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JEE Advanced · Physics · 24. Ray Optics

Two equilateral-triangular prisms \(\mathrm{P}_1\) and \(\mathrm{P}_2\) are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism \(P_1\) at an angle of incidence \(\theta\) such that the outgoing ray undergoes minimum deviation in prism \(P_2\). If the respective refractive indices of \(\mathrm{P}_1\) and \(\mathrm{P}_2\) are \(\sqrt{\frac{3}{2}}\) and \(\sqrt{3}\), then \(\theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{\beta}\right)\right]\), where the value of \(\beta\) is _______ .

  1. A 12
  2. B 24
  3. C 15
  4. D 10
Verified Solution

Answer & Solution

Correct Answer

(A) 12

Step-by-step Solution

Detailed explanation


At surface BC
\(\begin{aligned} & \sqrt{\frac{3}{2}} \sin \mathrm{r}_2=\sqrt{3} \sin 30 \\ & \sqrt{\frac{3}{2}} \sin \mathrm{r}_2=\frac{\sqrt{3}}{2} \\ & \operatorname{sinr}_2=\frac{1}{\sqrt{2}} \\ & \mathrm{r}_2=45^{\circ} \\ & \mathrm{r}_1=60^{\circ}-45^{\circ}=15^{\circ}\end{aligned}\)
At surface \(A B\)
\(\begin{array}{ll} 1 \sin \theta=\sqrt{\frac{3}{2}} \sin 15^{\circ} \\ \theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \frac{\pi}{12}\right] \\ \beta=12\end{array}\)
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