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