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A prism having refractive index $1.414$ and refracting angle $30^{\circ}$ has one of the refracting surfaces silvered. A beam of light incident on the other refracting surface will retrace its path, if the angle of incidence is
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The correct answer is:
$45^{\circ}$
At second surface, there is no refraction so
$$
\therefore \quad r_{1}=A=30^{\circ}
$$
From Snell's law
$\begin{aligned} n=& \frac{\sin i_{1}}{\sin r_{1}} \\ n=& \frac{\sin i_{1}}{\sin 30^{\circ}} \\ \sin i_{1} &=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} \\ \sin i_{1} &=\frac{1}{\sqrt{2}} \\ i_{1} &=45^{\circ} \end{aligned}$
$$
\therefore \quad r_{1}=A=30^{\circ}
$$
From Snell's law
$\begin{aligned} n=& \frac{\sin i_{1}}{\sin r_{1}} \\ n=& \frac{\sin i_{1}}{\sin 30^{\circ}} \\ \sin i_{1} &=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} \\ \sin i_{1} &=\frac{1}{\sqrt{2}} \\ i_{1} &=45^{\circ} \end{aligned}$
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