The prism, dispersion and the colours of the sky

A crystal on the window

A glass crystal hangs in a sunny window and throws small patches of red, green and blue on the wall. The glass has no colour of its own. The colours were in the sunlight all the time. A piece of glass with flat faces at an angle to each other splits white light into its colours. It is a prism, and in this lesson we find out how it bends light, why it separates the colours, and how the same idea makes a rainbow and a blue sky.

The prism

A prism is a block of glass with two flat faces that meet at an angle. The two faces are the refracting faces. The angle between them is the angle of the prism, $A$. The third side is the base.

A ray meets the first face with the angle of incidence $i$. It bends towards the normal and travels inside the glass, at the angle $r_1$ to the normal. At the second face it makes the angle $r_2$ with the normal there. It leaves at the angle of emergence $e$, bending away from the normal. The ray ends up turned towards the base. The angle between the first direction and the last is the angle of deviation $\delta$.

The angles of the prism and the deviation

The two normals meet at an angle $180^\circ - A$, because each normal is at a right angle to its face. In the small triangle formed by the two normals and the ray inside the glass, the three angles add to $180^\circ$:

At the first face the ray is turned by $i - r_1$. At the second it is turned by $e - r_2$. Both turns are towards the base, so they add:

Remember the two results: $r_1 + r_2 = A$ and $\delta = i + e - A$.

Worked example 1: following one ray

A prism of $A = 60^\circ$ and $n = 1.5$ is hit at $i = 45^\circ$. Find $\delta$.

1. First face. $\sin r_1 = \dfrac{\sin 45^\circ}{1.5} = 0.4714$, so $r_1 = 28.1^\circ$.
2. Inside. $r_2 = A - r_1 = 60^\circ - 28.1^\circ = 31.9^\circ$.
3. Second face. $\sin e = 1.5 \times \sin 31.9^\circ = 0.7923$, so $e = 52.4^\circ$.
4. Deviation. $\delta = i + e - A = 45^\circ + 52.4^\circ - 60^\circ = 37.4^\circ$.

Check. A ray must satisfy $n\sin r_2 \le 1$ to leave the glass. Here $1.5 \times 0.528 = 0.79$, which is below $1$, so the ray comes out.

How the deviation depends on the angle of incidence

Turn the prism and watch the ray on the wall. First the spot moves one way, then it stops and turns round. The graph shows this. The deviation falls, reaches a smallest value, then rises. The smallest value is the angle of minimum deviation, $\delta_m$. A bigger refractive index gives a bigger deviation, and the graph moves up.

At the minimum the ray goes through the prism symmetrically. We can see why. Light paths can be reversed. If $i$ gives the emergence $e$, then $e$ as the angle of incidence gives $i$ as the emergence, and the same deviation. There are two angles of incidence for each deviation, except at the minimum, where the two are the same. So at minimum deviation

The ray inside is parallel to the base. Put these into $\delta = i + e - A$: $\delta_m = 2i - A$, so $i = \dfrac{A + \delta_m}{2}$. Snell's law at the first face gives $\sin i = n\sin r_1 = n\sin\dfrac{A}{2}$. So

In a spectrometer we measure $A$ and $\delta_m$, then find $n$. This is the standard way to measure the refractive index of glass.

Worked example 2: a measurement

In a spectrometer experiment on a prism of $A = 60.0^\circ$, the minimum deviation is $40.0^\circ$. Find $n$ and the angle of incidence at minimum deviation.

1. The index. $n = \dfrac{\sin 50.0^\circ}{\sin 30.0^\circ} = \dfrac{0.7660}{0.5000} = 1.53$.
2. The angle. $i = \dfrac{A + \delta_m}{2} = \dfrac{60.0^\circ + 40.0^\circ}{2} = 50.0^\circ$, and $r_1 = r_2 = 30.0^\circ$.

Start with the defaults. Move $i$ along the curve. What is the lowest deviation you find, and what $i$ gives it? Compare with the readout. Then make the prism angle smaller and see what happens to the curve.

The thin prism

Many prisms in spectacles, cameras and binoculars are thin, with a small angle $A$. Then all the angles are small, and $\sin x \approx x$ for angles in radian. Snell's law becomes $i = nr_1$ and $e = nr_2$. So

This thin prism result holds for rays that are close to the base-parallel path, and it does not depend on $i$. A thin prism of $A = 6.0^\circ$ and $n = 1.52$ deviates a ray by $(1.52 - 1) \times 6.0^\circ = 3.1^\circ$.

Why white light spreads: dispersion

Newton passed a beam of sunlight through a prism and found it spread into a band of colours, the spectrum: red, orange, yellow, green, blue, indigo and violet. He then passed one colour through a second prism, and it did not split any more. So the colours are in the white light already. The prism only sorts them.

The prism sorts the colours because the refractive index depends on the wavelength. This effect is dispersion. For ordinary glass $n$ is larger for blue and violet light than for red. A larger $n$ gives a larger deviation, so violet is bent the most and red the least. A typical crown glass has these values:

• Red light (wavelength $656\ \text{nm}$): $n_r = 1.514$.
• Yellow light ($589\ \text{nm}$): $n_y = 1.517$.
• Blue light ($486\ \text{nm}$): $n_b = 1.522$.

The change is small, only about half a per cent from red to blue. A simple model by Cauchy fits it well. In it $n = a + \dfrac{b}{\lambda^2}$, where $a$ and $b$ are constants of the glass. For this crown glass $a \approx 1.504$ and $b \approx 0.0042\ \mu\text{m}^2$, with $\lambda$ in micrometre. It says that $n$ falls as the wavelength grows. In a vacuum every colour has the same speed. In glass, shorter waves are slowed more.

In a thin prism, each colour has its own deviation, $(n - 1)A$. The angle of dispersion is the difference between two colours, for example blue and red:

The mean deviation is that of the yellow light, $\delta_y = (n_y - 1)A$. The dispersive power of the glass is the ratio of the two:

It is a pure number that belongs to the glass and does not depend on the prism angle. NCERT writes it with the violet and red indices, $\omega = \dfrac{n_v - n_r}{n - 1}$. Textbooks elsewhere often use the blue and red lines, as we do here. The idea is the same.

Worked example 3: a thin crown prism

A thin prism of crown glass has $A = 8.0^\circ$. Use the indices in the list above.

1. Mean deviation. $\delta_y = (1.517 - 1) \times 8.0^\circ = 4.14^\circ$.
2. Angle of dispersion. $\delta_b - \delta_r = (1.522 - 1.514) \times 8.0^\circ = 0.064^\circ$.
3. Dispersive power. $\omega = \dfrac{0.008}{0.517} = 0.0155$.

The colours are spread over only $0.064^\circ$. This is why a thin prism gives a very narrow spectrum, and why we need a large prism angle, like $60^\circ$, to see the colours.

A flat sheet of glass with parallel faces does not make a spectrum. Each colour leaves in the direction it entered, only moved a little sideways. The colours leave together and the light is white again.

Why the sky is blue

The air is made of molecules that are much smaller than the wavelength of light. They take in light and send it out again in all directions. This is scattering. Lord Rayleigh showed that for particles much smaller than the wavelength, the scattered intensity is proportional to $\dfrac{1}{\lambda^4}$.

Blue light of $440\ \text{nm}$ and red light of $700\ \text{nm}$ differ by this ratio:

Blue is scattered about six times as much as red. When we look at a part of the sky away from the Sun, the light that reaches our eyes is light that was scattered. It is mostly blue. Violet is scattered even more, but sunlight has less violet, the upper air takes some of it in, and our eyes are less sensitive to it.

At sunrise and sunset the Sun's light has to travel a long path through the air. Most of the blue is scattered out of the beam before it reaches us. The light that remains is red and orange. Clouds look white because their drops are large compared with the wavelength, and they scatter all colours alike.

The rainbow

A rainbow needs sunshine behind you and raindrops in front. A ray of sunlight enters a drop and bends. It is reflected once at the back of the drop and leaves, bending again. The total deviation $D$ for one reflection is the sum of three turns: $(i - r)$ at entry, $180^\circ - 2r$ at the reflection, and $(i - r)$ at exit.

Rays that enter at different heights have different $i$. Look at the curve of $D$ against $i$ in the simulation below. It has a lowest point, the minimum deviation. Near it, many rays leave in nearly the same direction, so the light is bright. This is the angle at which we see the bow. Setting the slope to zero gives $\cos^2 i = \dfrac{n^2 - 1}{3}$.

For red light $n = 1.331$. Then $i = 59.5^\circ$, $r = 40.4^\circ$ and $D = 137.6^\circ$. The bow is seen $180^\circ - 137.6^\circ = 42.4^\circ$ from the direction opposite to the Sun. For violet light, $n = 1.344$ and the bow is at $40.5^\circ$. So red is on the outside and violet on the inside.

With two reflections inside the drop, $D = 360^\circ + 2i - 6r$. The minimum gives the secondary bow, at $D - 180^\circ = 50.4^\circ$ for red and $53.7^\circ$ for violet. It is fainter, because some light is lost at the second reflection. The colours are in the reverse order, red on the inside.

Start with $n = 1.331$ for red light. Read the two bow angles. Then change $n$ to $1.344$, the value for violet light. Which bow moves outward, and which inward?

Other effects of the atmosphere

• The halo. A thin cloud of ice can give a ring of light $22^\circ$ from the Sun. Ice crystals are six-sided, and two of their faces make a prism of $60^\circ$. For ice, $n = 1.31$. The minimum deviation is $2\sin^{-1}(1.31 \sin 30^\circ) - 60^\circ = 21.8^\circ$.
• Early sunrise. The air is denser near the ground, so a ray from the Sun bends as it comes down. We see the Sun a little before it is above the horizon. NCERT gives about two minutes.
• The mirage. On a hot road, the air near the ground is warm and less dense. Light from the sky bends upwards by total internal reflection, so we see the sky on the road. The first course of this chapter explains this.

Try it in Python

The program sends rays at nine angles of incidence through a prism of $60^\circ$ and $n = 1.5$. It prints the deviation each time and compares the smallest one with the formula.

Common mistakes

• Leaving out the $-A$. The deviation is $i + e - A$, not $i + e$.
• Using $r_1 = r_2$ for every ray. That holds only at minimum deviation. In general $r_1 + r_2 = A$.
• Forgetting the halves. The formula for $n$ uses $\dfrac{A + \delta_m}{2}$ and $\dfrac{A}{2}$.
• Using the thin prism formula for a large angle. $\delta = (n - 1)A$ needs a small $A$ and small angles.
• Thinking the prism adds the colours. It sorts colours that were in the white light, since $n$ is different for each.
• Mixing up the order of the colours. In the primary bow red is on the outside. In the secondary bow it is on the inside.

Remember these

• Prism: $r_1 + r_2 = A$ and $\delta = i + e - A$.
• At minimum deviation: $i = e$, $r_1 = r_2 = \dfrac{A}{2}$ and $n = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$.
• Thin prism: $\delta = (n - 1)A$.
• Dispersion: $n$ depends on the wavelength. Violet bends the most. Dispersive power $\omega = \dfrac{n_b - n_r}{n_y - 1}$.
• Scattering: intensity is proportional to $\dfrac{1}{\lambda^4}$, so the sky is blue and the sunset is red.
• Rainbow: one reflection gives the primary bow at about $42^\circ$ and two give the secondary bow at about $51^\circ$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A thin prism of angle $5.0^\circ$ is made of glass of $n = 1.6$. By what angle does it deviate a ray?

2. An equilateral prism ($A = 60^\circ$) has a minimum deviation of $30^\circ$. Find $n$.

3. Light of wavelength $400\ \text{nm}$ and light of $800\ \text{nm}$ meet the same small particles. How many times more is the first scattered than the second?

4. In a primary rainbow, red is on the outside edge. Why is the order in a secondary bow reversed?

5. Why does a plane sheet of glass with parallel faces not split white light into colours?

References

• NCERT Physics Part II (Class 12), Chapter 9, the sections on refraction through a prism, dispersion and natural phenomena due to sunlight.
• OpenStax University Physics Volume 3, Chapter 1, the section on dispersion.
• SCHOTT, optical glass data sheet for N-BK7, for the refractive indices of a crown glass.