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Optional unit · institution-dependent · Interactive lesson

Refraction and Snell's law

Every refraction question is really one question: how much slower is the light on the other side? Turn that speed ratio into an index, measure both angles from the normal, and a single equation hands you the new direction — forwards or backwards.

UPII-13.02 · Lesson 2 of 8 in this unit. Position is not a completion score.
Starting model: Refraction and Snell's lawA ray in air meets a flat boundary at 40° from the dashed normal and enters a medium of index 1.33, where it travels at 2.25 × 10⁸ m s⁻¹ and its 589 nm wavelength shrinks to 443 nm. Snell's law turns it to 28.9° from the normal, a deviation of 11.1°. The faint dashed ray below the boundary is what the wrong angle-ratio rule predicts.incident ray n₁ = 1.00 (air)θ₁ = 40° from the normalboundaryn₂ = 1.33θ₂ = 28.9°v₂ = 2.25 × 10⁸ m s⁻¹λ₂ = 443 nm (λ₀ = 589 nm)dashed: θ₁/n₂ = 30.1°
Start with a picture. Change the model in the lab below.

Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.

1

Build the model

A boundary changes the speed, and a ray that arrives obliquely pays for it by turning.

The key idea

Refraction is the change of direction a wave makes on crossing a boundary where its speed changes, and Snell's law fixes the new direction: n₁ sin θ₁ = n₂ sin θ₂, with both angles measured from the normal to the surface.

Example. A ray in air meeting water (n = 1.333) at 55° from the normal continues at sin θ₂ = sin 55°/1.333 = 0.6145, so θ₂ = 37.9° — bent towards the normal, because the light is slower in the water.

The complete physical model and assumptions

Light travels at c only in vacuum. Inside matter the wave drives the bound electrons, they re-radiate, and the superposition advances more slowly — at v = c/n, where the refractive index n is nothing more than that speed ratio. Now send a plane wave obliquely at a flat boundary. A wavefront is a line, and one end of it crosses into the slow medium before the other, so that end falls behind: the front pivots, and the ray, drawn perpendicular to the front, turns with it. Carrying the geometry through gives sin θ₁/sin θ₂ = v₁/v₂ = n₂/n₁, or n₁ sin θ₁ = n₂ sin θ₂ — Snell's law, with both angles measured from the normal and both rays lying in one plane with it. The equation is symmetric in its two labels, so the path is reversible: a ray sent back along the outgoing direction retraces its route exactly. What the law fixes is a direction and nothing else. It is silent on how much light reflects instead of transmitting, it treats n as one number when n in fact drifts with wavelength, and it speaks of rays, which describe the light well only while every aperture and feature the beam meets is far larger than λ.

1Light slows down, and n keeps the books

In vacuum every electromagnetic wave travels at c = 2.998 × 10⁸ m s⁻¹. Inside matter the oscillating field drives the bound electrons, they re-radiate, and the sum of the original wave and everything they emit advances more slowly. Call that speed v and define the refractive index as n = c/v. It is a pure number, it is at least 1 for ordinary transparent media, and it is the only property of the material that plain refraction cares about. Air is 1.0003, near enough to 1 for any bench work; water is 1.333, so light crosses a swimming pool at 2.25 × 10⁸ m s⁻¹; crown glass is about 1.52; diamond is 2.42, or 1.24 × 10⁸ m s⁻¹. Nothing here strains relativity: c is the ceiling on the speed of a signal, and n > 1 says the wave in the medium runs below that ceiling, not above it.

2The wavefront pivots: where Snell's law comes from

Take a plane wavefront — a line of crests — running at angle θ₁ to a flat boundary, so one end of it reaches the surface before the other does. Let the trailing end need a time Δt to arrive. In that Δt it covers v₁Δt in medium 1, while the end that crossed first has already run v₂Δt into medium 2. Both of those distances hang off the same stretch L of boundary between the two crossing points, so sin θ₁ = v₁Δt/L and sin θ₂ = v₂Δt/L. Divide the two, and Δt and L both vanish: sin θ₁/sin θ₂ = v₁/v₂ = n₂/n₁, which rearranges to n₁ sin θ₁ = n₂ sin θ₂. The front has pivoted because one end was held back, and the ray turns with it. Air into water at θ₁ = 40°: sin θ₂ = 0.6428/1.333 = 0.4822, so θ₂ = 28.8°. Fermat's principle reaches the identical equation by minimising the travel time instead of tracking wavefronts.

3Angles belong to the normal, and to one plane

Three conventions travel with the equation. Measure every angle from the normal — the line perpendicular to the surface at the point of incidence — never from the surface itself. That is not cosmetic: a ray described as 60° to a water surface is 30° to the normal, and sin 30° = 0.500 against sin 60° = 0.866, so confusing them scales the answer by 1.7. Second, the incident ray, the normal and the refracted ray all lie in one plane, the plane of incidence; refraction never throws a ray sideways out of it, because the boundary looks the same everywhere along itself. Third, normal incidence is the case that catches people out: θ₁ = 0 gives sin θ₂ = 0 and θ₂ = 0, so a ray entering a window face-on slows down and shortens its wavelength but does not bend at all. A speed change is necessary for bending; it is not sufficient.

4Which way, and how far: the bending is not proportional

Direction first. If medium 2 is the slower one — larger n — then sin θ₂ < sin θ₁ and the ray bends towards the normal; crossing the other way, into the faster medium, it bends away from it. The size of the bend is where intuition fails, because it is the sines that stand in proportion, not the angles. Take air into glass of n = 1.50. At θ₁ = 30° Snell gives θ₂ = 19.5°; double the incidence to θ₁ = 60° and θ₂ climbs only to 35.3°, well short of the 38.9° that doubling would demand, while the deviation θ₁ − θ₂ swells from 10.5° to 24.7°. Each extra degree of incidence buys less turning inside than the degree before it. Small angles are the exception worth keeping: for θ in radians sin θ ≈ θ, so n₁θ₁ ≈ n₂θ₂, and that paraxial form is the seed the refracting-surface and thin-lens equations later in this section grow from. The law also carries a ceiling: sin θ₁ ≤ 1 forces sin θ₂ ≤ n₁/n₂, so a ray arriving from air can never be more than 41.8° from the normal inside n = 1.50 glass, however grazing its approach.

5The law reads the same backwards

Swap the labels 1 and 2 in n₁ sin θ₁ = n₂ sin θ₂ and the same equation comes back. The path is therefore reversible: send a ray backwards along the refracted direction and it leaves along the original incident line, angle for angle. Two consequences. A ray crossing a slab with parallel faces is bent at entry and bent back by exactly as much at exit, because the second refraction is the first one run in reverse, so it emerges parallel to the direction it arrived on. What is not restored is position. The ray spent its time inside the slab on a tilted path, so it steps sideways by Δ = t sin(θ₁ − θ₂)/cos θ₂. A 5.0 mm window of n = 1.52 struck at 45° refracts to 27.7° inside and shifts the emerging ray by 1.7 mm — enough to matter when that slab sits in front of a detector.

6The frequency survives the crossing; the wavelength does not

Crests can neither pile up at the boundary nor disappear into it: whatever arrives each second must leave each second, so the frequency f is the same on both sides, fixed by the source and not by the material. Since v = fλ and v falls by the factor n, the wavelength has to fall with it: λ = λ₀/n. Sodium light of λ₀ = 589 nm runs at 442 nm inside water. Colour follows f rather than λ, which is why nothing underwater changes hue. One caveat closes this topic and opens a later one. Treating n as a single number is an approximation: in BK7 crown glass n is 1.5224 at 486 nm and 1.5143 at 656 nm, so blue is bent a little more than red by the same surface. That drift is dispersion, what a prism lives on, and this section's final topic takes it up alongside the point where the ray model itself gives out.

2

Key equations

Read the units, assumptions and limits before substituting.

Refractive index

n = c/v

One number stands in for the speed: water's n = 1.333 means v = 2.25 × 10⁸ m s⁻¹.

c = 2.998 × 10⁸ m s⁻¹ in vacuum, v the wave speed in the medium in m s⁻¹; n is a pure number

Snell's law

n₁ sin θ₁ = n₂ sin θ₂

Hand it the incoming direction and it returns the outgoing one.

θ₁ and θ₂ are measured from the normal; n₁ and n₂ are dimensionless indices

Snell's law written in speeds

sin θ₁ / sin θ₂ = v₁ / v₂

Two protractor readings give a speed ratio without mentioning c at all.

v₁ and v₂ are the wave speeds in the two media in m s⁻¹; the ratio equals n₂/n₁

Wavelength inside a medium

λ = λ₀/n

589 nm sodium light runs at 442 nm inside water, and is still yellow.

λ₀ the vacuum wavelength in m, λ the wavelength in the medium in m; f is unchanged

Apparent depth, near-normal view

d′ = d (n₂/n₁)

A 1.20 m pool viewed from the air above looks 0.900 m deep.

d the true depth in the medium of index n₁, d′ the apparent depth seen from the medium of index n₂, both in m

Lateral shift through a parallel slab

Δ = t sin(θ₁ − θ₂) / cos θ₂

The exit direction is restored, but the ray is offset sideways.

t the slab thickness in m, θ₂ the internal angle; Δ in m, measured across the ray
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

40 °
0 to 80 °; steps of 2.
1.33
1 to 2.42; steps of 0.01.

Try this. Hold n₂ at 1.52 and drag the incidence angle from 0 to 80°: the deviation grows from 0° to 39.6°, and the faint dashed ray, drawn at θ₁/n₂, drifts further from the true one as the angles leave the small-angle range.

Live results

REFRACTION ANGLE θ₂
28.9 °
DEVIATION θ₁ − θ₂
11.1 °
SPEED IN MEDIUM 2
2.25 × 10⁸ m s⁻¹
λ IN MEDIUM (λ₀ = 589 nm)
443 nm

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics II modelA ray in air meets a flat boundary at 40° from the dashed normal and enters a medium of index 1.33, where it travels at 2.25 × 10⁸ m s⁻¹ and its 589 nm wavelength shrinks to 443 nm. Snell's law turns it to 28.9° from the normal, a deviation of 11.1°. The faint dashed ray below the boundary is what the wrong angle-ratio rule predicts.incident ray n₁ = 1.00 (air)θ₁ = 40° from the normalboundaryn₂ = 1.33θ₂ = 28.9°v₂ = 2.25 × 10⁸ m s⁻¹λ₂ = 443 nm (λ₀ = 589 nm)dashed: θ₁/n₂ = 30.1°
A ray in air meets a flat boundary at 40° from the dashed normal and enters a medium of index 1.33, where it travels at 2.25 × 10⁸ m s⁻¹ and its 589 nm wavelength shrinks to 443 nm. Snell's law turns it to 28.9° from the normal, a deviation of 11.1°. The faint dashed ray below the boundary is what the wrong angle-ratio rule predicts.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Angle of incidence θ₁: 40 °; Index of medium 2, n₂: 1.33

Baseline 40; choose 0 to 80, in steps of 2.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The common wrong move is to divide the angles instead of their sines — to write θ₁/θ₂ = n₂/n₁, so that a ray entering glass from air comes out at θ₁/n. It survives because it is nearly right where most textbook diagrams are drawn: at 10° into n = 1.50 glass it predicts 6.67° against Snell's 6.65°. Push it and it breaks. At 70° it gives 46.7° where the true answer is 38.8°, and at 89° it gives 59.3° against a true 41.8°. Worse, it knows nothing of the limit: the angle it returns keeps growing with θ₁, whereas sin θ₁ ≤ 1 caps the refracted angle at asin(n₁/n₂), which is 41.8° for air into n = 1.50 glass however grazing the incoming ray. Take the sine first, every time; the proportional rule is only the small-angle approximation wearing a disguise.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

A helium-neon laser of vacuum wavelength 633 nm, travelling in air, meets the flat face of a crown-glass block (n = 1.52) at 38° from the normal. Find the angle of the refracted ray, the speed of the light inside the glass, and its wavelength there.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A 12.0 mm thick plate of flint glass (n = 1.58) is held horizontally with air above and below it. A narrow beam strikes the top face at 55.0° from the normal. Find the angle inside the glass, the angle at which the beam leaves the bottom face, and how far the emerging ray is displaced sideways from the line the incident ray would have followed.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

HardExample 3

A tank holds 42.0 cm of water (n = 1.333) with a 15.0 cm layer of oil (n = 1.47) floating on top. A ray in air strikes the oil surface at 58.0° from the normal. (a) Find its angle in the oil and then in the water. (b) Compare that with the angle it would have had if the oil were not there. (c) Viewed from straight above, how deep does the tank floor appear?

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A 589 nm laser beam travelling in air strikes a flat water surface (n = 1.333) at 55° from the normal. Which statement describes the beam inside the water?
7

Review & continue

Refraction is the change of direction a wave makes on crossing a boundary where its speed changes, and Snell's law fixes the new direction: n₁ sin θ₁ = n₂ sin θ₂, with both angles measured from the normal to the surface.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Optional supporting lessons

These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.