Refraction, Snell's law and real and apparent depth: WASSCE Physics

A spoon in a glass of water looks bent. A swimming pool looks shallower than it is. A fish seen from a canoe is not quite where it appears. All because light changes speed — and therefore direction — when it passes from one material into another.

Learning objectives

By the end of this lesson you should be able to:

define refraction and state the laws of refraction (Snell's law)
define refractive index and calculate it in several ways
trace a ray through a rectangular glass block
explain real and apparent depth and use $n = \dfrac{\text{real depth}}{\text{apparent depth}}$

What is refraction?

Definition

Refraction is the bending (change of direction) of light as it passes obliquely from one transparent medium into another, because its speed changes.

Light entering a denser medium (air → glass, air → water) slows down and bends towards the normal.
Light entering a less dense medium (glass → air) speeds up and bends away from the normal.
A ray striking the boundary along the normal (at 0°) passes straight through without bending, though it still changes speed.

The laws of refraction

Definition

The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
Snell's law: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media:

\dfrac{\sin i}{\sin r} = \text{constant} = n

\begin{aligned}&\dfrac{\sin i}{\sin r} \\ &= \text{constant} \\ &= n\end{aligned}

The constant $n$ is the refractive index of the second medium relative to the first.

Refractive index

n = \dfrac{\sin i}{\sin r} = \dfrac{\text{speed of light in vacuum (air)}}{\text{speed of light in the medium}} = \dfrac{\text{real depth}}{\text{apparent depth}} = \dfrac{1}{\sin c}

\begin{aligned}n &= \dfrac{\sin i}{\sin r} \\ &= \dfrac{\text{speed of light in vacuum (air)}}{\text{speed of light in the medium}} \\ &= \dfrac{\text{real depth}}{\text{apparent depth}} \\ &= \dfrac{1}{\sin c}\end{aligned}

( $c$ is the critical angle, covered next lesson.)

Medium	Refractive index
Air	1.00
Water	1.33
Glass (crown)	1.50
Diamond	2.42

Worked example

A ray of light strikes a glass block at an angle of incidence of 45°. The angle of refraction is 28°. Find the refractive index of the glass and the speed of light in it. (speed of light in air = 3.0 × 10⁸ m s⁻¹)

n = \dfrac{\sin 45°}{\sin 28°} = \dfrac{0.707}{0.469} = 1.51

\begin{aligned}n &= \dfrac{\sin 45°}{\sin 28°} \\ &= \dfrac{0.707}{0.469} \\ &= 1.51\end{aligned}

v = \dfrac{3.0 \times 10^{8}}{1.51} = 1.99 \times 10^{8}\ \text{m s}^{-1}

\begin{aligned}v &= \dfrac{3.0 \times 10^{8}}{1.51} \\ &= 1.99 \times 10^{8}\ \text{m s}^{-1}\end{aligned}

Worked example

Light passes from air into water (n = 1.33) at an angle of incidence of 60°. Find the angle of refraction.

\sin r = \dfrac{\sin 60°}{1.33} = \dfrac{0.866}{1.33} = 0.651 \Rightarrow r = 40.6°

\begin{gathered}\begin{aligned}\sin r &= \dfrac{\sin 60°}{1.33} \\ &= \dfrac{0.866}{1.33} \\ &= 0.651\end{aligned} \\[6pt] \Rightarrow r = 40.6°\end{gathered}

Refraction through a rectangular glass block

A ray enters a rectangular glass block, bends towards the normal inside, and leaves parallel to its original direction but shifted sideways

Figure 1. Refraction through a parallel-sided glass block. WASSCE Tutor diagram.

The ray bends towards the normal on entering and away from it on leaving. Because the two faces are parallel, the emergent ray is parallel to the incident ray but laterally displaced (shifted sideways). The thicker the block and the larger the angle of incidence, the greater the displacement.

Practical: verifying Snell's law with a glass block

Trace the outline of the block on paper. Draw a normal and an incident ray at angle $i$ ; fix two pins on it.
Looking through the opposite face, fix two more pins in line with the images of the first two.
Remove the block, join the points to draw the emergent ray, and join entry to exit to get the refracted ray. Measure $r$ .
Repeat for several values of $i$ . Plot $\sin i$ against $\sin r$ : a straight line through the origin; its gradient is $n$ .

Precautions: pins upright and well apart; look at the bottom of the pins to avoid parallax; draw thin, sharp lines.

Real and apparent depth

Light from an object under water bends away from the normal as it leaves the water, so the object appears nearer the surface than it really is.

n = \dfrac{\text{real depth}}{\text{apparent depth}}

\begin{aligned}n &= \dfrac{\text{real depth}}{\text{apparent depth}}\end{aligned}

Worked example

A coin lies at the bottom of a bucket of water 40 cm deep. How deep does it appear when viewed from directly above? (n of water = 4/3)

\text{apparent depth} = \dfrac{\text{real depth}}{n} = \dfrac{40}{4/3} = 30\ \text{cm}

\begin{aligned}&\text{apparent depth} \\ &= \dfrac{\text{real depth}}{n} \\ &= \dfrac{40}{4/3} \\ &= 30\ \text{cm}\end{aligned}

The coin seems to be raised by $40 - 30 = 10$ cm (the displacement).

In Sierra Leone

Fishermen spearing fish in the shallows of a river aim below where the fish appears, because refraction makes it look higher than it is. The same effect makes a stream look shallower than it is — a real danger for anyone wading in.

Key points

Refraction: bending due to a change of speed between media.
Into a denser medium → towards the normal; into a less dense medium → away from the normal.
Snell's law: $\dfrac{\sin i}{\sin r} = n$ (constant).
$n = \dfrac{c_{air}}{c_{medium}} = \dfrac{\text{real depth}}{\text{apparent depth}}$ .
A glass block shifts a ray sideways without changing its direction.

WASSCE examination focus

WAEC tests this

State the laws of refraction, calculate refractive index from angles, speeds or depths, and describe the glass-block (pin) experiment to verify Snell's law with a graph of $\sin i$ against $\sin r$ . Real-and-apparent-depth problems are frequent in objective papers.

Quick check

Why does light bend when it enters glass from air?
A ray enters glass with i = 30° and r = 19.5°. Find n.
A pool is 2 m deep. How deep does it appear to a person looking down (n = 4/3)?

Answers: 1. Its speed decreases in glass. 2. $\dfrac{\sin 30°}{\sin 19.5°} = \dfrac{0.5}{0.334} = 1.50$ . 3. $2 \div \tfrac{4}{3} = 1.5$ m.

Refraction, Snell's law and real and apparent depth

Learning objectives

What is refraction?

The laws of refraction

Refractive index

Refraction through a rectangular glass block

Practical: verifying Snell's law with a glass block

Real and apparent depth

Key points

WASSCE examination focus

Quick check

Keep going with refraction and total internal reflection

More lessons in Refraction and Total Internal Reflection