The refractive index of an equilateral prism is $\sqrt 2$. The angle of emergence under minimum deviation position of prism, in degree, is ___________.
Answer (integer)
45
Solution
Refractive index
<br/><br/>
$$
\begin{aligned}
& \mu=\frac{\sin \left(\frac{A+\delta \sin }{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{2}=\sin \frac{\left(\frac{60+\delta \min }{2}\right)}{\sin 30^{\circ}} \\\\
& \Rightarrow \frac{1}{2}=\sin \left(\frac{60+\delta \min }{2}\right) \Rightarrow 45^{\circ}=\frac{60+\delta \min }{2} \\\\
& \Rightarrow \delta \min =30^{\circ} \\\\
& \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \\\\
& \text { Here, } e=i \\\\
& \text { So, } \delta \mathrm{min}=2 \mathrm{e}-\mathrm{A} \Rightarrow 2 \mathrm{e}=\delta \min +\mathrm{A} \\\\
& e=\frac{\delta \min +A}{2}=\frac{30^{\circ}+60^{\circ}}{2}=45^{\circ}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Optics · Topic: Refraction and Snell's Law
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