Question:
If refractive index of a material of equilateral prism is $\sqrt{3}$, then angle of minimum deviation of the prism is}
If refractive index of a material of equilateral prism is $\sqrt{3}$, then angle of minimum deviation of the prism is}
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For an equilateral prism, $\mu = \frac{\sin\left(30^{\circ} + \frac{\delta_m}{2}\right)}{\sin 30^{\circ}}$.
Updated On: May 3, 2026
- $30^{\circ}$
- $45^{\circ}$
- $60^{\circ}$
- $75^{\circ}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For minimum deviation, $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}}$.
Step 2: Detailed Explanation:
Here $A = 60^{\circ}$, $\mu = \sqrt{3}$. So $\sqrt{3} = \frac{\sin\left(30^{\circ} + \frac{\delta_m}{2}\right)}{\sin 30^{\circ}} = \frac{\sin\left(30^{\circ} + \frac{\delta_m}{2}\right)}{1/2}$. Thus $\sin\left(30^{\circ} + \frac{\delta_m}{2}\right) = \frac{\sqrt{3}}{2}$, so $30^{\circ} + \frac{\delta_m}{2} = 60^{\circ}$, giving $\delta_m = 60^{\circ}$.
Step 3: Final Answer:
The angle of minimum deviation is $60^{\circ}$.
For minimum deviation, $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\frac{A}{2}}$.
Step 2: Detailed Explanation:
Here $A = 60^{\circ}$, $\mu = \sqrt{3}$. So $\sqrt{3} = \frac{\sin\left(30^{\circ} + \frac{\delta_m}{2}\right)}{\sin 30^{\circ}} = \frac{\sin\left(30^{\circ} + \frac{\delta_m}{2}\right)}{1/2}$. Thus $\sin\left(30^{\circ} + \frac{\delta_m}{2}\right) = \frac{\sqrt{3}}{2}$, so $30^{\circ} + \frac{\delta_m}{2} = 60^{\circ}$, giving $\delta_m = 60^{\circ}$.
Step 3: Final Answer:
The angle of minimum deviation is $60^{\circ}$.
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