute refractive index — The ratio of the speed of light in vacuum (or air) to the speed of light in a given medium; it has no unit.
Angle of deviation — The angle between the direction of the incident ray and the direction of the ray after it has been turned by refraction; measured in degrees.
Angle of emergence — The angle that the ray finally leaving a block or a prism makes with the normal at the point where it leaves; measured in degrees.
Angle of incidence — The angle that the incident ray makes with the normal at the point of incidence; measured in degrees.
Angle of prism — The angle between the two refracting surfaces of a prism, denoted by \(A\); measured in degrees.
Angle of refraction — The angle that the refracted ray makes with the normal at the point of incidence; measured in degrees.
Base of a prism — The surface of a prism opposite to its refracting edge, usually ground rough so that light does not pass through it.
Critical angle — The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \(90^\\circ\); measured in degrees, and different for every pair of media.
Emergent ray — The ray of light that finally comes out of a block or a prism into the surrounding medium.
Erecting prism — A total reflecting prism used so that an inverted object gives an erect image, without deviating the light from its path.
Frequency — The number of waves sent out by a source each second; measured in hertz (Hz). It does not change when light is refracted.
Homogeneous medium — A medium that is the same all through, in which light travels in a straight line at a constant speed.
Incident ray — The ray of light that travels up to the boundary surface and strikes it.
Lateral displacement — The perpendicular distance between the path of the emergent ray and the direction of the incident ray produced, after refraction through a parallel-sided block; measured in metre (m), and in practice usually in cm or mm.
Mirage — The appearance of a pool of water on a hot road or in a desert, produced by the total internal reflection of light in layers of air of different temperature.
Normal — The line drawn perpendicular to the boundary surface at the point of incidence, from which all angles are measured.
Optical density — A property of a medium fixed only by the speed of light in it; a medium in which light travels more slowly is optically denser. It has nothing to do with ordinary density.
Optical fibre — A very thin thread of glass, coated with glass of lower refractive index, along which a light signal travels by repeated total internal reflection without loss of energy.
Optically denser medium — The medium of a pair in which light travels more slowly.
Optically rarer medium — The medium of a pair in which light travels faster.
Position of minimum deviation — The position of a prism relative to the incident ray at which the deviation is least, where the angle of incidence equals the angle of emergence.
Principal section — The triangular cross section of a prism, cut perpendicular to its refracting edge, by which a prism is drawn in ray diagrams.
Prism — A transparent medium bounded by five plane surfaces with a triangular cross section.
Refracted ray — The ray of light that travels on into the second medium after crossing the boundary, in a direction different from that of the incident ray.
Refracting edge — The line along which the two refracting surfaces of a prism meet.
Refracting surfaces — The two polished plane surfaces of a prism, inclined to each other, through which light enters and leaves.
Refraction of light — The change in the direction of the path of light when it passes from one transparent medium into another; it is a surface phenomenon.
Refractive index — The ratio of the sine of the angle of incidence in the first medium to the sine of the angle of refraction in the second; being a ratio of two similar quantities, it has no unit.
Snell’s laws — The two laws of refraction: the incident ray, the refracted ray and the normal lie in one plane, and the ratio \(\\dfrac{\\sin i}{\\sin r}\) is constant for a given pair of media.
Total internal reflection — The return of the whole of the light into the denser medium when it strikes the boundary with a rarer medium at an angle of incidence greater than the critical angle.
Total reflecting prism — A prism with angles \(45^\\circ\), \(90^\\circ\), \(45^\\circ\), in which light incident normally on a face suffers total internal reflection inside the prism.
Virtual image — An image formed where rays of light only appear to come from, obtained by producing the rays backwards; it cannot be caught on a screen.
Wavelength — The distance between two successive similar points of a light wave; measured in metre (m). It decreases when light enters a denser medium.
4.21 Points To Ponder
A glass block lowered into a liquid of exactly the same refractive index becomes almost invisible. If a whole object could be made of such a material, how would you go about proving it was there at all, using only light? What would still give it away?
A prism deviates violet light the most, a glass slab displaces violet light the most, and a violet letter under a slab is raised the most. All three follow from one property of the medium. Which property is it, and could any one of these three effects be used to measure that property?
The Moon has no atmosphere. Would an astronaut standing there see the Sun a few minutes before it rose above the horizon? Would the stars twinkle? Give your reasoning for each.
Total internal reflection returns the whole of the light, while the best silvered mirror returns only a part of it. A mirror is a manufactured surface and a boundary between two media is not manufactured at all. Why should the untouched boundary do the better job?
Suppose the water in a deep pond grew steadily saltier from top to bottom, so that its refractive index increased continuously with depth instead of changing at a single surface. What would the path of a ray of light through such water look like, and would a coin on the bottom still appear to lie at one definite depth?
A critical angle exists for light going from water into air, but not for light going from air into water. Is there any arrangement at all — any angle, any surface — by which light travelling in air could be prevented from entering water? What does your answer say about why total internal reflection is called internal?
4.22 Objective Questions
- (MCQ) When a ray of light passes from water into glass, the speed of light: (1 mark)
- increases
- decreases
- remains the same
- first increases, then decreases
- (MCQ) The speed of light in a certain transparent medium is \(1.5 \\times 10^{8}\\ \\text{m s}^{-1}\). Taking the speed of light in air as \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), the refractive index of the medium is: (1 mark)
- 1.5
- 2.0
- 2.25
- 4.5
- (MCQ) For a given pair of media, the critical angle is the greatest for: (1 mark)
- violet light
- green light
- yellow light
- red light
- (MCQ) Which of these quantities remains unchanged when light is refracted from air into glass? (1 mark)
- Speed
- Wavelength
- Frequency
- Intensity
- (MCQ) A ray of light falls normally on one of the \(45^\\circ\) faces of a total reflecting prism. The angle through which it is deviated on emerging from the prism is: (1 mark)
- \(45^\\circ\)
- \(60^\\circ\)
- \(90^\\circ\)
- \(180^\\circ\)
- (MCQ) A coin lies at a real depth of \(12\\ \\text{cm}\) in water of refractive index \(\\dfrac{4}{3}\). Seen from vertically above, it appears to be at a depth of: (1 mark)
- \(6\\ \\text{cm}\)
- \(9\\ \\text{cm}\)
- \(12\\ \\text{cm}\)
- \(16\\ \\text{cm}\)
- (MCQ) Total internal reflection can take place when light travels from: (1 mark)
- air into glass at any angle of incidence
- glass into air at an angle of incidence less than the critical angle
- glass into air at an angle of incidence greater than the critical angle
- air into water at an angle of incidence greater than the critical angle
- (MCQ) In a \(30^\\circ\), \(90^\\circ\), \(60^\\circ\) glass prism of critical angle \(42^\\circ\), a ray incident normally on the face opposite the \(60^\\circ\) angle: (1 mark)
- is totally internally reflected and deviated through \(90^\\circ\)
- is refracted out of the prism at the second face
- emerges undeviated from the second face
- is absorbed inside the prism
- (Assertion–Reason) (1 mark) Assertion (A): In the position of minimum deviation for an equilateral prism, the refracted ray inside the prism is parallel to its base. Reason (R): In the position of minimum deviation, the angle of incidence is equal to the angle of emergence.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is false but R is true
- A is true but R is false
- (Assertion–Reason) (1 mark) Assertion (A): A crack inside a glass tumbler shines like a strip of silver. Reason (R): Light travelling in the glass meets the thin film of air in the crack at an angle of incidence greater than the critical angle.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is false but R is true
- A is true but R is false
- (Assertion–Reason) (1 mark) Assertion (A): An object placed in a rarer medium and viewed from a denser medium appears nearer than it really is. Reason (R): Light bends towards the normal when it passes from a rarer medium into a denser medium.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is false but R is true
- A is true but R is false
- (Match The Following) Match each value in Column A with the correct description in Column B. (4 marks)
| Column A | Column B |
|---|---|
| a. \(42^\\circ\) | i. refractive index of glass |
| b. \(49^\\circ\) | ii. critical angle for diamond and air |
| c. \(25^\\circ\) | iii. critical angle for glass and air |
| d. \(1.5\) | iv. critical angle for water and air |
- (Fill In The Blanks) (4 marks)
- Light bends ________ the normal when it passes from a rarer medium into a denser medium.
- The refractive index of a medium ________ as the temperature of the medium rises.
- In total internal reflection, ________ per cent of the light is reflected back into the denser medium.
- For a prism, the sum of the two angles of refraction inside it is equal to the ________.
- (True Or False) (4 marks)
- The frequency of light changes when it passes from air into water.
- A ray incident normally on a glass slab suffers no change of speed and no change of direction.
- The emergent ray from a parallel-sided glass block is parallel to the incident ray.
- A critical angle exists for light passing from air into glass.
- (Name The Following) (4 marks)
- The angle of incidence in a denser medium for which the angle of refraction in the rarer medium is \(90^\\circ\).
- The perpendicular distance between the emergent ray and the direction of the incident ray after passing through a glass block.
- The prism used in a slide projector to obtain an erect image without deviation.
- The phenomenon that makes a hot road appear wet from a distance.
- (Give One Word Or Phrase For) (3 marks)
- The medium in which light travels more slowly than in the medium it has come from.
- The triangular cross section of a prism used in its ray diagrams.
- The thin glass thread along which a light signal travels by repeated total internal reflection.
- (State The SI Unit Of) (4 marks)
- Wavelength of light
- Frequency of light
- Speed of light in a medium
- Refractive index of a medium
- (Odd One Out, And Give A Reason) (2 marks) Mirage on a hot road; the sparkle of a diamond; the twinkling of a star; the silvery shine of an empty test tube held in water.
4.23 Check Your Concepts
Define refraction of light, and state why it is called a surface phenomenon. (2 marks)
State the two laws of refraction of light, mentioning the condition under which the second law holds. (2 marks)
Define the absolute refractive index of a medium, and give a reason why it can never be less than 1. (2 marks)
A ray of light passes from glass into water. Give a reason for the direction in which it bends at the surface. (2 marks)
Give a reason why kerosene, though less dense than water, is said to be optically denser than water. (2 marks)
State the principle of reversibility of the path of light, and write the relation between \(*{1}\\mu*{2}\) and \(*{2}\\mu*{1}\) that follows from it. (2 marks)
Differentiate between the angle of deviation produced by a rectangular glass block and that produced by a prism, stating clearly the basis of the difference. (3 marks)
State how the refractive index of a medium depends on (i) the temperature of the medium, and (ii) the colour of the light, and hence state how the critical angle depends on each. (3 marks)
Explain, with one everyday example, why a ray of light incident normally on a boundary is not deviated although its speed changes there. (3 marks)
Why does a fish, seen from the bank of a pond, appear to be nearer to the surface than it really is? Explain with reference to the direction in which light bends as it leaves the water. (3 marks)
State the two essential conditions for total internal reflection, and explain why light passing from air into glass can never satisfy them, however large the angle of incidence. (3 marks)
What would happen to a beam of light entering normally through one \(45^\\circ\) face of a right-angled isosceles prism if the prism were made of a material whose critical angle was \(50^\\circ\)? Explain your answer. (3 marks)
Explain why a total reflecting prism is preferred to a plane mirror in a binocular. Give two separate reasons. (4 marks)
Describe how a mirage is formed on a hot road, mentioning the change in the refractive index of air with temperature and the condition satisfied at the layer where the light turns. (4 marks)
A student claims that the deviation produced by a prism can be reduced without limit by going on increasing the angle of incidence. Explain why this is wrong, and state what actually happens to the deviation as the angle of incidence is increased from a small value. (4 marks)
Explain how the laws of refraction can be verified, and the refractive index of glass determined, using a rectangular glass block, four pins and a protractor. State clearly what is measured, what is calculated, and what result confirms the second law. (5 marks)
An optical fibre carries a light signal for many kilometres, round bends, with very little loss. Explain the physics that makes this possible, naming the condition satisfied at the boundary inside the fibre and stating why so little energy is lost. (5 marks)
4.24 Diagrams, Graphs & Constructions
(Diagram) Draw a ray diagram to show the refraction of a ray of light passing from air into water at an oblique angle of incidence. Label the incident ray, the refracted ray, the reflected ray, the normal, the angle of incidence \(i\), the angle of refraction \(r\) and the angle of deviation \(\\delta\). (4 marks)
(Diagram) Redraw the path shown in Fig. 4.6-1 for a ray passing through a rectangular glass block, and on your diagram mark the angle of incidence, the angle of refraction, the angle of emergence and the lateral displacement. State which two rays in your diagram are parallel. (4 marks)
(Construction) A ray of light strikes the face of an equilateral glass prism of critical angle \(42^\\circ\) at right angles. Draw the complete path of the ray until it leaves the prism, marking the angle of incidence at every face it meets inside, and state the total deviation. (4 marks)
(Construction) Draw a labelled ray diagram of a total reflecting prism being used to turn a beam of light through \(180^\\circ\). Mark the angle of incidence at each face where total internal reflection occurs, and name one instrument in which this action is used. (4 marks)
(Diagram) Identify the labelled parts of a prism shown in Fig. 4.8-1: the refracting surfaces, the refracting edge, the base and the angle of prism. State which of these is usually ground rough, and why that does not affect the ray diagram. (3 marks)
(Graph) Fig. 4.9-2 shows the i–δ curve for a glass prism. State what is plotted on each axis and in what unit, describe how the deviation changes as the angle of incidence is increased from a small value, and explain how the graph shows that the minimum deviation occurs for one angle of incidence only. (4 marks)
(Graph) Sketch the shape of the graph you would expect if the angle of refraction \(r\) in glass were plotted against the angle of incidence \(i\) in air, for values of \(i\) from \(0^\\circ\) to \(80^\\circ\). Mark clearly which of the two angles is always the larger, and state the value of \(r\) when \(i = 0^\\circ\). (3 marks)
(Diagram) Draw a ray diagram to show why a coin lying at the bottom of a beaker of water appears raised when viewed from vertically above. Mark the real depth, the apparent depth and the position of the image. (4 marks)
4.25 Practice With Numericals
The speed of light in air is \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) and the refractive index of a certain glass is \(1.5\). Calculate the speed of light in that glass. (2 marks)
The speed of light in a transparent medium is \(2.4 \\times 10^{8}\\ \\text{m s}^{-1}\). Calculate the refractive index of the medium, taking the speed of light in air as \(3 \\times 10^{8}\\ \\text{m s}^{-1}\). (2 marks)
The refractive index of water with respect to air is \(\\dfrac{4}{3}\) and that of glass with respect to air is \(\\dfrac{3}{2}\). Calculate the refractive index of glass with respect to water. (2 marks)
The real depth of water in a tank is \(20\\ \\text{cm}\) and the refractive index of water is \(\\dfrac{4}{3}\). Calculate (i) the apparent depth of the tank seen from vertically above, and (ii) the shift of the image. (3 marks)
A mark is made on a sheet of paper and a glass slab of thickness \(9\\ \\text{cm}\) and refractive index \(1.5\) is placed over it. By what height does the mark appear to be raised when seen from vertically above? (2 marks)
Light of wavelength \(6000\\ \\text{Å}\) travelling in air enters a medium of refractive index \(1.5\). Taking the speed of light in air as \(3 \\times 10^{8}\\ \\text{m s}^{-1}\) and \(1\\ \\text{Å} = 10^{-10}\\ \\text{m}\), calculate (i) the frequency of the light in air, (ii) the speed of the light in the medium, and (iii) the wavelength of the light in the medium. (3 marks)
A ray of light passes from air into a liquid, the angle of incidence being \(45^\\circ\) and the angle of refraction \(30^\\circ\). Taking \(\\sin 45^\\circ = 0.707\) and \(\\sin 30^\\circ = 0.500\), calculate the refractive index of the liquid with respect to air. (2 marks)
The refractive index of a certain glass is \(1.5\). Taking \(\\sin 42^\\circ = 0.667\), calculate the critical angle for the glass–air pair. (2 marks)
The critical angle for a certain transparent medium and air is \(30^\\circ\). Taking \(\\sin 30^\\circ = 0.500\) and the speed of light in air as \(3 \\times 10^{8}\\ \\text{m s}^{-1}\), calculate (i) the refractive index of the medium, and (ii) the speed of light in it. (3 marks)
A ray of light suffers minimum deviation of \(40^\\circ\) while passing through an equilateral glass prism of refracting angle \(60^\\circ\). Calculate the angle of incidence in this position. (3 marks)
A ray of light is incident at an angle of \(50^\\circ\) on one face of a prism of refracting angle \(60^\\circ\), and emerges from the other face at an angle of \(45^\\circ\). Calculate the angle of deviation produced by the prism, and state whether the prism is in the position of minimum deviation. (3 marks)
A ray of light strikes a parallel-sided glass block of thickness \(6\\ \\text{cm}\) at an angle of incidence of \(45^\\circ\), the angle of refraction inside the glass being \(30^\\circ\). Taking \(\\sin 15^\\circ = 0.259\) and \(\\cos 30^\\circ = 0.866\), calculate the lateral displacement of the ray. (3 marks)