The big idea: Dip a straw into a glass of water and it looks snapped in half at the surface. That bending of light as it crosses from one material into another is refraction.
What to know
- A swimming pool looks shallower than it really is — that is refraction too.
- Light bends because it changes speed at the boundary.
- The normal is the dashed line drawn at 90° to the surface.
- Every angle is measured from the normal, not from the surface.
New word — refractive index: The refractive index n of a material tells you how much it slows light down.
A bigger n means slower light and more bending. We call a high-n material 'denser' (optically denser).
Vacuum/air ≈ 1.0 · water ≈ 1.3 · glass ≈ 1.5.
Entering the denser, slower glass the ray bends TOWARD the normal: θ₂ (30°) is smaller than θ₁ (50°). The dashed line is the normal — all angles are measured from it.
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Which way does it bend?: Into a denser (slower) medium → bends toward the normal (angle gets smaller).
Into a less-dense (faster) medium → bends away from the normal (angle gets bigger).
At a boundary: When a wave meets a boundary, part of it is reflected (angle of reflection = angle of incidence) and part is transmitted into the new medium, where it refracts if it arrives at an angle.
The frequency never changes at a boundary; the speed and wavelength do.
Refraction with wavefronts
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Snell's law links the two angles to the two refractive indices. The data booklet gives it as a set of ratios; the form you actually use rearranges to the line below.
- refractive index of medium 1 (no units)
- refractive index of medium 2 (no units)
- angle of incidence, measured from the normal (°)
- angle of refraction, measured from the normal (°)
Index also fixes the speed: The refractive index also tells you the light's speed in the material: n = c ÷ v.
So a high index means a slow speed. It follows from the booklet's form of Snell's law, n₁/n₂ = v₂/v₁, with n = 1 in a vacuum (where v = c).
- refractive index of the medium (no units)
- speed of light in a vacuum, 3.0 × 10⁸ m s⁻¹
- speed of light in the medium (m s⁻¹)
n = c ÷ v. Cover the quantity you want: c on top, n and v underneath. Two side by side → multiply; one above the other → divide.
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Any wave: Snell's law with speeds: For sound or water waves use the speeds: sin θ₁ / sin θ₂ = v₁ / v₂.
Total internal reflection happens going into the faster medium. Sound is faster in water (≈ 1500 m s⁻¹) than in air (≈ 340 m s⁻¹), so sound hitting water at more than about 13° from the normal cannot get in.
Through layers that get steadily faster, a wave bends further from the normal at each boundary until it is totally reflected: it can stay trapped near the slowest layer.
Light travels from air (n₁ = 1.0) into glass (n₂ = 1.5), hitting the surface at an angle of incidence of 50° to the normal. Find the angle of refraction in the glass.
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How this is tested — refraction questions split into two jobs:
Paper 1A
- A quick Snell's-law calculation — find an unknown angle or index.
- Sometimes through two layers in a row.
Paper 2
- Predict total internal reflection (TIR) — decide whether light escapes a denser medium or reflects back, using the critical angle.
The classic trap: Measuring an angle from the surface instead of from the normal, or forgetting TIR only happens going denser → less-dense.
Angle given to the surface, or to a wavefront?: If an angle is measured to the surface, the angle to the normal is 90° minus it.
The angle between a wavefront and the boundary equals the angle between the ray and the normal, so it goes straight into Snell's law.
Total internal reflection and the critical angle: Going from a denser medium to a less-dense one, the ray bends away from the normal. Make the angle big enough and the refracted ray would need to bend past 90° — it cannot, so all the light reflects back inside. That is total internal reflection (TIR).
The critical angle θc: The critical angle θc is the incidence angle at which the refraction angle is exactly 90°. Above θc you get TIR.
- critical angle (°) — the incidence angle giving a 90° refraction
- index of the denser medium the light starts in
- index of the less-dense medium it tries to enter
Going from the denser glass toward the less-dense air at a steep angle (60°, past the critical angle), no light escapes — it is all reflected back. This is total internal reflection.
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Light inside a glass block (n₁ = 1.5) meets the boundary with air (n₂ = 1.0). (a) Find the critical angle for this glass–air boundary. (b) A ray inside the same block hits the boundary at 55° to the normal — does it escape into the air, or is it totally internally reflected?
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