Find the minimum angle of deviation for a prism with angle A = 60° and $\mu$ = 1.5.
Answer & explanation
Correct answer: option 1
$ \mu = \frac{sin(\frac{A+\delta_m}{2})}{sin\frac{A}{2}} = \frac{sin(\frac{60^o+\delta_m}{2})} {sin30^o} $
$ sin^{-1}0.75 = \frac{60^o + \delta_m}{2}$
$\Rightarrow \delta_m = 2sin^{-1}0.75 - 60^o = 37^o$