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Physics Question on Refraction of Light

The least angle of deviation for a glass prism is equal to its refracting angle. The refractive index of glass is 1.51.5 . Then the angle of prism is

A

2 cos⁡−1(34)2\,\cos^{-1}\left( \frac{3}{4} \right)

B

sin⁡−1(34)\sin^{-1}\left( \frac{3}{4} \right)

C

2 sin⁡−1(32)2\,\sin^{-1}\left( \frac{3}{2} \right)

D

cos⁡−1(32)\cos^{-1}\left( \frac{3}{2} \right)

Answer

2 cos⁡−1(34)2\,\cos^{-1}\left( \frac{3}{4} \right)

Explanation

Solution

μ=sin⁡(A+δm)2sin⁡A2\mu=\frac{\sin \frac{\left(A+\delta_{m}\right)}{2}}{\sin \frac{A}{2}}
=sin⁡(A+A)/2sin⁡(A/2)=\frac{\sin (A+A) / 2}{\sin (A / 2)}
=2cos⁡A/2=2 \cos A / 2
or A=2cos⁡−1(μ2)A=2 \cos ^{-1}\left(\frac{\mu}{2}\right)
=2cos⁡−1(1.52)=2 \cos ^{-1}\left(\frac{1.5}{2}\right)
=2cos⁡−1(34)=2 \cos ^{-1}\left(\frac{3}{4}\right)