A ray is incident at an angle of incidence on one surface of a prism of small angle $A$ and emerges normally from the opposite surface. If the refractive index of the material of the prism is $\mu$, the angle of incidence $i$ is nearly equal to :

Hint: Here, we have to apply the concept of angle of deviation

Solution:

Step1: Draw free body diagram

 

 

Step2: Apply given conditions

According to free body diagram,

i=Angle of incident, A=Small angle of prism & µ=Refractive index of material
Assume, e= angle of emergence
given that, the incident ray emerges normally from the opposite surface,

Hence, $\angle e=0.....(1)$
Step3: Find angle of deviation

We know the angle of deviation ($\delta$) can be,

$\delta =\left(\mu -1\right)A.....(2)$
In case of prism, we know that, sum of small angle of prism & angle of deviation in a prism is equal to the sum of angle of incidence and angle of emergence. So, we get,
$\delta +A=i+e.....(3)$
Submit values from equation (1) & (2) in equation (3) we get,
$\left(\mu -1\right)A+A=i+0$
$\Rightarrow \mu A-A+A=i$

$\Rightarrow \mu A=i$ or $i=\mu A$
Hence, option (D) is correct.