Question

Find the angle of deviation produced by the prism for the situation shown in the figure :
$[\text{Use}si{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)={35}^{\circ }]$

• A. ${120}^{\circ }$ clockwise
• B. $0^{\circ }$
• C. ${30}^{\circ }$ clockwise
• D. ${15}^{\circ }$ clockwise

Answer 1

The correct option is B $0^{\circ }$



Applying Snell's law at surface $AB:$

$1\times \sin {60}^{\circ }=\sqrt{3}\times \sin r_1$

$\Rightarrow r_1={30}^{\circ }$

From the figure we can see that, after the first refraction, refracted ray will become parallel to the surface $AC$.
Therefore, the light ray will strike surface $BC$ first.

At surface $BC$ we get, $i={60}^{\circ }$,

${\theta }_C={\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx {35}^{\circ }$

Since,
$i>{\theta }_C,\text{TIR}$ occurs at $BC$.

Applying Snell's law at surface $AC$,

$\sqrt{3}\times \sin {30}^{\circ }=1\times \sin e\Rightarrow e={60}^{\circ }$

$\therefore$ Angle of deviation , $\delta =i+e-A$

Substituting the data obtained we get,

$\delta ={60}^{\circ }+{60}^{\circ }-{120}^{\circ }=0^{\circ }$.

Hence, option (b) is the correct answer.