JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2024
- Hard
- Numerical
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 _______ \(^\circ\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(60\)
Step-by-step Solution
Detailed explanation
\(\text { For } \delta_{\min }\)
\(\mathrm{i}=\mathrm{e}\)
\(\mathrm{r}_1=\mathrm{r}_2=\frac{\mathrm{A}}{2}\)
\(\frac{\delta_{\min }}{\mathrm{A}}=1\)
\(\frac{2 \mathrm{i}-\mathrm{A}}{\mathrm{A}}=1\)
\(2 \mathrm{i}=2 \mathrm{~A}\)
\(\mathrm{i}=\mathrm{A}\)
Snell's law
\(1 \times \sin i=\mu \sin r\)
\(\sin i=\mu \sin \left(\frac{A}{2}\right)\)
\(\sin A=\mu \sin \left(\frac{A}{2}\right)\)
\(2 \sin \frac{A}{2} \cos \frac{A}{2}=\sqrt{3} \sin \left(\frac{A}{2}\right)\)
\(\cos \left(\frac{A}{2}\right)=\frac{\sqrt{3}}{2}\)
\(\therefore \frac{A}{2}=30^{\circ}\)
\(\therefore A=60^{\circ}\)
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