Chapter 2 · The Human Eye and the Colourful World

2.2 Refraction through a Prism and Dispersion

A wedge of glass bends light twice and splits sunlight into a band of colours. Follow a ray through a prism, discover why violet bends more than red, and see how a sky full of raindrops paints a rainbow.

Hang a cut-glass ornament or a glass prism in a sunny window and, somewhere on the opposite wall, a little band of colour appears: red at one end, violet at the other, with orange, yellow, green and blue in between. Nothing in the glass is coloured. The colours were already in the sunlight, mixed together so completely that we see the mixture as plain white. The prism simply sorts them out.

In this section you will first follow a single ray of light through a triangular prism using the laws of refraction from Section 1.3, and see why it comes out bent towards the thick end of the prism. Then you will find out why white light spreads into a spectrum, how a second prism can put the colours back together, and how millions of falling raindrops do the same job on a grand scale to make a rainbow — always on the side of the sky opposite the Sun.

Refraction through a Triangular Glass Prism

A glass slab (Section 1.3) has two parallel faces, so light that enters one face leaves the other travelling in its original direction, only shifted sideways. A prism is different: its two faces are tilted towards each other, so the bends at the two faces add up instead of cancelling.

📘 Definition 2.5 — The triangular prism

A triangular prism is a block of transparent material with two triangular ends and three rectangular side faces. Light enters through one rectangular face and leaves through another; these two faces are the refracting surfaces.

The angle between the two refracting surfaces is the angle of the prism, $A$ (for the common equilateral prism, $A=60^\circ$). The edge where they meet is the apex, and the face opposite it is the base.

Trace a ray step by step. A ray $PE$ strikes the first face at $E$ with angle of incidence $i$. Going from air into glass (rarer to denser), it bends towards the normal and travels through the glass along $EF$ at angle of refraction $r_1$. At $F$ it meets the second face, now going from glass into air (denser to rarer), so it bends away from the normal and leaves as the emergent ray $FS$, making the angle of emergence $e$ with the normal there. Because the two normals are tilted towards each other, both bends turn the ray the same way — towards the base.

📘 Definition 2.6 — Angle of deviation

The angle of deviation, $\delta$, is the angle between the direction of the incident ray (extended forwards) and the direction of the emergent ray. It measures how much the prism has turned the light away from its original path.

📐 Result 2.3 — The path of light through a prism

For a ray passing through a triangular glass prism placed in air:

at the first face the ray bends towards the normal ($r_1\lt i$); at the second face it bends away from the normal;
Snell's law $n_1\sin i=n_2\sin r$ (Section 1.3) holds separately at each face;
the emergent ray is not parallel to the incident ray: it is deviated towards the base of the prism by the angle of deviation $\delta$.

The deviation depends on the angle of incidence, on the angle of the prism and on the refractive index of the glass.

⚠️ Drawing a prism ray diagram — what examiners look for

1. Normals at both faces. Draw a dashed normal at the point of entry and at the point of exit; mark $i$, $r_1$, $e$ from those normals, never from the face itself.

2. The right direction of bending. Inside the glass the ray must lean towards the first normal; on leaving it must lean away from the second normal. Both bends point towards the base.

3. Mark $\delta$ correctly. Extend the incident ray forwards as a dashed line and extend the emergent ray backwards; $\delta$ is the angle between those two lines where they cross inside or near the prism.

✏️ Example 1 — Light entering a prism face-on

A narrow beam of yellow light falls normally (at $90^\circ$ to the surface) on the first face of a glass prism of angle $A=30^\circ$ and refractive index $1.50$. Find the angle of emergence and the angle of deviation.

💡 Show solution▶

First face The beam travels along the normal, so $i=0^\circ$ and it goes straight into the glass without bending.

Second face The second face is tilted at $30^\circ$ to the first, so its normal is tilted at $30^\circ$ to the beam: the angle of incidence inside the glass is $30^\circ$.

Snell's law From glass to air: $1.50\sin30^\circ=1.00\sin e$, so $\sin e=0.750$ and $e=48.6^\circ$.

Deviation Inside the glass the beam was $30^\circ$ from the normal; outside it is $48.6^\circ$ from the same normal, and it was not turned at all at the first face. So $\delta=48.6^\circ-30^\circ=18.6^\circ$, towards the base.

Check $e\gt30^\circ$, as it must be when light passes from glass to air (it bends away from the normal). ✓

✏️ Example 2 — Slab or prism?

The same ray of light falls at $40^\circ$ on (a) a rectangular glass slab and (b) an equilateral glass prism. In which case is the emergent ray parallel to the incident ray? Explain the difference.

💡 Show solution▶

(a) Slab The two faces are parallel, so the two normals are parallel. The bend towards the normal at entry is exactly undone by the bend away from the normal at exit: $e=i=40^\circ$ and the emergent ray is parallel to the incident ray, only shifted sideways.

(b) Prism The faces meet at $60^\circ$, so the normals are tilted relative to each other. Both bends turn the ray towards the base and they add up; the emergent ray is deviated (for glass of index $1.50$, by about $38^\circ$).

Idea Refraction happens at both faces in both cases; what makes the difference is the angle between the faces.

A ray strikes the left face of an equilateral ($60^\circ$) crown-glass prism. Change the angle of incidence with the slider: the path is worked out from Snell's law at both faces, with the normals dashed. Choose white light to see it fan out into colours on the screen — violet ($n=1.532$) is bent more than red ($n=1.514$). The small graph shows how the deviation changes with the angle of incidence. Press Play to sweep the angle. (The spread of colours can be drawn four times wider so that you can see it; the readouts always give the true values.)

Light:angle of incidence:— spread ×4

$i$—

$r_1$ inside—

$e$—

deviation $\delta$—

red → violet spread—

Dispersion of White Light

Send a narrow beam of sunlight through a prism onto a white screen and you get not a white spot but a band of colours. In order from the least bent to the most bent they are red, orange, yellow, green, blue, indigo and violet — remembered by the acronym VIBGYOR read backwards (or forwards from violet).

📘 Definition 2.7 — Dispersion and spectrum

The splitting of white light into its component colours on passing through a prism (or any refracting medium) is called dispersion.

The band of coloured components of a beam of light is its spectrum. The spectrum of white light runs Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR); in a prism red is deviated least and violet most. Light of a single colour (one wavelength) is called monochromatic.

Why does it happen? Light of different colours has different wavelengths — red about $700\ \text{nm}$, violet about $400\ \text{nm}$. In a vacuum all colours travel at the same speed $c$, but inside glass violet light is slowed slightly more than red light. A larger slowing means a larger refractive index, and by Snell's law a larger refractive index means a sharper bend.

Colour	Red	Orange	Yellow	Green	Blue	Indigo	Violet
Typical wavelength	$700\ \text{nm}$	$610\ \text{nm}$	$580\ \text{nm}$	$530\ \text{nm}$	$470\ \text{nm}$	$440\ \text{nm}$	$400\ \text{nm}$
$n$ of crown glass	$1.514$	$1.516$	$1.517$	$1.519$	$1.524$	$1.528$	$1.532$

📐 Result 2.4 — Why a prism disperses white light

The refractive index of a transparent material is slightly different for different colours: it is largest for violet and smallest for red. So when white light is refracted,

every colour obeys Snell's law with its own refractive index;
violet light is bent (deviated) the most and red light the least;
the colours leave the prism in slightly different directions and spread into a spectrum.

Equivalently, red light travels fastest in glass and violet light slowest.

✏️ Example 3 — The race inside the glass

For a crown glass the refractive index is $1.514$ for red light and $1.532$ for violet light. Find the speed of each colour in the glass, taking $c=3.0\times10^8\ \text{m/s}$. Which colour is deviated more by a prism of this glass?

💡 Show solution▶

Formula $n=\dfrac cv$, so $v=\dfrac cn$.

Red $v_{\text{red}}=\dfrac{3.0\times10^8}{1.514}=1.98\times10^8\ \text{m/s}$.

Violet $v_{\text{violet}}=\dfrac{3.0\times10^8}{1.532}=1.96\times10^8\ \text{m/s}$.

Meaning Violet is slower in the glass (larger $n$), so it is bent more and deviated more. The speeds differ by only about $1\%$ — which is why the spectrum from a single prism is fairly narrow.

✏️ Example 4 — How wide is the spectrum?

White light falls normally on the first face of a $30^\circ$ prism of the glass in Example 3. Find the angles of deviation of red and violet light, the angle between them, and the width of the spectrum on a screen $2.0\ \text{m}$ from the prism.

💡 Show solution▶

Method As in Example 1, every colour enters undeviated and meets the second face at $30^\circ$ inside the glass; then $\sin e=n\sin30^\circ$ and $\delta=e-30^\circ$.

Red $\sin e=1.514\times0.500=0.757$, $e=49.2^\circ$, $\delta_{\text{red}}=19.2^\circ$.

Violet $\sin e=1.532\times0.500=0.766$, $e=50.0^\circ$, $\delta_{\text{violet}}=20.0^\circ$.

Spread The colours fan out over $20.0^\circ-19.2^\circ=0.8^\circ$. On a screen $2.0\ \text{m}$ away this is a band about $2.0\ \text{m}\times\tan0.8^\circ\approx0.028\ \text{m}=2.8\ \text{cm}$ wide (narrow beam assumed).

Check Violet is deviated more than red, as Result 2.4 says. ✓

Recombination: Putting the Colours Back Together

Isaac Newton was the first to show, around 1666, that the colours come from the sunlight and not from the glass. He let the spectrum from one prism fall on a narrow slit so that only one colour passed through, and sent that colour into a second prism: it was bent again, but it did not split into further colours. Then he placed a second, identical prism upside down next to the first — and the band of colours turned back into white light.

📐 Result 2.5 — Recombination of the spectrum

If the spectrum of white light from one prism passes through a second, identical prism placed in the inverted position (apex pointing the opposite way), the second prism bends each colour back by the same amount the first prism bent it. The colours come out parallel to one another and to the original beam, and they mix to give white light again.

This shows that white light is a mixture of the seven colours, and that a prism only separates the colours already present in it. A single colour cannot be split further.

The two prisms together act just like a glass slab with parallel faces: what one face does, the other undoes. The same idea lies behind a colour wheel painted with the seven colours, which looks whitish when it spins fast, and behind the pixels of a phone screen, which make "white" from tiny red, green and blue dots.

✏️ Example 5 — Newton's experiments, predicted

A narrow beam of white light passes through prism P₁ and spreads into a spectrum. Predict what comes out of prism P₂ in each case. (a) P₂ is identical to P₁ and placed the same way up. (b) P₂ is identical to P₁ and inverted, close to it. (c) A card with a narrow slit lets only the green part of the spectrum reach P₂.

💡 Show solution▶

(a) P₂ deviates every colour further in the same direction, violet still the most, so the spectrum becomes wider (about twice the angular spread).

(b) P₂ undoes the bending of P₁ for each colour; the colours come out parallel and overlap to give a beam of white light.

The Rainbow — Dispersion in Raindrops

A rainbow is a natural spectrum painted on the sky by countless water drops. You see one when the Sun is behind you and falling rain (or a fine spray) is in front of you — typically in the early morning or late afternoon after a shower. Each tiny drop works like a combination of a prism and a mirror.

📘 Definition 2.8 — The rainbow

A rainbow is a natural spectrum of sunlight that appears as an arc of colours in the sky, opposite the Sun, after rain. It is formed by three things happening in each water droplet:

refraction and dispersion as sunlight enters the drop;
internal reflection at the back surface of the drop;
refraction again as the light leaves the drop.

Light leaving the drops is concentrated at about $42^\circ$ from the direction of the incoming sunlight for red and about $41^\circ$ for violet, so the red band is on the outside of the arc and the violet band on the inside.

Why those angles? A ray that enters near the edge of a drop and a ray that enters near the middle leave at different angles, but a large bundle of rays all leave close to one particular angle — about $42.4^\circ$ for red light (water, $n=1.331$) and $40.6^\circ$ for violet ($n=1.343$). Light piles up at that angle, so the drop shines brightly in that direction. Interactive 2 lets you see the bundling happen.

Now picture yourself on the ground with the Sun behind you. Draw a line from the Sun through your head: it points to the antisolar point, where the shadow of your head falls. Every drop that lies $42.4^\circ$ away from that line sends red light into your eye; all such drops form a circle around the antisolar point. Drops at $40.6^\circ$ send you violet, a slightly smaller circle. The ground cuts off the lower part of the circle, so you see an arc. A drop higher up in the sky sends you red; a drop slightly lower sends you violet — so red is on top.

✏️ Example 6 — How high is the rainbow?

The Sun is $25^\circ$ above the horizon in the late afternoon. (a) In which direction should you look for a rainbow? (b) How high above the horizon are the tops of the red and violet bands? (c) Why is there never a rainbow at noon in summer, when the Sun is $70^\circ$ high?

💡 Show solution▶

(a) Opposite the Sun. A late-afternoon Sun is in the west, so the rainbow is in the eastern sky.

(b) The antisolar point is $25^\circ$ below the horizon. The top of the red circle is $42.4^\circ$ above that point: $42.4^\circ-25^\circ=17.4^\circ$ above the horizon. The violet top is at $40.6^\circ-25^\circ=15.6^\circ$.

(c) With the Sun at $70^\circ$ the antisolar point is $70^\circ$ below the horizon, and the whole $42^\circ$ circle lies below the horizon. There are no raindrops there to see, so no rainbow appears from the ground.

Check The higher the Sun, the lower the rainbow; a rainbow is possible only when the Sun is less than about $42^\circ$ high. ✓

Inside one drop: sunlight arrives from the left. Move the slider to choose where a ray strikes the drop (from the centre line to the edge) and follow it: refraction in, reflection at the back, refraction out. The graph shows the angle between the returning ray and the incoming sunlight — it rises to a maximum and turns back, so many rays leave near that angle: about $42.4^\circ$ for red, $40.6^\circ$ for violet. Tick many rays to see them bunch up. The view from the ground: the slider sets the Sun's height; see which drops send red and which send violet to your eye. Press Play to sweep.

View:where the ray hits:— many rays

ray strikes at—

red returns at—

violet returns at—

most rays leave at—

⚠️ Common mistakes

1. "The prism adds colour to the light." No — the colours are already in white light; the prism only separates them (Newton's inverted-prism experiment proves it).

2. Red bent most. Red has the longest wavelength and the smallest refractive index, so it is bent least. Violet is bent most.

3. Leaving out the reflection. A rainbow needs refraction with dispersion, internal reflection and refraction again. Answers that mention only refraction lose marks.

4. Looking towards the Sun. A rainbow is always on the opposite side of the sky from the Sun.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optional: the deviation formula and minimum deviation

In JEE Main, NEET and JEE Advanced 2026; the CBSE Class 10 syllabus treats deviation by a prism only qualitatively.

From the geometry of the triangle formed by the two normals and the ray inside the glass one can show that $r_1+r_2=A$, where $r_2$ is the angle of incidence at the second face, and that the deviation is

$$\delta=i+e-A .$$

Check with Example 1: $i=0^\circ$, $e=48.6^\circ$, $A=30^\circ$ give $\delta=18.6^\circ$. ✓ As Interactive 1's graph shows, $\delta$ first falls and then rises as $i$ increases; its smallest value, the angle of minimum deviation $\delta_m$, occurs when the ray passes through the prism symmetrically ($i=e$, $r_1=r_2=\tfrac A2$). Then

$$n=\frac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}.$$

For the $60^\circ$ crown-glass prism of Interactive 1 with yellow light ($n=1.517$), $\delta_m\approx38.7^\circ$ at $i\approx49.3^\circ$. You will meet these formulas properly in Class 12 Physics (Ray Optics).

Exercises

Refraction through a prism.

1 What is meant by the angle of a prism? What is its value for an equilateral prism?
2 Name the angles $i$, $r_1$, $e$ and $\delta$ in a prism ray diagram, saying from which line each is measured.
3 At which face does the ray bend towards the normal, and at which face away from it? Why?
4 Towards which part of the prism is the emergent ray deviated?
5 Why is the emergent ray from a glass slab parallel to the incident ray, but not the emergent ray from a prism?
6 A ray falls along the normal to one face of a prism. Is it bent at that face? Explain.
7 Describe the ray diagram for a ray passing through a triangular glass prism, with all the labels an examiner expects.
8 A prism made of a glass of larger refractive index replaces the first one; everything else is kept the same. Will the deviation be larger or smaller?

💡 Show solutions▶

1 The angle between the two refracting surfaces of the prism; $60^\circ$ for an equilateral prism.

2 $i$: incident ray and normal at the first face; $r_1$: refracted ray (inside) and the same normal; $e$: emergent ray and normal at the second face; $\delta$: between the incident ray extended forwards and the emergent ray extended backwards.

3 Towards the normal at the first face (air to glass, rarer to denser, light slows down); away from the normal at the second face (glass to air, denser to rarer, light speeds up).

4 Towards the base of the prism.

5 A slab's faces are parallel, so the bend at the second face exactly cancels the first. A prism's faces are inclined, so both bends turn the ray the same way and add up.

6 No: along the normal $i=0^\circ$, so $r=0^\circ$ and the ray goes straight in. It is bent only at the second face.

7 Triangle with apex at the top; incident ray to the first face with a dashed normal and $i$ marked; refracted ray inside bending towards that normal ($r_1$); a dashed normal at the second face with the emergent ray bending away from it ($e$); incident ray extended forwards and emergent ray extended backwards (dashed), with $\delta$ marked between them; arrows on all rays.

8 Larger: a larger refractive index means sharper bending at each face.

Dispersion and recombination. Take $c=3.0\times10^8\ \text{m/s}$.

1 Define dispersion of white light. Name the colours of the spectrum in order, starting from the one deviated least.
2 What is the cause of dispersion?
3 Which colour has (a) the longest wavelength, (b) the largest refractive index in glass, (c) the greatest speed in glass?
4 The refractive index of a glass is $1.52$ for yellow light. Find the speed of yellow light in it.
5 In a flint glass red light travels at $1.85\times10^8\ \text{m/s}$ and violet light at $1.80\times10^8\ \text{m/s}$. Find the two refractive indices. Which colour is deviated more?
6 How did Newton show that the colours come from the white light and not from the prism?
7 What is monochromatic light? What happens when it passes through a prism?
8 A disc painted with the seven colours of the spectrum looks nearly white when spun fast. Which property of white light does this show?

💡 Show solutions▶

1 The splitting of white light into its component colours by a prism. Least deviated first: red, orange, yellow, green, blue, indigo, violet.

2 The refractive index of glass is different for different colours (wavelengths), so each colour is bent through a different angle.

3 (a) Red; (b) violet; (c) red.

4 $v=\tfrac{c}{n}=\tfrac{3.0\times10^8}{1.52}=1.97\times10^8\ \text{m/s}$.

5 $n_{\text{red}}=\tfrac{3.0\times10^8}{1.85\times10^8}=1.62$; $n_{\text{violet}}=\tfrac{3.0\times10^8}{1.80\times10^8}=1.67$. Violet (larger $n$) is deviated more.

6 He placed a second identical prism inverted after the first: the colours recombined into white light. He also showed that a single colour passed through a second prism did not split further.

7 Light of a single colour (one wavelength). A prism deviates it towards the base but does not split it.

8 That white light is a mixture of the seven colours: the eye merges the quickly changing colours into white.

III

Numericals with Snell's law at the prism faces. In parts 1–4 the light falls normally on the first face, so it meets the second face at an angle equal to the angle of the prism $A$.

1 $A=30^\circ$, $n=1.60$. Find the angle of emergence and the deviation.
2 $A=40^\circ$, $n=1.50$. Find the angle of emergence and the deviation.
3 $A=20^\circ$, $n=1.50$. Find the angle of emergence and the deviation.
4 A ray falling normally on a $30^\circ$ prism emerges at $45^\circ$ to the normal of the second face. Find the refractive index of the glass.
5 Light strikes the first face of a prism ($n=1.50$) at $i=45^\circ$. Find the angle of refraction $r_1$ inside the glass.
6 White light falls normally on a $30^\circ$ prism with $n_{\text{red}}=1.51$ and $n_{\text{violet}}=1.53$. Find the deviations of red and violet light and the angular spread of the spectrum.
7 Using parts 2 and 3 and Example 1 (all with $n=1.50$), how does the deviation change as the angle of the prism increases (same glass, normal incidence)?

Check your answers: a prism at normal incidence

angle of prism $A$ (°):refractive index $n$:

$\sin e$—

$e$—

$\delta$—

💡 Show solutions▶

1 $\sin e=1.60\sin30^\circ=0.800$, $e=53.1^\circ$, $\delta=53.1^\circ-30^\circ=23.1^\circ$.

2 $\sin e=1.50\sin40^\circ=0.964$, $e=74.6^\circ$, $\delta=34.6^\circ$.

3 $\sin e=1.50\sin20^\circ=0.513$, $e=30.9^\circ$, $\delta=10.9^\circ$.

4 $n\sin30^\circ=\sin45^\circ$, so $n=\tfrac{0.707}{0.500}=1.41$.

5 $\sin r_1=\tfrac{\sin45^\circ}{1.50}=0.471$, $r_1=28.1^\circ$.

6 Red: $\sin e=0.755$, $e=49.0^\circ$, $\delta=19.0^\circ$. Violet: $\sin e=0.765$, $e=49.9^\circ$, $\delta=19.9^\circ$. Spread $\approx0.9^\circ$.

7 It increases: $10.9^\circ$ for $A=20^\circ$, $18.6^\circ$ for $30^\circ$ (Example 1), $34.6^\circ$ for $40^\circ$.

Rainbows and applications. Take the rainbow angles as $42.4^\circ$ (red) and $40.6^\circ$ (violet).

1 Name the three processes that happen inside a raindrop to make a rainbow.
2 Why is red on the outside of a rainbow and violet on the inside?
3 The Sun is $30^\circ$ above the horizon. How high above the horizon are the tops of the red and violet bands?
4 In the morning, in which part of the sky should you look for a rainbow? Why?
5 Why are rainbows usually seen in the morning or late afternoon and not around midday?
6 Garden experiment. Explain how to make a rainbow with a garden hose on a sunny day, and where to look for it.
7 Challenge. From an aeroplane, a rainbow can sometimes be seen as a complete circle. Explain why.

💡 Show solutions▶

1 Refraction with dispersion on entering, internal reflection at the back, refraction on leaving.

2 Red leaves the drops at a larger angle ($42.4^\circ$) from the antisolar direction than violet ($40.6^\circ$), so the drops that send red to your eye lie higher in the sky — on the outer edge of the arc.

3 Red: $42.4^\circ-30^\circ=12.4^\circ$; violet: $40.6^\circ-30^\circ=10.6^\circ$.

4 In the west: the morning Sun is in the east, and a rainbow is always opposite the Sun.

5 Around midday the Sun is usually higher than about $42^\circ$, so the whole rainbow circle lies below the horizon.

6 Stand with the Sun behind you and spray a fine mist in front of you. Look about $42^\circ$ away from the shadow of your head (roughly halfway between the shadow and straight ahead, upwards); the arc appears in the spray.

7 The rainbow is really a circle of radius about $42^\circ$ around the antisolar point. From the ground the earth hides the lower part; from high in the air there can be drops all round the antisolar point (below the plane too), so the whole circle can be seen.

🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board · 2 marks Draw a labelled ray diagram to show the refraction of a ray of monochromatic light through a triangular glass prism. Mark the angle of incidence, the angle of emergence and the angle of deviation.

💡 Show solution▶

Idea Two refractions: towards the normal at entry, away from the normal at exit; net turn towards the base.

Diagram Triangle $ABC$; incident ray $PE$ on face $AB$, dashed normal at $E$, angle $i$; refracted ray $EF$ bending towards that normal; dashed normal at $F$ on face $AC$; emergent ray $FS$ bending away from it, angle $e$; $PE$ extended forwards and $FS$ extended backwards (dashed), with $\delta$ marked where they meet.

Answer Award points: both normals, correct bending at both faces, $i$, $e$, $\delta$ marked, arrows. Exam tip: a missing arrow or normal is the commonest way to lose a mark here.

E2. Board · 3 marks What is meant by dispersion of white light? Why do the different colours deviate through different angles in a prism? Which colour is deviated the least?

💡 Show solution▶

Definition The splitting of white light into its seven component colours on passing through a prism.

Reason The refractive index of glass is different for each colour; light of each colour travels at a different speed in glass, so each bends through a different angle.

Answer Red is deviated least (smallest refractive index), violet most. Exam tip: "different speeds in glass" or "different refractive indices" is the key phrase — just "different wavelengths" is incomplete.

E3. Board · Case study After a morning shower at a cricket ground, spectators see a bright rainbow. The Sun is $20^\circ$ above the horizon. (i) In which direction must the spectators be facing, and what three processes in the raindrops produce the rainbow? [2 marks] (ii) Using a rainbow angle of $42^\circ$ for red light, how high above the horizon is the top of the red band? [1 mark] (iii) Which colour is on the inner edge of the bow? [1 mark]

💡 Show solution▶

(i) With their backs to the Sun — towards the west in the morning. Processes: refraction and dispersion on entering the drop, internal reflection inside, refraction on leaving.

(ii) $42^\circ-20^\circ=22^\circ$ above the horizon.

(iii) Violet. Exam tip: "Sun behind the observer" earns the direction mark even without naming west.

E4. NEET When white light passes through a glass prism, the colour for which the glass has the greatest refractive index, and the colour that travels fastest in the glass, are respectively (A) red, violet (B) violet, red (C) violet, violet (D) red, red

💡 Show solution▶

Idea Violet has the largest $n$; since $v=\tfrac cn$, the colour with the smallest $n$ (red) is fastest.

Answer (B). Exam tip: larger $n$ ⟺ slower ⟺ bent more.

E5. JEE Main A ray falls normally on one face of a prism of angle $30^\circ$ and leaves the other face at $60^\circ$ to its normal. The refractive index of the prism is (A) $1.41$ (B) $1.50$ (C) $1.73$ (D) $2.00$

💡 Show solution▶

Working No bending at the first face, so the ray meets the second face at $30^\circ$ inside the glass: $n\sin30^\circ=\sin60^\circ$, $n=\tfrac{0.866}{0.500}=1.73$ $(=\sqrt3)$.

Answer (C). Exam tip: at normal incidence the angle inside at the second face always equals the prism angle $A$.

E6. Board · 1 mark Assertion (A): A rectangular glass slab does not produce a spread-out spectrum of white light the way a prism does. Reason (R): In a slab the two refracting faces are parallel, so every colour emerges parallel to the incident beam and the colours overlap. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

💡 Show solution▶

Idea Inside the slab the colours separate slightly, but the second face bends each one back by the same amount, so they leave parallel and overlap (only tinted edges appear).

Answer (a). Exam tip: the same reasoning explains Newton's recombination with an inverted prism.

🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. Is the rainbow angle really about $42^\circ$ — and would a rainbow made with a different liquid be bigger or smaller?

Why it matters. Measuring how strongly a material disperses light is everyday work in optics. Camera and telescope designers in Japan, Germany and the USA choose glasses with low dispersion so that the colours of an image do not separate into coloured fringes; jewellers value diamonds because their high dispersion makes them "fire" with colour; and atmospheric scientists use the exact angles and colours of rainbows and halos to estimate the size and shape of water drops and ice crystals in clouds.

What to do.

On a sunny late afternoon, stand with your back to the Sun and spray a fine mist from a garden hose or a spray bottle until you see a bow.
Ask a partner to hold a long stick pointing from your eye to the shadow of your head (the antisolar point), and a second stick from your eye to the red edge of the bow. Measure the angle between the sticks with a large protractor (or use a phone clinometer app for the two directions).
Repeat five times and take the mean. Compare it with $42.4^\circ$.
Use Interactive 2's idea: for water $n=1.33$ the bunching angle is about $42^\circ$. Ask: for a liquid with a larger refractive index, would the angle be larger or smaller? Predict, then check your prediction with the instructor note.

Tools: a garden hose with a mist nozzle or a spray bottle, two metre sticks, a large protractor or phone clinometer, a notebook.

Hand in: a sketch of your set-up, your five measurements with the mean and spread, and a paragraph answering the research question, with possible sources of error.

Going further. Look for a fainter second rainbow outside the first. In which order are its colours? Find out how two internal reflections inside the drop explain it.

Instructor note: students usually measure $40$–$43^\circ$; the main errors are judging the edge of a diffuse bow and locating the shadow of the head. For a liquid with $n=1.36$ the bunching angle drops to about $38.3^\circ$ (a smaller bow), so a larger refractive index gives a smaller rainbow. The secondary bow (two internal reflections) appears at about $50$–$53^\circ$ (red inside, at about $50^\circ$) with the colour order reversed. Never let students look directly at the Sun.

🧪 Self-check quiz

0/5 correct

Q1. When a ray of light passes through a triangular glass prism, the emergent ray is

Q2. In the spectrum of white light formed by a prism, the colour deviated the most is

Q3. Dispersion of white light by a prism happens because

Q4. A second identical prism placed upside down after the first prism

Q5. To see a rainbow, an observer must stand with

← 2.1 The Human Eye and Defects of Vision 2.3 Atmospheric Refraction and Scattering of Light →