Light travels from air into a block. The recorded pairs (angle of incidence i in degrees, angle of refraction r in degrees) are (0, 0), (15, 10), (30, 19), (45, 28), (60, 35). (a) Describe the angle pattern. [2] (b) For each non-zero pair calculate n = sin i / sin r and find a representative value. [5] (c) Use that value to estimate the critical angle for light going from the block to air. [3] (d) Evaluate whether the evidence supports one constant refractive index and propose two improvements. [4]

Question

Accepted Answer

The refracted angle increases more slowly than the incident angle. The non-zero n values are about 1.50, 1.54, 1.51 and 1.51, giving n ≈ 1.51 and a critical angle sin⁻¹(1/1.51) ≈ 41.5°. (a) Refraction is towards the normal because r < i for every non-zero reading. Both angles increase, but equal angle increments do not produce equal refracted-angle increments. (b) The ratios sin i/sin r are approximately 1.49, 1.54, 1.51 and 1.51 for 15°, 30°, 45° and 60°. The 0° pair cannot be used because it gives 0/0. (b) A representative mean is about 1.51. The small scatter is plausible for degree-scale angle readings and supports, but does not prove, a constant index for this wavelength and material. (c) At the block–air critical angle, sin c = n_air/n_block ≈ 1/1.51. Therefore c ≈ sin⁻¹(0.662) = 41.5°. (d) Repeat independently, use a narrow laser line, draw a thin normal and measure from the normal rather than the surface. Plot sin i against sin r; a straight line through the origin tests the model more directly. (d) Wavelength and block homogeneity are uncontrolled in the claim of one universal value, so repeat with stated colours and several entry positions before generalising.

Test whether a transparent block has one refractive index

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Incident and refracted angles

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Incident and refracted angles — supplied values
Angle of incidence / °	Angle of refraction / °
0	0
15	10
30	19
45	28
60	35