Question:
The angle of minimum deviation for an incident light ray on an equilateral prism is equal to its refracting angle. The refractive index of its material is
The angle of minimum deviation for an incident light ray on an equilateral prism is equal to its refracting angle. The refractive index of its material is
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When $\delta_m = A$ for an equilateral prism, $\mu = \sqrt{3}$.
Updated On: Apr 10, 2026
- $\frac{1}{\sqrt{2}}$
- $\sqrt{3}$
- $\frac{\sqrt{3}}{2}$
- $\frac{3}{2}$
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The Correct Option is B
Solution and Explanation
Step 1: Given Data
For an equilateral prism, the angle of prism $A = 60^{\circ}$. The problem states the angle of minimum deviation $\delta_m = A = 60^{\circ}$.
Step 2: Formula
The refractive index is given by $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Step 3: Calculation
$\mu = \frac{\sin(\frac{60^{\circ} + 60^{\circ}}{2})}{\sin(\frac{60^{\circ}}{2})} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
Final Answer: (b)
For an equilateral prism, the angle of prism $A = 60^{\circ}$. The problem states the angle of minimum deviation $\delta_m = A = 60^{\circ}$.
Step 2: Formula
The refractive index is given by $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Step 3: Calculation
$\mu = \frac{\sin(\frac{60^{\circ} + 60^{\circ}}{2})}{\sin(\frac{60^{\circ}}{2})} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
Final Answer: (b)
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