Question

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

• A. $si{n}^{-1}\left(\frac{\mu }{2}\right)$
• B. $si{n}^{-1}\sqrt{\frac{\mu -1}{2}}$
• C. $2co{s}^{-1}\left(\frac{\mu }{2}\right)$
• D. $2co{s}^{-1}\left(\frac{\mu }{8}\right)$

Answer 1

Answer: (C)

$2co{s}^{-1}\left(\frac{\mu }{2}\right)$

We can use the formula: $\mu =\frac{sin\left(\frac{A+D}{2}\right)}{sin\left(\frac{A}{2}\right)}$ where $D$ is angle of minimum deviation and $A$ is angle of prism.

According to given condition $A=D$

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

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

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

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

$\therefore$ $A=2{\cos }^{-1}\left(\frac{\mu }{2}\right)$ which is the required solution.