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A ray of light is incident at E at $$45^\circ$$ as it passes from air in to a medium across boundary AB. It then is incident at F on boundary CD and emerges in to air. Sum of the angles at E and F in the medium is $$60^\circ$$. Find the deviation (in degrees) of the ray at F. Refractive index of the medium with respect to air is $$\sqrt{2}$$.
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Correct Answer: 15
At E, $$\sin r_1 = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1}{2}$$, so $$r_1 = 30^\circ$$. The angles in the medium add up to $$60^\circ$$, so the angle of incidence at F is $$30^\circ$$. Then $$\sin e = \sqrt{2} \sin 30^\circ = \frac{1}{\sqrt{2}}$$, so $$e = 45^\circ$$, and the deviation at F is $$45^\circ - 30^\circ = 15^\circ$$.
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