A monochromatic ray travels inside a transparent block of refractive index 1.44 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 27°. (c) Explain what happens at incidence 50°. Neglect absorption.

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Critical angle 43.98°; refracted angle 40.82°. 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(1/1.44) = 43.98°. This construction is possible because light is travelling from a higher to a lower refractive index. For incidence 27°, Snell's law gives sin θ_air = 1.44 sin 27°, so θ_air = 40.82°. 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 2

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