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(C) According to Snell's law,$\sin r = \frac{\sin i}{\mu} = \frac{\sin 60^{\circ}}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$.Therefore,th

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$\frac{40}{\sqrt{3}}$

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(C) According to Snell's law,$\sin r = \frac{\sin i}{\mu} = \frac{\sin 60^{\circ}}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$.Therefore,the angle of refraction $r = 30^{\circ}$.The lateral shift $d$ produced by a glass slab of thickness $b$ is given by the formula:$d = b \cdot \frac{\sin(i - r)}{\cos r}$Substituting the given values:$d = 0.04 \cdot \frac{\sin(60^{\circ} - 30^{\circ})}{\cos 30^{\circ}}$$d = 0.04 \cdot \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = 0.04 \cdot 	an 30^{\circ}$$d = 0.04 \cdot \frac{1}{\sqrt{3}} 	ext{ m}$To convert the result into $	ext{mm}$,we multiply by $1000$:$d = \frac{0.04 	imes 1000}{\sqrt{3}} 	ext{ mm} = \frac{40}{\sqrt{3}} 	ext{ mm}$.

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