Refraction and total internal reflection

A spoon that bends in water

Put a spoon in a glass of water and look from the side. The spoon seems to break at the surface. A swimming pool looks shallower than it is. A diamond flashes brighter than a piece of glass of the same shape. In each case light changes direction at a boundary between two materials. This lesson explains how, and why a diamond is special.

Refraction

When a ray of light goes from one material into another, it usually changes direction at the boundary. This is refraction. The cause is the speed. Light travels at $c = 3.00 \times 10^{8}\ \text{m/s}$ in a vacuum, and more slowly in a material.

The refractive index $n$ of a material tells us how much it slows light:

Here $v$ is the speed of light in the material. The index has no unit, and it is never less than $1$. Some typical values for yellow light follow.

These are standard values. Glass varies from one kind to another between about $1.5$ and $1.9$, and the index of every material changes a little with the colour of the light. Some books write the Greek letter $\mu$ for the refractive index. A-Level, IB, AP and the newer NCERT books write $n$, and so do we.

A material with a larger $n$ is called optically denser. This has nothing to do with its mass density. Some oils are optically denser than water and yet float on it.

What changes and what does not. The wave goes from medium 1 into medium 2. The frequency $f$ does not change, because the wave crests arrive at the boundary at a fixed rate and must leave at the same rate. The speed changes from $v_1$ to $v_2$. Since $v = f\lambda$, the wavelength changes in the same ratio. In a material of index $n$ the wavelength is $\lambda = \dfrac{\lambda_0}{n}$, where $\lambda_0$ is the wavelength in a vacuum.

Example 1: sodium light in water. Sodium light has a wavelength of $589\ \text{nm}$ in air. In water ($n = 1.33$) its wavelength is $\dfrac{589\ \text{nm}}{1.33} = 443\ \text{nm}$. Its speed is $\dfrac{3.00 \times 10^{8}\ \text{m/s}}{1.33} = 2.26 \times 10^{8}\ \text{m/s}$. The frequency stays at $\dfrac{3.00 \times 10^{8}\ \text{m/s}}{589 \times 10^{-9}\ \text{m}} = 5.09 \times 10^{14}\ \text{Hz}$. The colour you see depends on the frequency, so the light is still yellow under water.

Snell's law

Think of a ray that meets a boundary at an angle. We measure the angle of incidence $\theta_1$ and the angle of refraction $\theta_2$ from the normal. Snell's law says:

The incident ray, the refracted ray and the normal lie in one plane. When light goes into a denser material ($n_2 > n_1$), $\sin\theta_2$ is smaller than $\sin\theta_1$, so the ray bends toward the normal. When it goes into a rarer material, it bends away from the normal.

The ratio $\dfrac{n_2}{n_1}$ is the refractive index of medium 2 with respect to medium 1, also written $n_{21}$. It is also equal to $\dfrac{v_1}{v_2}$, because $n = \dfrac{c}{v}$.

Example 2: air into glass. A ray in air meets crown glass ($n = 1.50$) at an angle of incidence of $50°$.

1. We want $\theta_2$, so $\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{1.00 \times \sin 50°}{1.50} = \dfrac{0.766}{1.50} = 0.511$.
2. So $\theta_2 = 30.7°$.

The ray turned toward the normal. Check. A ray along the normal has $\theta_1 = 0°$, so $\sin\theta_2 = 0$ and it goes straight on. This agrees with the picture that a ray at right angles to a boundary does not bend.

Apparent depth

A pool of depth $D$ looks shallower when you look straight down. A ray from a coin at the bottom bends away from the normal as it leaves the water. Your eye follows it back in a straight line and puts the coin higher than it is.

For rays close to the vertical, the angles are small, so $\sin\theta \approx \tan\theta$. Call the horizontal distance from the coin to the point where the ray leaves the water $x$. Then $x = D\tan r$, and $x = D'\tan i$ for the apparent depth $D'$. With $\sin i = n\sin r$ we get

Example 3. A pool is $1.60\ \text{m}$ deep, so $D = 1.60\ \text{m}$. Its apparent depth is $\dfrac{1.60\ \text{m}}{1.33} = 1.20\ \text{m}$. A swimmer looking up sees the opposite: a bird above the pool seems higher than it is, by the factor $n$.

Light through a glass slab

A ray goes into a parallel-sided slab and comes out again.

At the first face the ray bends toward the normal. At the second face it bends away by exactly the same angle. The emergent ray is parallel to the incident ray, but it is shifted sideways. The shift for a slab of thickness $t$ is

Example 4. A glass block of thickness $t = 4.0\ \text{cm}$ and $n = 1.50$ has a ray at $i = 45°$. Then $\sin r = \dfrac{\sin 45°}{1.50} = 0.471$, so $r = 28.1°$. The shift is $d = \dfrac{4.0\ \text{cm} \times \sin 16.9°}{\cos 28.1°} = 1.32\ \text{cm}$.

If you look straight through the slab at a mark, the mark seems to rise by $t\left(1 - \dfrac{1}{n}\right)$. For this block, $4.0\ \text{cm} \times \dfrac{1}{3} = 1.33\ \text{cm}$.

Begin with glass above and air below, and an angle of $30°$. Slowly raise the angle of incidence. At what angle does the refracted ray vanish? Compare this angle with the critical angle in the readouts.

Total internal reflection

Now send light from a denser material to a rarer one, from glass to air. The ray bends away from the normal, so the angle of refraction is larger than the angle of incidence. Raise the angle of incidence. At one value, the angle of refraction becomes $90°$ and the refracted ray goes along the surface. This angle is the critical angle $C$. Put $\theta_2 = 90°$ in Snell's law: $n_1\sin C = n_2$, so

For a material in air, $\sin C = \dfrac{1}{n}$.

For an angle of incidence larger than $C$, there is no refracted ray at all. All the light is reflected back into the denser material. This is total internal reflection. It needs two things.

• The light must travel from a denser material to a rarer one.
• The angle of incidence must be larger than the critical angle.

Critical angles in air: water $48.8°$, crown glass $41.8°$, flint glass $38.1°$ and diamond $24.4°$. A material with a larger $n$ has a smaller critical angle.

Total internal reflection at work

The diamond. The critical angle of diamond is only $24.4°$. Light that enters a cut diamond meets the inside faces at angles larger than this, and it is reflected again and again. It leaves through the top. The jeweller cuts the faces to send as much light as possible back up to the eye. A piece of glass has a larger critical angle, so more light leaks out of the back, and it looks dull.

The optical fibre. A glass fibre as thin as a hair has a core of index $n_1$ inside a cladding of a slightly smaller index $n_2$. Light that meets the boundary of the two at more than the critical angle is totally reflected and travels along the fibre, even when the fibre bends.

Light must enter the end of the fibre at a small enough angle $\theta$ to the axis. A ray refracts at the end face, so $\sin\theta = n_1\sin r$. It meets the side wall at the angle $90° - r$, and it is trapped when $\sin(90° - r) \ge \dfrac{n_2}{n_1}$. Put these together, and the largest angle, the acceptance angle $\theta_a$, satisfies

For a core of $1.50$ and a cladding of $1.45$ the right side is $\sqrt{1.50^2 - 1.45^2} = 0.384$, so $\theta_a = 22.6°$. Fibres carry telephone calls, the internet and television, and doctors use them in endoscopes to look inside the body.

The mirage. On a hot day the air close to a road is hotter and thinner than the air above it, so its refractive index is a little smaller. A ray from the sky going down meets layers of smaller and smaller $n$. It bends away from the normal each time, until at some layer the angle of incidence is larger than the critical angle. The ray turns upward and reaches your eye. You follow it back in a straight line, and you seem to see the sky on the road, like a pool of water.

The totally reflecting prism. A prism with angles $45°$, $45°$ and $90°$ and made of glass with $n = 1.50$ has a critical angle of $41.8°$. A ray that goes into one short face at right angles meets the long face at $45°$, which is larger than $C$, and it is totally reflected.

The ray is turned through $90°$ (left) or, if it goes in through the long face, through $180°$ (right). Periscopes and binoculars use such prisms, because a prism reflects all the light and a silvered mirror does not. The glass must have $n > \sqrt{2} = 1.41$, so that $C < 45°$.

Practise with new numbers

Try it in Python

The program sends a ray at $30°$ from air into four materials, and finds the critical angle for light that tries to leave each of them for air.

Common mistakes

• Measuring angles from the surface. In Snell's law every angle is from the normal.
• Taking the sine of the wrong angle. The index $n_1$ goes with the angle in medium 1, and $n_2$ with the angle in medium 2.
• Looking for total internal reflection when light goes from air to glass. It happens only from a denser to a rarer material.
• Dividing the angle, not its sine, by $n$. Snell's law is about sines.
• Forgetting that the apparent depth is real depth divided by $n$. An object in a denser material looks nearer than it is.
• Saying that the frequency of light changes in glass. The speed and the wavelength change. The frequency does not.

Remember these

• Refractive index: $n = \dfrac{c}{v}$. In a material the wavelength is $\dfrac{\lambda_0}{n}$ and the frequency does not change.
• Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$.
• Apparent depth for a view straight down: real depth divided by $n$. A slab shifts a ray sideways by $d = \dfrac{t\sin(i - r)}{\cos r}$.
• Critical angle: $\sin C = \dfrac{n_2}{n_1}$. Total internal reflection needs a denser to a rarer material and $i > C$.
• Uses: the sparkle of a diamond, the optical fibre with $\sin\theta_a = \sqrt{n_1^2 - n_2^2}$, the mirage, and the totally reflecting prism.

Check yourself

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

1. Find the speed of light in diamond, which has $n = 2.42$.

2. A ray in air meets water ($n = 1.33$) at $45°$ to the normal. Find the angle of refraction.

3. Why does a fish in a pond see a bird above the water higher than the bird really is?

4. Can light going from water into air at an angle of incidence of $60°$ be totally reflected? Can light going from air into water at $60°$?

5. Why does a diamond sparkle more than a glass imitation of the same shape?

References

• NCERT Physics Part II (Class 12), Chapter 9, the sections on refraction and total internal reflection. Some books round the index of crown glass to $1.5$, and others print $1.52$.
• OpenStax University Physics Volume 3, Chapter 1, the sections on refraction and total internal reflection.
• OpenStax College Physics 2e, Chapter 25, the sections on the law of refraction and total internal reflection.