The big idea: Refraction is the bending of light (or any wave) as it crosses from one material into another.
It bends because the light 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.
[Diagram: phys-ray] - Available in full study mode
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).
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. This one is given too.
- 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)
[Diagram: phys-formula-triangle] - Available in full study mode
Worked example — refraction into glass
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.
Solution
- Start with the given Snell's law:
- Put in the numbers (n₁ = 1.0, θ₁ = 50°, n₂ = 1.5):
- Rearrange to make sin θ₂ the subject:
- Take the inverse sine — keep it as an angle:
Final answer
θ₂ = 31° — smaller than 50°, so the ray bent toward the normal, as expected for entering denser 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 / Paper 1A: predict total internal reflection (TIR) — decide whether light escapes a denser medium or reflects back, using the critical angle.
Classic trap: measuring an angle from the surface instead of from the normal, or forgetting TIR only happens going denser → less dense.
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 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
[Diagram: phys-ray] - Available in full study mode
IB-style question — (a) critical angle
Light inside a glass block (n₁ = 1.5) meets the boundary with air (n₂ = 1.0). Find the critical angle for this glass–air boundary.
Solution
- Start with the critical-angle equation (Snell's law with θ₂ = 90°):
- Put in the numbers (less-dense air on top, denser glass below):
- Take the inverse sine:
Final answer
θc = 42°. Any ray hitting this boundary at more than 42° from the normal is totally internally reflected.
IB-style question — (b) does it escape?
A ray inside the same glass block hits the glass–air boundary at 55° to the normal. Does the ray escape into the air, or is it totally internally reflected?
Solution
- Compare the angle of incidence with the critical angle from part (a):
- The incidence angle is bigger than the critical angle:
- Above the critical angle, no light escapes — it all reflects:
Final answer
It is totally internally reflected — 55° is past the 42° critical angle, so none of the light escapes into the air.