A glass slab of refractive index 1.5 is cut in the shape of a parallelepiped ABCD as shown in the figure. Surface BC is polished to form a perfect reflector. Light is incident at surface AB from inside at an angle /3. Which one of the following is the correct angle ABC ( ) such that the reflected light retraces its path?
A. /6
For the reflected light to retrace its path, it must strike the polished surface BC normally (perpendicularly). This means the angle between the reflected ray and surface BC must be 90^ . Assuming the given angle /3 (60^ ) is the angle the incident ray makes with the surface AB, by the law of reflection, the reflected ray will also make an angle of 60^ with the surface AB. Consider the triangle formed by the point of incidence on AB, the point of normal incidence on BC, and the vertex B. The sum of the internal angles in this triangle must be 180^ . The angle between the reflected ray and surface AB is 60^ . The angle between the reflected ray and surface BC is 90^ . Therefore, the angle (angle ABC) is: = 180^ - (90^ + 60^ ) = 30^ Converting 30^ to radians, we get = /6.
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