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

- For a prism at minimum deviation, the refractive index is given by: \[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
- At minimum deviation condition:
- angle of incidence = angle of emergence
- hence the path inside the prism is symmetric
- Given: \[ D = 60^\circ \]
- Using prism relation: \[ A + D = 2i \quad \text{and} \quad A = 2r \]
- For minimum deviation: \[ i = r \Rightarrow A + D = 2A \]
- Substitute: \[ A + 60^\circ = 2A \]
- Solve: \[ A = 60^\circ \]
- Hence, the angle of prism is: 60°