Question

A prism of refractive index $\sqrt{2}$ has refracting angle of ${60}^o$. At what angle a ray must be incident one it so that it suffers a minimum deviation?

• A. ${45}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${180}^o$

Answer 1

The correct option is A ${45}^o$

In the position of minimum deviation angle of incidence is equal to angle of emergence.
Let a ray of monochromatic light PQ be incident on face AB. PQRS is path of light ray.
where i is angle of incidence r angle of refraction, r angle of incidence and i angle of emergence.
In position of minimum deviation
$i^{\prime }=i,r^{\prime }=r,\delta =\delta m$
$2r=A$ or $r=\frac{A}{2}$
Given $A={60}^o$,
$r=\frac{{60}^o}{2}={30}^o$
Also from Snell's law
$\mu =\frac{\sin i}{\sin r}=\frac{\sin i}{\sin {30}^o}$
$\sqrt{2}=\frac{\sin i}{\sin {30}^o}$
$\Rightarrow \sin i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow i={45}^o$.