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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)

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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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