Question

A part of a prism has been shown in the figure. If the angle of incidence at the surface $\text{I}$ of the prism is ${45}^{\circ }$, then the light ray is incident on the surface $\text{II}$ at an angle of

• A. ${30}^{\circ }$
• B. ${20}^{\circ }$
• C. ${40}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is B ${20}^{\circ }$

From the given section of prism, the angle of prism $\left(A\right)$ is,


$A={180}^{\circ }-{60}^{\circ }-{70}^{\circ }={50}^{\circ }$
$[$from geometry$]$
From Snell's law at surface $\text{I}$,
${\mu }_a\sin i={\mu }_d\sin r$
$\Rightarrow 1\times \sin {45}^{\circ }=\sqrt{2}\sin r$
$\Rightarrow \sin r=\frac{1}{2}$
$\therefore r=r_1={30}^{\circ }$
At surface $\text{II}$, the angle of incidence is $r_2$.
We know,
$r_1+r_2=A$
$\Rightarrow {30}^{\circ }+r_2={50}^{\circ }$
$\therefore r_2={50}^{\circ }-{30}^{\circ }={20}^{\circ }$

Why this question? Bottom line: Whenever a section of prism is given which does not look like a prism, then first complete the prism and find the prism angle $A$.