Question 246 of 295

The refractive index of prism is $\mu=\sqrt{3}$ and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _______.

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Answer: 60

For $\delta_{\min }$
$\begin{aligned}
& \mathrm{i}=\mathrm{e} \\
& \mathrm{r}_1=\mathrm{r}_2=\frac{\mathrm{A}}{2} \\
& \frac{\delta_{\min }}{\mathrm{A}}=1 \\
& \frac{2 \mathrm{i}-\mathrm{A}}{\mathrm{A}}=1 \\
& \\
& 2 \mathrm{i}=2 \mathrm{~A} \\
& \mathrm{i}=\mathrm{A}
\end{aligned}$
Snell's law
$\begin{aligned}
& 1 \times \sin i=\mu \sin r \\
& \sin i=\mu \sin \left(\frac{A}{2}\right) \\
& \sin A=\mu \sin \left(\frac{A}{2}\right) \\
& 2 \sin \frac{A}{2} \cos \frac{A}{2}=\sqrt{3} \sin \left(\frac{A}{2}\right) \\
& \cos \left(\frac{A}{2}\right)=\frac{\sqrt{3}}{2} \\
& \therefore \frac{A}{2}=30^{\circ} \\
& \therefore A=60^{\circ}
\end{aligned}$

JEE Main Physics practice question on Ray Optics (Prism and Dispersion of Light), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.