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An equiangular glass prism of refractive index $1.6$ is kept fully immersed in water of refractive index $4 / 3$, for a certain ray of monochromatic light. What is the closest value for the angle of minimum deviation of the light ray in this setup? (Take, $\sin 37^{\circ}=0.6$ )
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The correct answer is:
$14^{\circ}$
For equiangular glass prism, $A=60^{\circ}$
Refractive index of prism, $\mu_{p}=1.6$
Refractive index of water, $\mu_{v}=\frac{4}{3}$
We know that,
$=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin A / 2}$
$\frac{\mu_{p}}{\mu_{w}}=\frac{\sin \left(\frac{60+\delta_{m}}{2}\right)}{\sin 60 / 2}$
$\Rightarrow \quad \frac{1.6}{4 / 3}=\frac{\sin \left(\frac{60+\delta_{m}}{2}\right)}{\sin 30^{\circ}}$
$\Rightarrow \sin \left(\frac{A+\delta_{m}}{2}\right)=\frac{1.6 \times 3}{4} \times \sin 30^{\circ}$
$\Rightarrow \sin \left(\frac{A+\delta_{m}}{2}\right)=0.6$
$\Rightarrow \quad \sin \frac{A+\delta_{m}}{2}=\sin 37^{\circ} \quad[\because \sin 37=0.6]$
$$
\begin{aligned}
\Rightarrow & & \frac{A+\delta_{m}}{2} &=37^{\circ} \\
\Rightarrow & & 60+\delta_{m} &=2 \times 37=74^{\circ} \\
\Rightarrow & \delta_{m} &=74-60=14^{\circ}
\end{aligned}
$$
Refractive index of prism, $\mu_{p}=1.6$
Refractive index of water, $\mu_{v}=\frac{4}{3}$
We know that,
$=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin A / 2}$
$\frac{\mu_{p}}{\mu_{w}}=\frac{\sin \left(\frac{60+\delta_{m}}{2}\right)}{\sin 60 / 2}$
$\Rightarrow \quad \frac{1.6}{4 / 3}=\frac{\sin \left(\frac{60+\delta_{m}}{2}\right)}{\sin 30^{\circ}}$
$\Rightarrow \sin \left(\frac{A+\delta_{m}}{2}\right)=\frac{1.6 \times 3}{4} \times \sin 30^{\circ}$
$\Rightarrow \sin \left(\frac{A+\delta_{m}}{2}\right)=0.6$
$\Rightarrow \quad \sin \frac{A+\delta_{m}}{2}=\sin 37^{\circ} \quad[\because \sin 37=0.6]$
$$
\begin{aligned}
\Rightarrow & & \frac{A+\delta_{m}}{2} &=37^{\circ} \\
\Rightarrow & & 60+\delta_{m} &=2 \times 37=74^{\circ} \\
\Rightarrow & \delta_{m} &=74-60=14^{\circ}
\end{aligned}
$$
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