Solveeit Logo

Question

Physics Question on Refraction of Light

A prism is made up of material of refractive index 3\sqrt{3}. The angle of prism is AA. If the angle of minimum deviation is equal to the angle of the prism, then the value of AA is

A

3030^{\circ}

B

4545^{\circ}

C

6060^{\circ}

D

7575^{\circ}

Answer

6060^{\circ}

Explanation

Solution

Given M=3M =\sqrt{3} and δm=A\delta_m = A Now, μ=sin(A+δm2)sin(A2)\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} =sin(A+A2)sin(A2)=sinAsinA2=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin A}{\sin \frac{A}{2}} =2sin(A2)cos(A2)sin(A2)=\frac{2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)} =2cos(A2)=2 \cos \left(\frac{A}{2}\right) Now. 2cos(42)=μ=32 \cos \left(\frac{4}{2}\right)=\mu=\sqrt{3} or cos(A2)32\cos \left(\frac{A}{2}\right) \Rightarrow \frac{\sqrt{3}}{2} =A2=30=\frac{A}{2}=30^{\circ} or A=60A=60^{\circ}