Question

A prism having angle \(\theta\) and refractive index of the material of prism ‘n’ are related by \[ \theta = 2\sin^{-1} \left( \frac{1}{\sqrt{n^2 + 1}} \right) \] - If the angle of minimum deviation is \(60^\circ\), then the angle of the prism is

• A. \(54^\circ\)
• B. \(60^\circ\)
• C. \(30^\circ\)
• D. \(45^\circ\)

Hint:

For prism-related problems, remember the key formula: \[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin(A/2)} \] This helps in determining the relationship between the angle of the prism and the minimum deviation.

Answer 1

The Correct Option is B

Step 1: Understanding the relation For a prism, the relation between the refractive index (\(n\)), the prism angle (\(\theta\)), and the angle of minimum deviation (\(D\)) is given by: \[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin(A/2)} \] where: 
- \( A \) is the angle of the prism. 
- \( D \) is the angle of minimum deviation. 

Step 2: Substituting given values Given \( D = 60^\circ \), the equation becomes: \[ n = \frac{\sin\left(\frac{A + 60^\circ}{2}\right)}{\sin(A/2)} \] We also have the formula: \[ A = 2 \sin^{-1} \left( \frac{1}{\sqrt{n^2 + 1}} \right) \] By solving for \( A \), we obtain: \[ A = 60^\circ \] Thus, the correct answer is \( \mathbf{(2)} \ 60^\circ \).