Question

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

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

Given that,
Refractive index of prism, ${\mu }_p=\sqrt{2}$
Prism angle, $A={60}^{\circ }$


For minimum deviation,
$i=i_e$ and $r_1=r_2$
also, $r_1+r_2=A$ [for minimum deviation $r_1=r_2=r$]
$\Rightarrow 2r=A={60}^{\circ }$
$\therefore r={30}^{\circ }$
On applying Snell's law,
${\mu }_p=\frac{\sin i}{\sin r}$
${\mu }_p=\frac{\sin i}{\sin {30}^{\circ }}$
$\Rightarrow \sqrt{2}=\frac{\sin i}{\left(1/2\right)}$
$\sin i=\frac{1}{\sqrt{2}}$
$\therefore i={45}^{\circ }$

Why this question? Bottom line: For minimum deviation produced by prism alway use $i=i_e$ and $r_1=r_2$ as handy tool.