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 𝜃 such that the outgoing ray undergoes minimum deviation in prism \(P_2\). If the respective refractive indices of \(P_1\) and\( P_2\) are \(√ 3 /2\) and \(√3\), then \(\theta  = sin{−1}[\sqrt \frac{ 3}{ 2} sin ( \frac{\pi}{B} )],\) where the value of \(\beta\) is ______.

Answer 1

Correct Answer: 12

Refractive Index and Angle Analysis for the Prism 

For the second prism, the relationship between the refractive index \( n_2 \), the angle of refraction \( r_2 \), and the angle \( \theta \) is:

\[ n_2 \sin r_2 = \sin \theta, \quad r_2 = \frac{A}{2} \] where \( A \) is the angle of the prism.

Using the condition for minimum deviation:

\[ \sin \theta = n_2 \sin \frac{A}{2} \]

Given that \( \sin \theta = \frac{\sqrt{5}}{2} \), we have:

For the first prism:

\[ \sin \theta = \frac{\sqrt{5}}{2} \]

From the refractive index relationship:

\[ n_1 \sin i = n_2 \sin r_2 \] where \( i \) is the angle of incidence. Substituting for \( \sin r_2 \), we find:

 

\[ \sin i = n_1 \sin \theta = \frac{3}{2} \cdot \frac{\sqrt{5}}{2} \] Simplifying: \[ \sin i = \frac{3}{2} \cdot \sin \frac{\pi}{12} \]

Equating and solving, we determine:

\[ \theta = \sin^{-1} \left( \frac{3}{2} \cdot \sin \frac{\pi}{12} \right) \]

Final Answer:

• The value of \( \beta \) is: \( \beta = 12 \)