Question

Two light rays $1$ and $2$ are incident on two faces $AC$ and $AB$ of an isosceles prism as shown in the figure. The rays emerge from the side $BC$. Then :

• A. minimum deviation of ray $1$ > minimum deviation of ray $2$
• B. minimum deviation of ray $1$ < minimum deviation of ray $2$
• C. minimum deviation of ray $1$ = minimum deviation of ray $2$
• D. minimum deviation cannot be compared.

Answer 1

The correct option is C minimum deviation of ray $1$ = minimum deviation of ray $2$

For the isosceles prism shown in figure,
$\angle B=\angle C={75}^{\circ }$

The angle of prism for ray 1 $=A_1=\angle ACB={75}^{\circ }$
The angle of prism for ray 2 $=A_2=\angle ABC={75}^{\circ }$
So, $A_1=A_2={75}^{\circ }$

From the formula of minimum deviation :

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

Here, $A$ and $\mu$ is same for both the rays.
$\Rightarrow$ minimum deviation of ray $1$ = minimum deviation of ray $2$
So,
$\Rightarrow {\left({\delta }_m\right)}_1={\left({\delta }_m\right)}_2$

Why this question? Angle of prism $\left(A\right)$ is the angle contained by two refracting faces of the prism as per the direction of incident ray.