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A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be $40^{\circ}$. What is the refractive index of the material of the prism? The refracting angle of the prism is $60^{\circ}$. If the prism is placed in water (refractive index $1.33$ ), predict the new angle of minimum deviation of a parallel beam of light.
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Verified Answer
The angle of minimum deviation, $\delta_{\mathrm{m}}=40^{\circ}, \mathrm{A}=60^{\circ}, \mu=$ ?
$$
\begin{aligned}
\mu_{\mathrm{ga}} &=\frac{\sin \frac{\left(\mathrm{A}+\delta_{\mathrm{m}}\right)}{2}}{\sin \mathrm{A} / 2}=\frac{\sin \frac{\left(60^{\circ}+4\right.}{2}}{\sin \frac{60^{\circ}}{2}} \\
\Rightarrow \mu_{\mathrm{ga}}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}}=\frac{0.766}{0.500}=1.53
\end{aligned}
$$
When the prism is placed in water, $\mu_{\mathrm{wa}}=1.33, \mu_{\mathrm{ga}}=1.53$
$$
\begin{aligned}
&\therefore \mu_{\mathrm{gw}}=\frac{\mu_{\mathrm{ga}}}{\mu_{\mathrm{wa}}}=\frac{1.53}{1.33}=1.50 \\
&\mu_{\mathrm{gw}}=\frac{\sin \frac{\left(\mathrm{A}+\mathrm{D}_{\mathrm{m}}\right)}{2}}{\sin \frac{\mathrm{A}}{2}} \text { (where } \mathrm{D}_{\mathrm{m}} \text { is the angle of minimum }
\end{aligned}
$$
$1.150=\frac{\sin \frac{\left(60^{\circ}+D_m\right)}{2}}{\sin \frac{60^{\circ}}{2}}$
$\therefore \sin \left(\frac{60^{\circ}+D_m}{2}\right)=1.150 \times \sin 30^{\circ}=1.150 \times \frac{1}{2}=0.575$.
or, $\sin \left(\frac{60^{\circ}+D_m}{2}\right) \simeq \sin 35^{\circ}$
$\therefore \frac{60^{\circ}+D_m}{2}=35^{\circ} \Rightarrow 60^{\circ}+D_m=70^{\circ}$
$\therefore D_m=70^{\circ}-60^{\circ}=10^{\circ}$
$$
\begin{aligned}
\mu_{\mathrm{ga}} &=\frac{\sin \frac{\left(\mathrm{A}+\delta_{\mathrm{m}}\right)}{2}}{\sin \mathrm{A} / 2}=\frac{\sin \frac{\left(60^{\circ}+4\right.}{2}}{\sin \frac{60^{\circ}}{2}} \\
\Rightarrow \mu_{\mathrm{ga}}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}}=\frac{0.766}{0.500}=1.53
\end{aligned}
$$
When the prism is placed in water, $\mu_{\mathrm{wa}}=1.33, \mu_{\mathrm{ga}}=1.53$
$$
\begin{aligned}
&\therefore \mu_{\mathrm{gw}}=\frac{\mu_{\mathrm{ga}}}{\mu_{\mathrm{wa}}}=\frac{1.53}{1.33}=1.50 \\
&\mu_{\mathrm{gw}}=\frac{\sin \frac{\left(\mathrm{A}+\mathrm{D}_{\mathrm{m}}\right)}{2}}{\sin \frac{\mathrm{A}}{2}} \text { (where } \mathrm{D}_{\mathrm{m}} \text { is the angle of minimum }
\end{aligned}
$$
$1.150=\frac{\sin \frac{\left(60^{\circ}+D_m\right)}{2}}{\sin \frac{60^{\circ}}{2}}$
$\therefore \sin \left(\frac{60^{\circ}+D_m}{2}\right)=1.150 \times \sin 30^{\circ}=1.150 \times \frac{1}{2}=0.575$.
or, $\sin \left(\frac{60^{\circ}+D_m}{2}\right) \simeq \sin 35^{\circ}$
$\therefore \frac{60^{\circ}+D_m}{2}=35^{\circ} \Rightarrow 60^{\circ}+D_m=70^{\circ}$
$\therefore D_m=70^{\circ}-60^{\circ}=10^{\circ}$
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