For a prism kept in air it is found that for an angle of incidence 60°, the angle of prism 'A', angle of deviation '$\delta$' and angle of emergence 'e' become equal. Then the refractive index of the prism is
Answer & explanation
Correct answer: option 1
Given i = 60° and A = $\delta$ = e
$\delta=i+e-A \Rightarrow \delta=i$ (∵ e = A)
$\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}}$ Here angle of deviation $(\delta)$ is minimum (∵ i = e)
$\mu=\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \left(60^{\circ} / 2\right)}=1.732$