A ray of light is incident on an isosceles right-angled prism made of a material of critical angle $$42^\circ$$. Find the distance of point of emergence of the ray from the prism if $$AB = d$$ and $$AP = L$$
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The ray enters face $$AC$$ normally at $$P$$, so it goes straight on and meets face $$AB$$ at an angle of incidence of $$45^\circ$$. Since $$45^\circ > 42^\circ$$ it is totally internally reflected and travels parallel to $$AC$$, at a height $$L$$ above it, to face $$BC$$, where the angle of incidence is again $$45^\circ$$, so it is totally reflected once more and travels perpendicular to $$AC$$. It therefore leaves through $$AC$$ normally, at the point $$Q$$ that is the mirror image of $$P$$ about the perpendicular from $$B$$, so $$CQ = L$$. Since $$AC = \sqrt{2}\,d$$, the ray emerges at a distance $$AQ = \sqrt{2}\,d - L$$ from $$A$$ (that is, $$\sqrt{2}\,d - 2L$$ from $$P$$), travelling parallel to the incident ray but reversed.