Question

For a prism kept in air, it is found that for an angle of incidence ${60}^{\circ }$, the angle of refraction ${}^{\prime }{A}^{\prime }$, angle of deviation ${}^{\prime }{\delta }^{\prime }$, and angle of emergence ${}^{\prime }{e}^{\prime }$ become equal. The minimum angle of incidence of a ray that will be transmitted through the prism is

• A. $1.73$
• B. $1.15$
• C. $1.5$
• D. $1.33$

Answer 1

Answer: (A)

$1.73$

Given $i={60}^{\circ },A=\delta =e$
$\delta =i+e-A\Rightarrow \delta =1$
$(\because e=A)$ and $(\delta =i=e$
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
Here, angle of deviation is minimum ($\because i=e)$
$\mu =\frac{\sin \left(\frac{{60}^{\circ }+{60}^{\circ }}{2}\right)}{\sin \left({60}^{\circ }/2\right)}=\sqrt{3}$.