How light bends at boundaries, refractive index, critical angle and total internal reflection
AQA A-Level Physics 3
📐Define refractive index as n = c/v
📏Apply Snell's law: n₁ sin θ₁ = n₂ sin θ₂
🔁Derive and use sin C = 1/n for the critical angle
💡Explain total internal reflection and its conditions
🌐Describe applications: optical fibres and prisms
🔬Carry out the required practical to measure refractive index
Refractive Index
Refractive index (n): A measure of how much a medium slows light compared to a vacuum. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium.
n = c / v
c = speed of light in vacuum = 3.00 × 10⁸ m s⁻¹
v = speed of light in the medium (m s⁻¹)
n has no units and is always ≥ 1
Common values: air ≈ 1.000 (effectively 1), water ≈ 1.33, glass ≈ 1.5, diamond ≈ 2.42.
A higher refractive index means light travels more slowly in that medium. The frequency of light does not change when it crosses a boundary — only the speed and wavelength change. Since v = fλ, if v decreases, λ decreases proportionally.
Snell's Law
Refraction: The change in direction of a wave when it crosses a boundary between two media, caused by the change in wave speed.
When light travels from one medium to another, it bends. If it enters a denser medium (higher n), it bends towards the normal; if it enters a less dense medium (lower n), it bends away from the normal.
Snell's Law:
n₁ sin θ₁ = n₂ sin θ₂
n₁ = refractive index of incident medium
θ₁ = angle of incidence (measured from normal)
n₂ = refractive index of refracting medium
θ₂ = angle of refraction (measured from normal)
All angles are measured from the normal (the line perpendicular to the boundary at the point of incidence), NOT from the surface. If n₁ < n₂ (entering denser medium): θ₂ < θ₁ (bends towards normal). If n₁ > n₂ (entering less dense medium): θ₂ > θ₁ (bends away from normal).
A common mistake is to measure angles from the surface rather than the normal. Always draw the normal first.
Critical Angle and Total Internal Reflection
Critical angle (C): The angle of incidence in the denser medium at which the refracted ray travels along the boundary (angle of refraction = 90°). Above this angle, total internal reflection occurs.
Total internal reflection (TIR) occurs when:
Light travels from a denser medium to a less dense medium (n₁ > n₂)
The angle of incidence exceeds the critical angle (θ₁ > C)
Applying Snell's law at the critical angle (θ₂ = 90°, n₂ = 1 for air):
n sin C = 1 × sin 90° = 1
sin C = 1/n
More generally (for any two media):
sin C = n₂/n₁
At angles below C: refraction occurs (light exits, though some is reflected). At angle = C: refracted ray goes along the boundary. At angles above C: total internal reflection — all light reflects back into the denser medium (no refraction at all).
Applications
Optical Fibres: Glass fibre with a core (higher n) surrounded by cladding (lower n). Light enters at one end and undergoes total internal reflection at every internal boundary, travelling along the fibre with minimal loss. Used in communications (signal) and endoscopes (image transmission).
Cladding is needed so that the core-cladding boundary provides TIR. Without cladding, fibres touching each other would leak light. The cladding must have a lower refractive index than the core.
Prisms: A 45-90-45° glass prism (n ≈ 1.5, C ≈ 42°) can deflect light by exactly 90° or 180° using TIR. Light enters perpendicular to one face (no refraction) and hits the hypotenuse at 45° > C → TIR. This is more efficient than a mirror. Used in periscopes, binoculars and retroreflectors.
A ray of light travels from air (n = 1.00) into glass (n = 1.52) with an angle of incidence of 35°. Calculate the angle of refraction.
1Apply Snell's law: n₁ sin θ₁ = n₂ sin θ₂
21.00 × sin 35° = 1.52 × sin θ₂
3sin θ₂ = sin 35° / 1.52 = 0.5736 / 1.52 = 0.3773
4θ₂ = arcsin(0.3773) = 22.1°
Angle of refraction = 22.1° (bends towards the normal, as expected entering a denser medium)
Calculate the critical angle for a glass–air boundary where the refractive index of glass is 1.47.
1sin C = 1/n = 1/1.47 = 0.6803
2C = arcsin(0.6803) = 42.8°
Critical angle = 42.8°. Any light hitting this glass–air boundary at an angle greater than 42.8° will undergo TIR.
The speed of light in a transparent material is 1.97 × 10⁸ m s⁻¹. Calculate the refractive index and the critical angle for a boundary with air.
1n = c/v = 3.00 × 10⁸ / 1.97 × 10⁸ = 1.523
2sin C = 1/n = 1/1.523 = 0.6566
3C = arcsin(0.6566) = 41.0°
n = 1.52; Critical angle = 41.0°
Light travels from water (n = 1.33) into a glass block (n = 1.60) at 40°. Find the refracted angle. Does TIR occur?
1n₁ sin θ₁ = n₂ sin θ₂ → 1.33 × sin 40° = 1.60 × sin θ₂
3TIR cannot occur: light goes from less dense (n=1.33) to more dense (n=1.60). TIR only occurs going from dense to less dense.
θ₂ = 32.3°. No TIR — light is entering a denser medium.
1. The refractive index of a medium is 1.6. What is the speed of light in this medium?
v = c/n = 3.00 × 10⁸ / 1.6 = 1.875 × 10⁸ m s⁻¹.
2. Light hits a glass–air boundary from inside the glass (n = 1.5) at 50°. The critical angle is 41.8°. What happens?
50° > 41.8° (critical angle), and light is going from a denser medium to air — both conditions for TIR are met.
3. A ray in air hits a water surface (n = 1.33) at 60° to the normal. Calculate the angle of refraction.
sin θ₂ = (1.00 × sin 60°)/1.33 = 0.8660/1.33 = 0.651 → θ₂ = arcsin(0.651) = 40.6°. Light bends towards the normal on entering water.
4. Which property of light changes when it enters a denser medium?
Frequency is set by the source and does not change. Since v = fλ and f is constant, a decrease in speed v causes a proportional decrease in wavelength λ.
5. Calculate the critical angle (to 1 d.p.) for a diamond–air boundary. Diamond has n = 2.42.
1. An optical fibre has a core with n = 1.55 and cladding with n = 1.42. Calculate the critical angle at the core–cladding boundary. A ray enters the fibre along the axis at 0°. Will TIR occur at the core–cladding boundary?
sin C = n_cladding / n_core = 1.42/1.55 = 0.9161. C = arcsin(0.9161) = 66.4°. A ray travelling along the axis hits the boundary at 90° to the surface, which is 0° to the normal — well below the critical angle, so TIR does NOT occur. For TIR, rays must hit the boundary at angles greater than 66.4° to the normal (i.e. shallow grazing angles relative to the boundary surface). Rays that enter the fibre nearly parallel to the axis (small entry angles) hit the core–cladding boundary at large angles to the normal — but not necessarily greater than 66.4°. Rays that enter at larger angles to the axis hit the boundary at larger angles to the normal and may exceed C.
2. A glass block (n = 1.50) is submerged in water (n = 1.33). (a) Calculate the critical angle at the glass–water boundary. (b) Compare this with the glass–air critical angle and explain why optical fibres use cladding with a refractive index just below the core rather than air.
(a) sin C = n_water/n_glass = 1.33/1.50 = 0.8867. C = arcsin(0.8867) = 62.2°. (b) Glass–air critical angle: sin C = 1/1.50 = 0.667. C = 41.8°. The glass–water boundary has a larger critical angle (62.2° vs 41.8°). This means light needs to hit the boundary at a greater angle to the normal to achieve TIR at a glass–water boundary — so fewer rays are totally internally reflected. Optical fibres use cladding just below the core n (rather than air) to: (1) protect the fibre surface from scratches that would scatter light, (2) keep adjacent fibres from touching and leaking light between them, and (3) still achieve TIR at the core–cladding boundary (even though the critical angle is larger, carefully designed entry angles ensure TIR throughout).
3. A ray of light undergoes refraction at an air–glass (n = 1.50) interface. The refracted ray is at 20° to the normal inside the glass. Calculate the angle of incidence in air. Then calculate what would happen if this ray subsequently hit a glass–air boundary at the same angle of 20°.
For air–glass: n_air sin θ₁ = n_glass sin θ₂ → 1.00 × sin θ₁ = 1.50 × sin 20° → sin θ₁ = 1.50 × 0.342 = 0.513 → θ₁ = 30.9°. Critical angle for glass–air: sin C = 1/1.50 = 0.667 → C = 41.8°. The ray hits the glass–air boundary at 20°, which is less than 41.8° — so refraction occurs and the ray exits the glass at θ = arcsin(1.50 × sin 20°) = arcsin(0.513) = 30.9° (same as the original incident angle, as expected by reversibility of light).
Required Practical Investigating Refraction — Measuring Refractive Index
Objective: Determine the refractive index of glass using a rectangular glass block and a ray box.
Equipment
Rectangular glass block (known n ≈ 1.5)
Ray box with single-slit accessory and power supply
Plain white paper and pencil
Protractor and ruler
Pins (optional alternative to ray box)
Method
Place the glass block on white paper. Draw around its outline. Mark the normal at the mid-point of one face.
Direct a ray from the ray box at the surface at a chosen angle of incidence θ₁ (e.g. 10°, 20°, 30°, 40°, 50°, 60°).
Mark the incident ray and the emergent ray on the other side. Remove the block and join the entry and exit points with a straight line — this is the ray inside the block.
Measure the angle of refraction θ₂ inside the block using the normal drawn at the entry face.
Record θ₁ and θ₂ in a table. Calculate sin θ₁ and sin θ₂ for each pair.
Plot a graph of sin θ₁ (y-axis) vs sin θ₂ (x-axis). The gradient = n (refractive index of glass).
Analysis
θ₁ / °
θ₂ / °
sin θ₁
sin θ₂
10
6.6
0.174
0.115
30
19.5
0.500
0.333
50
30.7
0.766
0.511
60
35.2
0.866
0.577
Gradient of sin θ₁ vs sin θ₂ = 0.866/0.577 ≈ 1.50 = n.