IGCSE CP4b – Semi-circular Slab | Edexcel International GCSE Physics

Theory — TIR and Critical Angle

The semi-circular slab is the perfect tool for investigating total internal reflection because the curved surface eliminates any refraction on entry.

Why Use a Semi-circular Slab?

When a ray enters the curved face of a semi-circular block aimed at the centre, it always hits the curved surface along the radius — perpendicular to the surface. This means the angle of incidence at the curved face is always 0°, so no refraction occurs on entry. The ray travels straight to the flat face.

This is the key advantage: you can set any angle of incidence at the flat face precisely, without the curved surface interfering. The flat face is where all the interesting physics happens.

Total Internal Reflection (TIR)

At the flat face, the ray travels from glass (denser) to air (less dense). Snell's Law gives:

n sin θ = 1.00 × sin θ_exit

As θ increases, sin θ_exit increases. When sin θ_exit would exceed 1.00, no refracted ray is possible — all light reflects back into the glass. This is Total Internal Reflection (TIR).

Critical Angle θ_c

The critical angle is the angle of incidence at which the refracted ray would travel along the boundary (θ_exit = 90°). Above θ_c, TIR occurs.

n sin θ_c = sin 90° = 1 sin θ_c = 1/n → θ_c = arcsin(1/n)

For glass (n=1.50): θ_c = arcsin(1/1.50) = arcsin(0.667) = 41.8°

Verifying sin θ_c = 1/n with a Graph

For different materials (different n), measure the critical angle θ_c. Then plot sin θ_c (y-axis) against 1/n (x-axis). From sin θ_c = 1/n:

sin θ_c = (1) × (1/n)

This is y = mx with gradient = 1 and passes through the origin. A straight line through origin with gradient = 1 confirms the relationship sin θ_c = 1/n.

Applications of TIR

Optical fibres — light signals travel along glass fibres by TIR. Used in internet broadband cables and medical endoscopes
Prism binoculars — right-angle prisms use TIR to fold the light path and give an upright image
Diamond cutting — diamonds are cut at angles that maximise TIR, making them sparkle (n=2.42 gives θ_c=24.4° — very small, so TIR occurs easily)

Procedure

Finding the critical angle for a semi-circular glass block and verifying sin θ_c = 1/n.

Equipment

Semi-circular glass block · Ray box · Plain white paper · Protractor · Ruler · Sharp pencil · Darkened room

1

Set up the semi-circular block

Place the flat face of the block horizontally on paper. Draw around the block. Mark the centre of the flat face — this is where all rays will hit. Draw the normal at this point (vertical line perpendicular to the flat face).

💡 The centre point of the flat face is critical — all rays must pass through this exact point. If the ray misses, it will refract at the curved surface and your results will be wrong.

2

Direct ray through curved face at centre

Aim the ray box so the ray enters the curved face and passes through the centre of the flat face. At small angles, you will see both a refracted ray exiting the flat face and a weak reflected ray inside the block.

💡 Check that the ray enters perpendicular to the curved surface — it should hit the glass without bending. If it bends on entry, it is not aimed at the centre.

3

Slowly increase the angle

Rotate the ray box gradually, increasing the angle of incidence at the flat face. Watch the refracted ray: it bends further from the normal (toward the flat face surface). At the critical angle, the refracted ray travels along the flat face.

4

Find the critical angle

Continue increasing the angle past the critical angle — the refracted ray disappears completely and only the internal reflected ray remains. The critical angle is the angle of incidence at which the refracted ray just disappears. Mark the ray positions carefully and measure θ_c with the protractor from the normal.

💡 It can help to approach the critical angle from both sides — increase past it, then decrease back to it — and take the average of the two readings.

5

Repeat with different materials and plot the graph

Different semi-circular blocks (glass, perspex, water-filled semicircle) give different critical angles. For each: measure θ_c, calculate sin θ_c and 1/n. Plot sin θ_c vs 1/n — the gradient should equal 1, confirming sin θ_c = 1/n.

      🔴 Ray enters curved face with no refraction (aimed at centre). Adjust angle at the flat face. Watch for TIR above the critical angle. Record θ_c for each material.
    

Material

Angle of Incidence (at flat face)

θ from normal 30°

0°85°

Readings

Angle θ (flat face)30°

Critical angle θ_c—

sin θ_c—

1/n—

StatusRefraction

✓ Refraction occurring

Data Table

Critical angle measured for each material. Plot sin θ_c vs 1/n — gradient should equal 1.

Material	n	1/n	θ_c measured / °	sin θ_c	Expected θ_c / °
No readings yet. Use the Simulation tab — find the critical angle for each material and record it.

Graph — sin θ_c vs 1/n

If sin θ_c = 1/n then plotting sin θ_c against 1/n gives a straight line through the origin with gradient = 1.

Gradient of graph

— Expected: exactly 1.000

Results

Gradient—

R²—

% error in gradient—

Interpretation

Record at least 3 materials to plot the graph.

Questions

Question 1

Explain why using a semi-circular glass block is better than a rectangular block for investigating total internal reflection. Why does the ray not refract when it enters the curved face?

With a rectangular block, the ray refracts at both the entry face and the exit face — this makes it difficult to control exactly what angle the ray hits the flat face at. With the semi-circular block, the ray is aimed at the centre point of the flat face through the curved surface. Because the ray travels along the radius toward the centre, it hits the curved surface perpendicularly (the angle of incidence at the curved surface is 0°). Since sin 0° = 0, Snell's Law gives sin θᵣ = 0, so there is no refraction. The ray travels straight through to the flat face, where all the interesting physics occurs. This gives complete control over the angle of incidence at the flat face.

Question 2

The critical angle for a glass block is found to be 42°. Calculate the refractive index of the glass. Then calculate what the critical angle would be for diamond (n = 2.42).

From sin θ_c = 1/n: n = 1/sin θ_c = 1/sin 42° = 1/0.6691 = 1.49. This is the refractive index of the glass. For diamond: sin θ_c = 1/n = 1/2.42 = 0.4132. θ_c = arcsin(0.4132) = 24.4°. The very small critical angle means that light inside a diamond undergoes TIR for almost any angle — this is why diamonds sparkle so brilliantly. Light that enters a diamond bounces around inside many times before exiting, producing the characteristic brilliance.

Question 3

A student plots sin θ_c (y-axis) against 1/n (x-axis) for five different materials and gets a straight line through the origin with gradient 0.97. What does this gradient tell us? Is the result good? Suggest one reason why the gradient might not be exactly 1.00.

The gradient represents the value of sin θ_c × n. Since the relationship is sin θ_c = 1/n, rearranging gives n × sin θ_c = 1, so the gradient should be exactly 1.00. A gradient of 0.97 means there is a 3% error — this is a good result for a school experiment (within ±5% is generally acceptable). Reason for gradient ≠ 1.00: The critical angle is difficult to measure precisely because the refracted ray does not disappear suddenly — it dims gradually as the angle approaches θ_c. If the student records the angle slightly before the ray fully disappears, θ_c will be measured as slightly too small, making sin θ_c too small, and the gradient less than 1. Repeating the measurement and approaching from both sides (increasing then decreasing the angle) and averaging gives a more accurate result.