A glass slab of refractive index 1.5 is cut in the shape of a parallelepiped ABCD. Surface BC is polished to form a perfect reflector. Light is incident at surface AB from inside at an angle π/3.

Which one of the following is the correct angle ABC (θ) such that the reflected light retraces its path?A.π/6B.π/3C.π/4D.5π/12

Question

A glass slab of refractive index 1.5 is cut in the shape of a parallelepiped ABCD. Surface BC is polished to form a perfect reflector. Light is incident at surface AB from inside at an angle π/3. Which one of the following is the correct angle ABC (θ) such that the reflected light retraces its path?

Accepted Answer

π/3. The ray hits face AB from inside at 60° from the normal. This is more than the critical angle for glass of refractive index 1.5, which is about 41.8°. So the ray undergoes total internal reflection at AB, not simple refraction. By the law of reflection, the ray leaves AB at 60° from the normal too. That means it leaves AB at 30° from the surface AB itself. For the light to retrace its whole path, the ray must hit mirror BC exactly along its normal, so it bounces straight back the way it came. Now look at the small triangle formed by vertex B, the point where the ray leaves AB, and the point where it hits BC. The angle at B is angle ABC, call it θ. The angle at the point on AB is 30°. The angle at the point on BC is 90°, since the ray meets BC perpendicularly. Angles in a triangle add up to 180°. So θ + 30° + 90° = 180°, which gives θ = 60° = π/3. So option (b) is correct.

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