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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 _______ \(^\circ\).

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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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