Refraction - Physics for Pilot Assessments

Light bends when it crosses from one transparent material into another. That single sentence covers a straw that looks snapped at the waterline, a swimming pool that is deeper than it looks, the working of every lens and every pair of spectacles, and the rainbow. This lesson explains why the bending happens, establishes which way it goes — which is the most heavily tested fact in the module and is purely a matter of direction — and puts a number on how much light slows in a given material. It builds on two things you already have: that light travels more slowly inside a material than in a vacuum, from Lesson 1, and that angles in optics are measured from the normal, from Lesson 2.

What changes at a boundary, and what does not

When light arrives at the surface of a block of glass, three things happen at once. A little of it reflects, obeying the law of reflection exactly as a mirror would — this is why you can see faint reflections in a window. Some is absorbed, which is why a thick enough sheet of glass is not perfectly clear. And the rest crosses into the glass and carries on, at the lower speed the material imposes. It is that last portion this lesson is about, and the word for what happens to it is refraction: the change of direction that comes with the change of speed.

Be clear about which quantities change and which do not, because items exploit the difference. The speed changes: down on entering a denser material, back up on leaving it. The wavelength changes with it, in the same proportion. The frequency does not change at all — the wave on the far side of the boundary is driven by the wave on the near side, so it must oscillate the same number of times per second. And the colour does not change either, because your eye responds to frequency: light that looked red in air still looks red under water. A frequency that survives the crossing is why.

Why a change of speed turns the beam

The bending is not a separate rule bolted onto the slowing; it is a direct consequence of it, and the reason becomes obvious once you stop thinking about a thin ray and think about the wavefront instead. A wavefront is the line joining all the points of the wave that are in step with each other — the crest, stretched out sideways — and it lies at right angles to the direction of travel. Now let a wavefront arrive at a boundary at an angle rather than square on. One end of it reaches the glass first and immediately slows down, while the other end is still in air and still going at full speed. The fast end therefore gets ahead, the front pivots, and since the direction of travel stays perpendicular to the front, the beam has turned.

A column of soldiers marching in ranks across a field onto soft sand makes the same point mechanically. If a rank reaches the sand square on, everyone slows together and the column carries straight on. If it reaches the boundary at an angle, the soldier at one end hits the sand and shortens their stride while the one at the far end is still striding out on firm ground, so the rank swings round and the whole column changes direction. Nothing was steered; the turn came entirely from one side slowing before the other. Take away the angle and there is no turn at all — light arriving exactly along the normal slows down and keeps going in a straight line, which is a case items like to include as a check.

Which way it bends

Now the direction, which is the fact worth more marks than anything else in this lesson. Going into a slower material — air into glass, air into water — the ray bends towards the normal, so the angle inside is smaller than the angle outside. Coming out into a faster material — glass into air, water into air — it bends away from the normal, so the angle outside is larger. The marching analogy gives you a way to recover it rather than memorise it: the end that slows first falls behind, so the front swings towards the boundary's perpendicular on the way in, and the reverse on the way out.

Entering the glass the ray bends toward the normal; leaving it, it bends away by exactly the same amount, so the emerging ray is parallel to the original one. The dashed line is the path it would have followed had the glass not been there: the two surfaces cancel each other's turn and leave only a sideways displacement.

The figure makes a second point worth having, because it is a favourite of diagram-based items. A parallel-sided block bends the ray twice, in opposite senses and by equal amounts, so the ray comes out travelling in exactly the direction it went in. What it does not do is come out along the same line — it is shifted sideways, and the thicker the block or the steeper the angle, the bigger the shift. That is why looking through a thick window at an angle displaces the view slightly without tilting it. A prism, whose faces are not parallel, does not get that cancellation, which is why it deviates a beam for good; that is Lesson 7.

Refractive index: putting a number on it

How much a material slows light, and therefore how strongly it bends it, is captured by a single number: the refractive index. It is defined as the speed of light in a vacuum divided by the speed of light in the material, so it says directly how many times slower the light is going.

Refractive index

n = \frac{c}{v}

$n$: refractive index — a bare ratio, no units
$c$: the speed of light in a vacuum, 3 × 10⁸ metres per second (m/s)
$v$: the speed of light in the material, in metres per second (m/s)

Two consequences fall out of that definition before any arithmetic. Because it is a speed divided by a speed, the refractive index has no units at all — it is a bare ratio, like the coefficient of friction. And because nothing travels faster in a material than it does in a vacuum, v is always smaller than c, so n is always greater than 1. An option offering a refractive index of 0.8 is offering something impossible. A vacuum has n = 1 exactly, by definition; air is 1.0003, which everybody rounds to 1; water is about 1.33; ordinary glass about 1.5; diamond about 2.4. Higher index means slower light and stronger bending, and it is diamond's exceptionally high index that gives cut stones their glitter.

Example 2 — the same relationship the other way round

Water has a refractive index of about 1.33, which is four thirds. How fast does light travel in water? Here the index is given and the speed is missing, so the definition has to be rearranged.

Start from $n = \frac{c}{v}$ and multiply both sides by v: that gives n v = c.
Now divide both sides by n, which leaves v alone: $v = \frac{c}{n}$
Substitute, using the exact fraction to keep the arithmetic mental: $v = \frac{3 \times 10^8}{4/3}$
Dividing by four thirds is the same as multiplying by three quarters, so v = (3 × 10⁸) × 3/4 = 2.25 × 10⁸ metres per second.
Check the direction of the answer: it must come out smaller than c, and 2.25 × 10⁸ is smaller than 3 × 10⁸. Had you multiplied by n instead of dividing, you would have got 4 × 10⁸ — faster than light, which is the signal that the operation went the wrong way round.

How much it bends: the exact relationship

The refractive index does more than record a speed; it also fixes the angles. The relationship, known as Snell's law, is that the refractive index of the first material multiplied by the sine of the angle in the first material equals the refractive index of the second multiplied by the sine of the angle in the second. The sine of an angle is a ratio between 0 and 1 that grows as the angle grows from zero to ninety degrees; you will not be asked to compute one, but you need to know it increases with the angle, because that is what makes the law readable.

Snell's law

n_1 \sin\theta_1 = n_2 \sin\theta_2

$n_1$: refractive index of the material the light starts in — a bare ratio
$\theta_1$: angle in the first material, measured from the normal, in degrees (°)
$n_2$: refractive index of the material the light enters — a bare ratio
$\theta_2$: angle in the second material, measured from the normal, in degrees (°)

Read it as a see-saw and the direction of bending stops being something to remember. The two products have to stay equal, so if the index goes up as the light crosses the boundary, the sine of the angle must come down to compensate — and a smaller sine means a smaller angle, which is the ray moving towards the normal. If the index goes down, the sine and therefore the angle must go up, which is the ray moving away. Light going from air into glass has n going from 1 to 1.5, so the angle must shrink; light going the other way has n falling, so the angle must grow.

Example 3 — reading a numeric case off Snell's law

A ray inside glass of index 1.5 reaches the surface at 30° to the normal and passes out into air. What angle does it make with the normal outside?

Label the materials in the order the light meets them: material 1 is the glass, so n₁ = 1.5 and θ₁ = 30°; material 2 is the air, so n₂ = 1.
Substitute into n₁ sin θ₁ = n₂ sin θ₂: $1.5 \times \sin 30^\circ = 1 \times \sin\theta_2$
The sine of 30° is exactly 0.5, one of the few worth knowing, so the left-hand side is 1.5 × 0.5 = 0.75.
So sin θ₂ = 0.75, and the angle whose sine is 0.75 is about 49°.
Check the sense rather than the trigonometry: the light left a denser material for a less dense one, so the angle had to grow, and 49° is larger than 30°. That is the check worth doing, because items here test the direction far more often than the number.

Why a pencil in water looks broken

Stand a pencil in a glass of water and the part below the surface appears to be at a different angle from the part above, so the pencil looks bent or snapped at the waterline. Nothing has happened to the pencil. Light leaving the submerged part travels up through the water, meets the surface, and crosses into air — from a denser medium to a less dense one, so it bends away from the normal. Your eye receives rays that are steeper than they were when they left the pencil, and it does what it always does: assumes light travelled in a straight line and traces them back along their final direction. That traced-back position is displaced from where the pencil really is, so the submerged part appears shifted, and the visible discontinuity sits exactly at the surface where the bending happens.

The same effect explains why water is always shallower than it looks from above. Rays leaving the bottom of a pool bend away from the normal as they break the surface, so they reach your eye as if they had come from a point higher up, and the bottom appears raised. A pool a metre and a half deep looks about a metre and a bit deep, and the shift is larger the more steeply you look across the surface rather than straight down. It is also why a coin at the bottom of an opaque cup can come into view when you pour water in — light from the coin that would have missed the rim is bent over it.

For the common case of looking straight down into water, the shrink has a clean value: the apparent depth is the real depth divided by the refractive index. The rule comes from tracing the near-vertical rays your eye actually uses when you look straight down, and it is worth having because it turns the qualitative "shallower than it looks" into a number an item can ask for.

Apparent depth, viewed from directly above

d_{app} = \frac{d_{real}}{n}

$d_{app}$: apparent depth — how deep the water seems to be, in metres (m)
$d_{real}$: real depth of the water, in metres (m)
$n$: refractive index of the water or other material — a bare ratio, no units

One honest caveat, because it is the difference between the model and reality: the refractive index of a material is not perfectly constant, but varies slightly with the wavelength of the light passing through it. For all the arithmetic in this lesson that variation is small enough to ignore. It is not ignorable when you want to know why a prism produces colours, and Lesson 7 comes back to it as the actual cause of dispersion.

Why it matters in aviation

Bending at a boundary is what makes a lens possible, and it is the mechanism inside a precision approach path indicator. Each light unit beside the runway contains optics that split its output, so the part of the beam leaving above the intended approach slope emerges white and the part leaving below it emerges red. Nothing changes at the aircraft: a pilot who is high intercepts the white part of the beam and one who is low intercepts the red, so the colour shown is a direct readout of position relative to the slope, produced entirely by refraction inside a fixed piece of glass.

Refraction also happens in the air itself, with no single surface to cross. Air density falls with height and changes with temperature, so light crossing the atmosphere is bent gradually rather than abruptly. That raises the apparent position of objects near the horizon slightly above where straight-line geometry alone would put them, and it is the cause of the shimmering that distorts the far end of a runway on a hot day: light passing through pockets of hot, less dense air near the surface is bent by continually changing amounts, so the image wavers rather than holding still.

What the test asks

The dominant distractor is the direction of bending, offered exactly backwards: a ray entering water or glass shown bending away from the normal. Fix it with the mechanism rather than the picture — the side that slows first falls behind, so the front pivots towards the normal on the way in. The second recurring trap is claiming that the light's frequency or colour changes as it enters water; the speed and wavelength change, the frequency and the colour do not. Expect an item on the pencil in water whose wrong option blames a distortion of the glass or the water's magnifying power rather than refraction at the surface. And in the numeric items, watch for a refractive index quoted as less than 1, or for $\frac{c}{v}$ inverted so the answer comes out faster than light.