Reflection and refraction of waves — Physics

Before this

Wave properties
Refraction of light · coming soon

This leads to

A pool that looks shallower than it is

Stand at the edge of a swimming pool and look down at the bottom. A pool that is 2.0 m deep looks only about 1.5 m deep. A straw standing in a glass of water looks bent at the surface. A coin at the bottom of an empty cup, just hidden by the rim, comes into view when someone pours water in.

Nothing has moved. The pool floor, the straw and the coin are exactly where they were. What has changed is the path of the light coming from them. In air, light travels in straight lines. In water, it also travels in straight lines. But at the surface, where it crosses from one into the other, it changes direction.

Why should a wave change direction just because it crosses into a new material? This lecture answers that question with one idea: a change of speed. By the end, you will be able to calculate the new direction exactly, and you will see why the pool looks shallower.

Reflection: the angle in equals the angle out

Start with the simpler thing a wave can do at a boundary: bounce back. This is reflection.

To describe directions at a boundary, physicists draw a line at right angles to the surface, at the point where the wave arrives. It is called the normal. Every angle in this lecture is measured from the normal, not from the surface. The reason is practical: a surface can be curved, but at any point it has exactly one normal.

A ray reflecting from a flat mirror. The angle of incidence

i

and the angle of reflection

r

are both measured from the normal (dashed), and they are equal.

The law of reflection

The angle of incidence equals the angle of reflection: $i = r$ .

The incident ray, the reflected ray and the normal all lie in the same flat plane.

This law holds for every kind of wave: light from a mirror, sound from a wall as an echo, water waves from a harbour wall. A reflected wave stays in the same medium, so its speed does not change. Its frequency and wavelength do not change either.

Reflection is often only partial. When light meets a glass window head-on, about 4% of it reflects and the rest goes through. That is why you can see a faint reflection of yourself in a shop window. The part that goes through is where things get interesting.

Refraction starts with a change of speed

When a wave crosses into a new medium, its speed usually changes. Light travels more slowly in water than in air, and more slowly still in glass. The change of direction that can come with it is called refraction.

Here is a picture that shows why a change of speed turns a wave. Imagine a row of people walking side by side, arms linked, from a hard path onto soft sand at an angle. The person at one end reaches the sand first and slows down. The others are still on the path, still walking fast. For a moment, one end of the row is slow and the other end is fast. So the row swings round, like a door on a hinge, until everyone is on the sand. Then they all walk on in a straight line again, but in a new direction.

Now imagine the same row walking straight at the sand, head-on. Everyone reaches the sand at the same moment and slows down together. Nothing swings, so the direction does not change. A wave that meets a boundary along the normal slows down or speeds up, but it does not bend.

A wave is like that row. Its wavefronts are the lines joining the crests, and the wave travels at right angles to them. When one end of a wavefront crosses into a slower medium first, the wavefront swings round. So the wave turns towards the normal when it slows down, and away from the normal when it speeds up.

What stays the same at the boundary

The last lecture showed that two sounds of the same frequency have different wavelengths in different media. It did not show why the frequency itself cannot change at a boundary. Here is the reason.

Think of the crests arriving at the boundary. Suppose 500 crests reach it every second from one side. Every crest that arrives must leave on the other side, because crests cannot pile up at the boundary, and new ones cannot appear from nowhere. So 500 crests leave every second on the other side as well. The frequency of a wave is set by its source, and it is the same in every medium the wave passes through.

Now use $v = f λ$ . If the frequency stays the same and the speed falls, the wavelength must fall in the same proportion. So in a slower medium, the crests are closer together.

λ = \frac{v}{f}

Stop and think

Red light goes from air into glass, where it travels more slowly. Which of these change: its speed, its frequency, its wavelength, its colour?

Show the answer

The speed and the wavelength change. The frequency does not, for the reason above. And the colour does not change either, because the colour of light is set by its frequency. A red torch shone into a fish tank still looks red to a fish, even though the wavelength in the water is shorter. If colour were set by wavelength, every object would change colour when you opened your eyes under water. It does not.

Why the angles obey a sine law

Now the change of direction can be calculated exactly. This is the hard part of the lecture. Spend time here, and follow the picture.

Straight wavefronts (blue) crossing into a slower medium. Each wavefront in medium 1 continues into medium 2 from the same point on the boundary. So neighbouring wavefronts meet the boundary the same distance

d

apart on both sides. The wavelength is

λ_{1}

above and the shorter

λ_{2}

below, and the ray turns from

θ_{1}

to the smaller

θ_{2}

Take two neighbouring wavefronts. They meet the boundary at two points, a distance $d$ apart. A wavefront cannot break at the boundary, so the same two points belong to the wavefronts on both sides.

In medium 1, the wavefronts are $λ_{1}$ apart, and they make an angle $θ_{1}$ with the boundary. This is the same angle that the ray makes with the normal, because both have been turned through a right angle. A right-angled triangle with hypotenuse $d$ gives $λ_{1} = d sin θ_{1}$ . In medium 2, the same reasoning gives $λ_{2} = d sin θ_{2}$ .

Divide one by the other. The $d$ cancels:

\frac{sin θ _{1}}{sin θ _{2}} = \frac{λ _{1}}{λ _{2}}

The frequency is the same on both sides, so the wavelengths are in the same ratio as the speeds, $λ_{1} : λ_{2} = v_{1} : v_{2}$ .

The law of refraction, for any wave

$\frac{sin θ _{1}}{sin θ _{2}} = \frac{v _{1}}{v _{2}}$ , where $θ_{1}$ and $θ_{2}$ are measured from the normal, in the media where the speeds are $v_{1}$ and $v_{2}$ .

A wave that slows down turns towards the normal. A wave that speeds up turns away from it.

That is it. That is the whole idea. The sine law is not a rule handed down from nowhere. It is the geometry of wavefronts that cannot break, carried by a frequency that cannot change.

Slower is not the same as denser

Many books say that light bends towards the normal when it enters a "denser" medium. For light in water and glass, that happens to work. But the law above says nothing about density. It is about speed. Test it on a different wave.

Predict, then reveal — sound going into water

A sound wave travels from air into a lake, at 10° to the normal. Water is about 800 times denser than air. Sound travels at 343 m/s in air and about 1500 m/s in water. Which way does the sound turn?

Sound speeds up when it goes from air into water, so it turns away from the normal. At 10° in air, it travels at about 49° in the water.

Sound from air into water, at 10° to the normal. It speeds up, so it turns away from the normal, to about 49°.

So "denser" can mislead you. For light, physicists use the phrase optically denser, which means only "light travels more slowly in it". It has nothing to do with mass. Always ask which medium is slower for this wave.

Explore the law on six pairs of media below. Watch the ticks across each ray: they are the wavefronts, and they get closer together in the slower medium.

Explore — a wave at a boundary

Choose two media. Move the angle of incidence. Read the angle of refraction, and watch the spacing of the wavefronts change at the boundary. Try glass into air at a large angle.

Media:

Angle of incidence: 40°

The angle of refraction for light going from air into three materials, from $sin θ_{2} = sin θ_{1} \div n$ .

Angle in air	Into water (n = 1.33)	Into glass (n = 1.50)	Into diamond (n = 2.42)
0°	0°	0°	0°
20°	≈ 14.9°	≈ 13.2°	≈ 8.1°
40°	≈ 28.9°	≈ 25.4°	≈ 15.4°
60°	≈ 40.6°	≈ 35.3°	≈ 21.0°
80°	≈ 47.8°	≈ 41.0°	≈ 24.0°

The larger the refractive index, the more the light turns towards the normal.

What should you have seen? Whatever the wave, the rule is the same: slower means towards the normal, faster means away. The wavefronts are always closer together on the slower side. And with glass into air at a large angle, the refracted wave disappears. The wave cannot leave the glass at all. That surprising case is the subject of the next lecture.

Refractive index: one number for each material

For light, it is useful to give each material a single number that says how much it slows light down. This is its refractive index, $n$ :

n = \frac{c}{v}

Here $c = 3.0 \times 1 0^{8} m/s$ is the speed of light in a vacuum, and $v$ is its speed in the material. The refractive index has no unit, because it is a ratio of two speeds.

Material	Refractive index n
Air	1.0003 (≈ 1.00)
Water	1.33
Window glass	1.52 (other glasses about 1.5 to 1.9)
Diamond	2.42

Put $v = c / n$ into the law of refraction. The ratio of speeds $v_{1} / v_{2}$ becomes $n_{2} / n_{1}$ , and the law takes the form you will see most often for light. It is called Snell's law:

n_{1} sin θ_{1} = n_{2} sin θ_{2}

Write it this way every time, with medium 1 the one the light comes from. Then put in what you know and solve. This one form works in both directions, into a material and out of it, so there is nothing else to learn.

For visible light in ordinary materials, light is slower than in a vacuum, so $n$ is greater than 1. A larger $n$ means slower light, and more turning towards the normal.

The law has a long history. The Persian scientist Ibn Sahl used it in Baghdad in 984 to design lenses. The Dutch mathematician Willebrord Snellius found it again in 1621 but never published it, and René Descartes published it in 1637. It carries Snellius's name, though he was not the first.

Worked examples

Worked Example 1 — Into glass

Light in air meets a glass block of refractive index 1.50 at an angle of incidence of 40°. Find the angle of refraction.

Medium 1 is air ( $n_{1} = 1.00$ ) and medium 2 is glass ( $n_{2} = 1.50$ ).

1.00 \times sin 40° = 1.50 \times sin θ_{2}, so sin θ_{2} = 0.6428 \div 1.50 \approx 0.4285, so θ_{2} \approx 25.4° .

Check the direction: the light slows down in glass, so it should turn towards the normal. 25.4° is smaller than 40°. ✓

Worked Example 2 — Out of glass

Light inside the same glass meets the surface at 20° to the normal. Find the angle at which it leaves into the air.

Now medium 1 is glass ( $n_{1} = 1.50$ ) and medium 2 is air ( $n_{2} = 1.00$ ).

1.50 \times sin 20° = 1.00 \times sin θ_{2}, so sin θ_{2} = 1.50 \times 0.3420 \approx 0.513, so θ_{2} \approx 30.9° .

Check: the light speeds up into air, so it turns away from the normal. 30.9° is larger than 20°. ✓ The same equation did both examples. Only the choice of medium 1 changed.

Worked Example 3 — Identify a liquid

A beam of light passes from air into an unknown liquid. The angle of incidence is 50° and the angle of refraction is 35°. Find the refractive index of the liquid.

1.00 \times sin 50° = n_{2} \times sin 35°, so n_{2} = 0.7660 \div 0.5736 \approx 1.34 .

This is close to water's 1.33. But careful: an angle measured to the nearest degree cannot tell 1.33 from 1.34. If the true angle were 35.2°, the answer would be 1.33. So the honest conclusion is "consistent with water", not "it is water". A different liquid with $n = 1.34$ would give the same reading.

Worked Example 4 — Light inside a diamond

Red light of wavelength 650 nm in air enters a diamond ( $n = 2.42$ ). Find its speed, frequency and wavelength inside the diamond.

Speed: $v = c / n = 3.0 \times 1 0^{8} \div 2.42 \approx 1.24 \times 1 0^{8} m/s$ , about 41% of its speed in air.

Frequency, from the air: $f = c / λ = 3.0 \times 1 0^{8} \div (650 \times 1 0^{- 9}) \approx 4.6 \times 1 0^{14} Hz$ . It is the same inside the diamond.

Wavelength in the diamond: $λ = v / f$ , which is the air wavelength divided by $n$ : 650 nm ÷ 2.42 ≈ 269 nm.

Check: $v = f λ$ inside the diamond gives $4.6 \times 1 0^{14} \times 269 \times 1 0^{- 9} \approx 1.24 \times 1 0^{8} m/s$ . ✓

Worked Example 5 — Sound entering a lake

Sound from a boat's engine travels through the air (343 m/s) and meets the surface of a lake (about 1500 m/s) at 10° to the normal. Find its direction in the water.

Snell's law with $n$ is only for light. For sound, use the general law with speeds:

sin θ_{2} = \frac{v _{2}}{v _{1}} sin θ_{1} = \frac{1500}{343} \times sin 10° \approx 0.759, so θ_{2} \approx 49° .

A small angle in air becomes a large one in water, because the sound speeds up so much. If the angle in air were more than about 13°, $sin θ_{2}$ would come out bigger than 1. No sound would enter the water at all. You will see why that can happen in the next lecture.

Three classic mistakes

Trap 1 — Measuring the angle from the surface

Every angle in reflection and refraction is measured from the normal. If a question says a ray meets the surface "at 30° to the surface", the angle of incidence is 90° − 30° = 60°. Draw the normal first, every time.

Trap 2 — "The frequency changes too"

At a boundary, the speed and the wavelength change together, and the frequency stays the same. The frequency is set by the source. If a question gives the frequency in one medium, it is also the frequency in the other.

Trap 3 — "Denser always means towards the normal"

The direction depends on speed, not on density. Light slows down in water, so it turns towards the normal. Sound speeds up in water, so it turns away. For light, "optically denser" means only "slower for light".

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Light passes from air into water ( $n = 1.33$ ) at an angle of incidence of 30°. Find the angle of refraction.

Hint 1
Write $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ with air as medium 1.

Hint 2
$sin θ_{2} = sin 30° \div 1.33 = 0.5 \div 1.33$ .

Full answer
$1.00 \times sin 30° = 1.33 sin θ_{2}$ , so $sin θ_{2} = 0.5 \div 1.33 \approx 0.376$ , and $θ_{2} \approx 22.1°$ . Check: light slows down in water, so it turns towards the normal, and 22.1° is less than 30°. ✓
Yellow-green light of wavelength 600 nm in air enters glass of refractive index 1.50. Find its speed, wavelength and frequency in the glass.

Hint 1
Use $v = c / n$ . The frequency is the same in both media.

Hint 2
Find $f$ from the air values: $f = c / λ$ .

Full answer
Speed: $3.0 \times 1 0^{8} \div 1.50 = 2.0 \times 1 0^{8} m/s$ . Frequency: $3.0 \times 1 0^{8} \div (600 \times 1 0^{- 9}) = 5.0 \times 1 0^{14} Hz$ , in air and in glass. Wavelength in glass: $2.0 \times 1 0^{8} \div (5.0 \times 1 0^{14}) = 4.0 \times 1 0^{- 7} m = 400 nm$ . Check: 600 nm ÷ 1.50 = 400 nm. ✓
Water waves with a frequency of 2.0 Hz travel at 0.50 m/s in deep water. They cross into shallow water, where they travel at 0.30 m/s, meeting the boundary at 40° to the normal. Find the new direction and the wavelength on each side.

Hint 1
These are not light waves, so use $sin θ_{1} / sin θ_{2} = v_{1} / v_{2}$ .

Hint 2
$sin θ_{2} = (0.30 \div 0.50) \times sin 40°$ . For the wavelengths, use $λ = v / f$ .

Full answer
$sin θ_{2} = 0.6 \times 0.6428 \approx 0.386$ , so $θ_{2} \approx 22.7°$ , towards the normal because the waves slow down. Wavelengths: deep $0.50 \div 2.0 = 0.25 m$ ; shallow $0.30 \div 2.0 = 0.15 m$ . Check: the wavelengths are in the ratio 0.15 : 0.25 = 0.6, the same as the speeds. ✓ This is why waves near a beach turn to arrive almost square-on to the shore: the water gets shallower, and they keep slowing down.
Explain the hook. A pool is 2.0 m deep, and you look straight down into it. Light from a point on the floor leaves the water at a small angle $θ_{2}$ to the normal and reaches your eye at a slightly larger angle $θ_{1}$ in the air. For small angles, $sin θ \approx tan θ$ . Show that the pool appears to be about 2.0 m ÷ 1.33 deep.

Hint 1
Draw the ray from the floor to the surface, a small sideways distance $x$ from directly above the point. Your eye traces the ray in air straight back to where it seems to come from.

Hint 2
$tan θ_{2} = x / (real depth)$ and $tan θ_{1} = x / (apparent depth)$ . Then use Snell's law with sines replaced by tangents.

Full answer
The ray leaves the water at the point a distance $x$ to the side. In the water, $tan θ_{2} = x /2.0$ . Your eye traces the ray in air back in a straight line, to a point at the apparent depth $h$ , so $tan θ_{1} = x / h$ . Snell's law, from water to air: $1.33 sin θ_{2} = 1.00 sin θ_{1}$ . For small angles, replace each sine by its tangent: $1.33 \times (x /2.0) = x / h$ . The $x$ cancels, so $h = 2.0 \div 1.33 \approx 1.5 m$ . The light speeds up as it leaves, so it turns away from the normal, and the floor seems closer than it is. Check: the apparent depth is less than the real depth, as in the hook. ✓ Looking at a slant, the effect is even larger, which is one reason people misjudge how deep water is.

The bigger idea

Here is a different way to see the same law. Imagine a lifeguard on a beach who must reach a swimmer in the sea. Running on sand is faster than swimming. The quickest route is not a straight line: it is better to run a little further along the beach and swim a shorter distance. The French mathematician Pierre de Fermat showed that light crossing a boundary takes exactly this kind of quickest route, and that the quickest route obeys the sine law. Two very different arguments, wavefronts and least time, give the same answer.

Refraction is also how you see. The curved front surface of your eye, the cornea, does about two-thirds of the eye's focusing, and the lens behind it does the rest and adjusts it (Wikipedia, "Cornea"). Cameras, glasses and microscopes all shape glass to control refraction. And when light inside glass meets the surface at too large an angle, as in the explorer, it cannot get out at all. That is the idea behind optical fibres, and the subject of the next lecture.

Key concepts in this lesson

All angles in reflection and refraction are measured from the normal, the line at right angles to the boundary.
In reflection, the angle of incidence equals the angle of reflection, and the speed, frequency and wavelength do not change.
At a boundary, the frequency of a wave stays the same, so its wavelength changes in proportion to its speed.
For any wave, $sin θ_{1} / sin θ_{2} = v_{1} / v_{2}$ , so a wave that slows down turns towards the normal.
The refractive index is $n = c / v$ , and for light the law becomes Snell's law, $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ .
The direction of turning depends on speed, not density, which is why sound turns away from the normal when it enters water.