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, $\mu=\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$=\frac{\sin A}{\sin \left(\frac{A}{2}\right)}$
$=\frac{2 \sin \left(\frac{A}{2}\right) \cdot \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\frac{3}{2}=2 \cos \left(\frac{A}{2}\right)$
$\cos \left(\frac{A}{2}\right)=\frac{3}{4}$
$\frac{A}{2}=\cos ^{-1}\left(\frac{3}{4}\right)$
$A=2 \cos ^{-1}\left(\frac{3}{4}\right)$