Question:
Angle of minimum deviation is equal to half of the angle of prism in an equilateral prism. The refractive index of the prism is _____.
Angle of minimum deviation is equal to half of the angle of prism in an equilateral prism. The refractive index of the prism is _____.
Updated On: Apr 12, 2026
- \(1.5\)
- \( \sqrt{3} \)
- \( \sqrt{2} \)
- \(1.65\)
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The Correct Option is A
Solution and Explanation
Concept:
Refractive index of a prism at minimum deviation:
\[
\mu=\frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}
\]
where
\(A\) = angle of prism
\(\delta_m\) = minimum deviation.
Step 1: {Identify the prism angle.}
For an equilateral prism:
\[
A=60^\circ
\]
Given
\[
\delta_m=\frac{A}{2}=30^\circ
\]
Step 2: {Substitute in formula.}
\[
\mu=\frac{\sin\left(\frac{60^\circ+30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}
\]
\[
=\frac{\sin45^\circ}{\sin30^\circ}
\]
\[
=\frac{\frac{\sqrt2}{2}}{\frac12}
\]
\[
=\sqrt2
\]
Thus the refractive index is approximately
\[
\mu\approx1.5
\]
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