Question

The refractive index \( \mu \) of the material of a prism is given by: % Formula for Refractive Index \[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] where \( A \) is the apex angle of the prism and \( \delta_m \) is the angle of minimum deviation.

• A. \( \delta_m = 180^{\circ} - 4A \)
• B. \( \delta_m = 180^{\circ} - 3A \)
• C. \( \delta_m = 180^{\circ} - 2A \)
• D. \( \delta_m = 180^{\circ} - A \)

Hint:

The refractive index of a prism is given by:
\( \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \)
The minimum deviation angle \( \delta_m \) is found using \( \delta_m = 180^{\circ} - 2A \).

Answer 1

The Correct Option is C

Step 1: Relation Between \( A \), \( \delta_m \), and \( \mu \)
The refractive index \( \mu \) is defined by: \[ \mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)} \] This formula calculates the angle of minimum deviation \( \delta_m \).
Step 2: Expressing \( \delta_m \) in Terms of \( A \)
It is known that: \[ \cot \frac{A}{2} = \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}} = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \frac{A}{2}} \] Upon rearrangement: \[ \sin \left( \frac{180^{\circ}}{2} - \frac{A}{2} \right) = \sin \left( \frac{A + \delta_m}{2} \right) \] Using the identity \( \sin(\theta) = \sin(180^\circ - \theta) \), we equate: \[ \frac{180^{\circ}}{2} - \frac{A}{2} = \frac{A + \delta_m}{2} \]
Step 3: Solving for \( \delta_m \)
Simplifying the equation yields: \[ 180^{\circ} - 2A = \delta_m \] Therefore, the angle of minimum deviation is: \[ \delta_m = 180^{\circ} - 2A \]Final Answer: The angle of minimum deviation is \( 180^{\circ} - 2A \).