Question

The angle of minimum deviation for a glass prism is equal to its refracting angle. The refractive index of glass is $1.5$. Then the angle of prism is

• A. $ 2{{\cos }^{-1}}(3/4) $
• B. $ {{\sin }^{-1}}(3/4) $
• C. $ 2{{\sin }^{-1}}(3/2) $
• D. $ {{\cos }^{-1}}(3/2) $

Answer 1

Answer: (A)

According to question, the angle of minimum deviation $\delta_{m}=$ Refracting angle $(A)$
Refractive index, $\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)} $
$=\frac{\sin A}{\sin \left(\frac{A}{2}\right)} $
$=\frac{2 \sin \left(\frac{A}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)} $
$\frac{3}{2}=2 \cos \left(\frac{A}{2}\right) $
$\cos \left(\frac{A}{2}\right)=\frac{3}{4}$
$\frac{A}{2}=\cos ^{-1}\left(\frac{3}{4}\right) $
$A=2 \cos ^{-1}\left(\frac{3}{4}\right)$