A monochromatic ray travels inside a transparent block of refractive index 1.52 toward a block–air surface. Air has refractive index 1.00. Angles are measured from the normal. (a) Find the critical angle. (b) Find the refracted angle for incidence 31°. (c) Explain what happens at incidence 50°. Neglect absorption.

Question

Accepted Answer

Critical angle 41.14°; refracted angle 51.52°. Incidence 50° gives total internal reflection. At the critical angle, the transmitted ray would travel along the boundary, so its angle to the normal is 90°. Snell's law therefore gives n_block sin θ_c = n_air. Hence θ_c = sin⁻¹(1/1.52) = 41.14°. This construction is possible because light is travelling from a higher to a lower refractive index. For incidence 31°, Snell's law gives sin θ_air = 1.52 sin 31°, so θ_air = 51.52°. The ray bends away from the normal. At 50° the incident angle exceeds the calculated critical angle. A propagating refracted ray is impossible in the ray model; all incident energy is reflected when absorption is neglected. The computed refracted angle exceeds the incident angle but is below 90°, consistent with an incidence below the critical angle. Do not interchange angles measured from the surface with angles measured from the normal.

Critical angle and a refracted ray · set 6

What a complete response should do

How did this one feel?