Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is

Question

Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is ______. [2026]

Smart Achievers · Accepted Answer

(3)

$For equilateral prism, A = 60 °$

$Minimum deviation δ_{m} = 2 i - A = \frac{A}{2}$

$μ_{p} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} = \frac{\sin \frac{3 A}{4}}{\sin \frac{A}{2}}$

$μ_{p} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \sqrt{2}$

Solution

Q.
Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is ______. [2026]

1 $1.5$

2 $\sqrt{3}$

3 $\sqrt{2}$

4 $1.65$

Ans.
(3)

$For equilateral prism, A = 60 °$

$Minimum deviation δ_{m} = 2 i - A = \frac{A}{2}$

$μ_{p} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} = \frac{\sin \frac{3 A}{4}}{\sin \frac{A}{2}}$

$μ_{p} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \sqrt{2}$

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Solution

Q. Angle of minimum deviation is equal to the half of the angle of prism in an equilateral prism. The refractive index of the prism is ______. [2026]

1 1.5

2 3

3 2

4 1.65

Ans. (3) For equilateral prism, A=60° Minimum deviation δm=2i-A=A2 μp=sin(A+δm2)sin(A2)=sin3A4sinA2 μp=1212=2