JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 24)
The refractive index of an equilateral prism is $$\sqrt 2 $$. The angle of emergence under minimum deviation position of prism, in degree, is ___________.
Answer
45
Explanation
Refractive index
$$ \begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta \sin }{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{2}=\sin \frac{\left(\frac{60+\delta \min }{2}\right)}{\sin 30^{\circ}} \\\\ & \Rightarrow \frac{1}{2}=\sin \left(\frac{60+\delta \min }{2}\right) \Rightarrow 45^{\circ}=\frac{60+\delta \min }{2} \\\\ & \Rightarrow \delta \min =30^{\circ} \\\\ & \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \\\\ & \text { Here, } e=i \\\\ & \text { So, } \delta \mathrm{min}=2 \mathrm{e}-\mathrm{A} \Rightarrow 2 \mathrm{e}=\delta \min +\mathrm{A} \\\\ & e=\frac{\delta \min +A}{2}=\frac{30^{\circ}+60^{\circ}}{2}=45^{\circ} \end{aligned} $$
$$ \begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta \sin }{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{2}=\sin \frac{\left(\frac{60+\delta \min }{2}\right)}{\sin 30^{\circ}} \\\\ & \Rightarrow \frac{1}{2}=\sin \left(\frac{60+\delta \min }{2}\right) \Rightarrow 45^{\circ}=\frac{60+\delta \min }{2} \\\\ & \Rightarrow \delta \min =30^{\circ} \\\\ & \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \\\\ & \text { Here, } e=i \\\\ & \text { So, } \delta \mathrm{min}=2 \mathrm{e}-\mathrm{A} \Rightarrow 2 \mathrm{e}=\delta \min +\mathrm{A} \\\\ & e=\frac{\delta \min +A}{2}=\frac{30^{\circ}+60^{\circ}}{2}=45^{\circ} \end{aligned} $$
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