IGCSE CP4a – Rectangular Glass Slab | Edexcel International GCSE Physics

Theory — Refraction

Light changes direction when it crosses the boundary between two media of different optical densities.

What is Refraction?

When light passes from one medium to another it changes speed, causing it to change direction. This is refraction.

Air → Glass: light slows down → bends towards the normal → θᵣ < θᵢ
Glass → Air: light speeds up → bends away from the normal → θᵣ > θᵢ
Ray along the normal (θᵢ = 0°): passes straight through with no bending

The Normal — Most Important Concept

The normal is a line drawn perpendicular to the surface at the point where the ray meets the boundary. For a horizontal glass surface, the normal is vertical.

All angles are measured from the normal — never from the surface. Measuring from the surface gives 90° minus the correct angle, which is wrong.

Snell's Law

n₁ sin θᵢ = n₂ sin θᵣ

n₁ = refractive index of first medium · θᵢ = angle of incidence from normal · n₂ = refractive index of second medium · θᵣ = angle of refraction from normal

For a parallel-sided glass block: the ray refracts when entering the top surface and refracts again when leaving the bottom surface. Because the surfaces are parallel, the exit angle equals the original angle of incidence — so the emergent ray is parallel to the incident ray, but laterally displaced.

Correct Protractor Use

Place the protractor centre exactly on the point where the ray meets the surface
Align the protractor baseline along the normal — for a horizontal surface this means the baseline is vertical
Measure the angle between the ray and the normal on each side
Never measure from the surface itself — this gives the wrong angle

Procedure

Tracing rays through a glass block placed flat on paper, measuring angles at the top surface.

Equipment

Rectangular glass block · Ray box with single slit · Plain white paper · Ruler · Sharp pencil · Protractor · Darkened room or blackout board

Place block flat and outline it

Place the glass block flat on the paper with its largest face down. Draw around it. Mark the midpoint of the top (long) face — this is where your ray will enter.

💡 The ray enters through the top surface. The normal at the top surface is vertical (perpendicular to the horizontal glass surface).

Direct the ray at the top surface

Aim the ray box so the ray hits the marked point on the top surface. Mark two dots on the incoming ray before it hits the block. The ray box should be above the block.

💡 Start with a small angle (about 10–15°) so both the incident and refracted rays are clearly visible.

Mark the emergent ray at the bottom

The refracted ray travels through the glass and exits from the bottom surface. Mark two dots on the ray that emerges below the block. Remove the block and join all the dots to draw the complete ray path.

Draw the normal and measure θᵢ and θᵣ

At the top entry point: draw the normal vertically (perpendicular to the horizontal top surface). Place the protractor centre on the entry point. Align the baseline of the protractor along the NORMAL (vertical). Measure θᵢ between the incident ray and the normal, and θᵣ between the refracted ray inside the block and the normal.

💡 The normal is vertical for a horizontal surface. Keep the protractor baseline vertical — if your baseline is horizontal (along the surface) you are doing it wrong and will get 90° minus the real angle.

Repeat for 6 different angles

Repeat for θᵢ = 10°, 20°, 30°, 40°, 50°, 60°. Record θᵢ, θᵣ, sin θᵢ and sin θᵣ each time. Plot sin θᵢ vs sin θᵣ in CP5 to find n.

      🔦 Drag the ray source (yellow dot above block) to change the angle of incidence. The protractor baseline is along the vertical normal — angles measured from it, not from the surface.
    

Material

Angle of Incidence

θᵢ from normal 30°

0°80°

Readings

θᵢ (angle of incidence)30.0°

θᵣ (angle of refraction)19.5°

Exit angle (= θᵢ)30.0°

Snell's Law

n (material)1.50

sin θᵢ0.5000

sin θᵣ0.3333

n = sinθᵢ / sinθᵣ1.500

n₁sinθᵢ = n₂sinθᵣ?✓

Display

Questions

Question 1

A ray hits the top surface of a glass block (n = 1.5) at 40° to the normal. Calculate the angle of refraction inside the glass. In which direction does the ray bend, and why?

Using Snell's Law: n₁ sin θᵢ = n₂ sin θᵣ → 1.00 × sin 40° = 1.50 × sin θᵣ → sin θᵣ = 0.6428/1.50 = 0.4285 → θᵣ = 25.4°. The ray bends towards the normal (from 40° to 25.4°) because it is entering a more optically dense medium (glass). Light slows down in glass, and slowing down causes bending towards the normal.

Question 2

Explain why the emergent ray leaving the bottom of the glass block is parallel to the original incident ray. Why is it displaced sideways?

When the ray hits the bottom surface (glass → air), Snell's Law applies in reverse: n₂ sin θᵣ = n₁ sin θ_exit. Since the top and bottom surfaces are parallel, the angle inside the glass at the bottom equals the refraction angle from the top. Applying Snell's Law gives θ_exit = θᵢ (the original angle). So the emergent ray makes the same angle with the vertical as the incident ray — they are parallel. The lateral displacement occurs because the ray travelled through the glass at a different angle (θᵣ < θᵢ), shifting the exit point sideways relative to where it would have been if the glass were not there.

Question 3

A student measures θᵢ = 50° and θᵣ = 30° for a glass block. Calculate the refractive index of the glass. The student forgot to draw the normal first and measured the angles from the glass surface instead. What angles would they have written down, and would their calculated n be correct?

n = sin θᵢ / sin θᵣ = sin 50° / sin 30° = 0.766/0.500 = 1.53. If angles are measured from the surface instead of the normal: θᵢ_wrong = 90° − 50° = 40°, θᵣ_wrong = 90° − 30° = 60°. Then n_wrong = sin 40° / sin 60° = 0.643/0.866 = 0.74. This gives n < 1, which is physically impossible for glass (n must be ≥ 1 for any real material). This is a clear sign the angles were measured from the surface, not the normal. Drawing the normal first and measuring from it always gives n > 1 for denser materials.