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A prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ}$. At what angle a ray must be incident on it so that it suffers a minimum deviation.
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Verified Answer
The correct answer is:
$45^{\circ}$
The relation for refractive index of prism is
$\mu=\frac{\sin i}{\sin r}$ $\ldots(1)$
The condition for minimum deviation is
$r=\frac{A}{2}=\frac{60^{\circ}}{2}=30^{\circ}$
Putting the given value of $\mu=\sqrt{2}$ and $r=30^{\circ}$ in eq (1), we get
$\begin{aligned} \sqrt{2} & =\frac{\sin i}{\sin 30^{\circ}} \\ \text { So, } \quad \sin i & =\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} \\ \sin i & =\sin 45^{\circ} \\ i & =45^{\circ}\end{aligned}$
$\mu=\frac{\sin i}{\sin r}$ $\ldots(1)$
The condition for minimum deviation is
$r=\frac{A}{2}=\frac{60^{\circ}}{2}=30^{\circ}$
Putting the given value of $\mu=\sqrt{2}$ and $r=30^{\circ}$ in eq (1), we get
$\begin{aligned} \sqrt{2} & =\frac{\sin i}{\sin 30^{\circ}} \\ \text { So, } \quad \sin i & =\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} \\ \sin i & =\sin 45^{\circ} \\ i & =45^{\circ}\end{aligned}$
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