Question

The angle of a prism is ${60}^o$. When light is incident at an angle of ${60}^o$ on the prism, the angle of emergence is ${40}^o$. The angle of incidence i for which the light ray will deviate the least is such that.

• A. $i<{40}^o$
• B. ${40}^o<i<{50}^o$
• C. ${50}^o<i<{60}^o$
• D. $i>{60}^o$

Answer 1

Answer: (B)

${40}^o<i<{50}^o$

Observing the figure, we have angle of the prism given as $A={60}^o$

Incidence angle ${\theta }_1={60}^o$ and emergence angle ${\theta }_4={40}^o$

Let the first refracted angle be ${\theta }_2=r$ and refractive index of the prism be $\mu$

From figure we get ${\theta }_3=A-{\theta }_2=A-r$

Applying Snell's law at two refraction points, $\mu =\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin {\theta }_4}{\sin {\theta }_3}$

i.e., $\frac{\sin {60}^o}{\sin r}=\frac{\sin {40}^o}{\sin ({60}^o-r)}$

$\sin r\approx 0.5719$

$\therefore \mu =\frac{\sin {60}^o}{\sin r}\approx 1.5719$

Incidence angle for minimum deviation is given by $\sin \left(i_m\right)=\mu \sin \left(A/2\right)\approx 1.5719/2=0.7572$

$i_m\approx {49.2139}^o$