Question

A prism of refractive index $\mu$ and angle of prism $A$ is placed in the position of minimum angle of deviation. If minimum angle of deviation is also $A$, then $A$ in terms of refractive index is

• A. ${\sin }^{-1}\left(\sqrt{\frac{\mu -1}{2}}\right)$
• B. $2{\cos }^{-1}\left(\frac{\mu }{2}\right)$
• C. ${\sin }^{-1}\left(\frac{\mu }{2}\right)$
• D. ${\cos }^{-1}\left(\frac{\mu }{2}\right)$

Answer 1

The correct option is B $2{\cos }^{-1}\left(\frac{\mu }{2}\right)$

The formula relating the refractive index $\mu$, Angle of prism $A$ and angle of minimum deviation ${\delta }_{min}$ is

$\mu =\frac{\sin \left(\frac{A+{\delta }_{min}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Given that, ${\delta }_{min}=A$

$\Rightarrow \mu =\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow \mu =\frac{\sin A}{\sin \frac{A}{2}}=\frac{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow \mu =2\cos \frac{A}{2}$

$\Rightarrow A=2{\cos }^{-1}\left(\frac{\mu }{2}\right)$

Hence, option $\left(A\right)$ is correct.