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A ray of light passing from air in to a medium is incident at $$53^\circ$$. Angle of refraction in the medium is $$37^\circ$$. Find the sine of angle of refraction if the angle of incidence is changed to $$37^\circ$$. (Use $$\sin 37^\circ = \frac{3}{5}$$ and $$\sin 53^\circ = \frac{4}{5}$$)
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Correct Answer: 0.45
The refractive index is $$\mu = \frac{\sin 53^\circ}{\sin 37^\circ} = \frac{4/5}{3/5} = \frac{4}{3}$$. For incidence at $$37^\circ$$, $$\sin r = \frac{\sin 37^\circ}{\mu} = \frac{3}{5} \times \frac{3}{4} = \frac{9}{20} = 0.45$$.
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