Question

Two equilateral-triangular prisms \(P_1\) and \(P_2\) 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 \(P_1\) and \(P_2\) are \(\sqrt{3}/\sqrt{2})\) and \(\sqrt{3}\), then \(\theta = \sin^{-1}\left(\sqrt{3}/\sqrt{2}) \sin\left(\frac{\pi}{\beta}\right)\right)\), where the value of \(\beta\) is \_\_\_\_.
\includegraphics[width=0.5\linewidth]{p11.png}

Hint:

For minimum deviation in prisms, use Snell's law and geometry to relate angles and refractive indices.

Answer 1

For the second prism: \[ n_2 \sin r_2 = \sin\theta, \quad r_2 = \frac{A}{2}. \] Using minimum deviation: \[ \sin\theta = n_2 \sin\left(\frac{A}{2}\right), \quad \sin\theta = \sqrt{5} \cdot \frac{1}{2}. \] For the first prism: \[ n_1 \sin i = n_2 \sin r_2. \] \[ \sin i = \frac{\sqrt{3}}{2} \cdot \sqrt{5} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} \cdot \sin\left(\frac{\pi}{\beta}\right). \] Equating: \[ \theta = \sin^{-1}\left(\frac{\sqrt{3}}{2} \sin\left(\frac{\pi}{12}\right)\right), \quad \beta = 12. \]