JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2021
- Hard
- Numerical
A ray of light passing through a prism \((\mu=\sqrt{3})\) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is ..... . (in degrees)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(60\)
Step-by-step Solution
Detailed explanation
At minimum deviation \(r_{1}=r_{2}=\frac{A}{2}\)
Also given \({i}=2 {r}_{1}={A}\)
Now \(1 \sin i =\sqrt{3} \sin r_{1}\)
\(1 \sin A=\sqrt{3} \sin \frac{A}{2}\)
\(2 \sin \frac{A}{2} \cos \frac{A}{2}=\sqrt{3} \sin \frac{A}{2}\)
\(\cos \frac{A}{2}=\frac{\sqrt{3}}{2} \Rightarrow \frac{A}{2}=30^{\circ}\)
\(A=60^{\circ}\)
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