Question

A ray of light is incident on a prism $ABC$ of refractive index $\sqrt{3}$ as shown in figure.
(a) Find the angle of incidence for which the deviation of light ray by the prism $ABC$ is minimum.
(b) By what angle the second prism must be rotated, so that the final ray suffer net minimum deviation.

• A. ${30}^o$, ${60}^o$
• B. ${60}^o$, ${30}^o$
• C. ${60}^o$, ${60}^o$
• D. None of these

Answer 1

The correct option is C ${60}^o$, ${60}^o$

Minimum deviation

$i_1=i_2=i$ and $r_1=r_2=r$

and $r_1+r_2=A$

$r=A/2={30}^{\circ }$

Using Snell's law,

$\frac{sin{i}_1}{sin{r}_1}=\mu$

$sin{i}_1=\sqrt{3}sin{30}^{\circ }$

$sin{i}_1=\frac{\sqrt{3}}{2}$

$i_1={60}^{\circ }$

When the prism is rotated anticlockwise by ${60}^{\circ }$ the combined prism will become a slab as shown in the figure 2. In this case the net deviation by the prism will be zero as the deviation by the two prism will be equal and opposite.