Question

The angle of minimum deviation for a prism is ${{40}^{\circ }}$ and the angle of the prism is ${{60}^{\circ }}$. The angle of incidence in this position will be:
$A.\text{ 3}{{0}^{\circ }}$
$B.\text{ 6}{{0}^{\circ }}$
$C.\text{ 5}{{0}^{\circ }}$
$D.\text{ 10}{{0}^{\circ }}$

Answer 1

Hint: The angle of minimum deviation is the minimum value of the angle of deviation when a ray of light passes through a prism. Substitute the value of angle of minimum deviation and angle of prism in the standard equation showing the relationship among angle of incidence, angle of minimum deviation, and the angle of prism.

Formula used: $i=\dfrac{A+{{d}_{m}}}{2}$

Complete step by step answer:

In the above figure, ABC represents the principle section of a glass prism having $\angle A$ as its refraction angle. Let us consider a ray EF is incident on the face AB at point F where $N'FN$ is normal and $\angle i$ is the angle of incidence. As we can see that the refraction takes place from air to glass, the refracted ray FG bends towards the normal such that $\angle {{r}_{1}}$ is the angle of refraction.
Again, the ray FG is incident on the face AC at point G where $N''GN$ is normal and $\angle {{r}_{2}}$ is the angle of incidence. Since the refraction now takes place from denser to a rarer medium, the emergent ray GH bends away from the normal and goes along GH such that $\angle e$ is the angle of emergence.
Now, if the prism was not there then the incident ray EF would have proceeded straight, but due to the presence of the prism it changes its path along the direction thereby making $\angle KDH$ gives the angle of deviation (the angle through which the incident ray deviates while passing through the prism).
Here we are dealing with the angle of minimum deviation which is the minimum value of angle of deviation when a ray of light passes through a prism. In minimum deviation position, $i=e$ and so
${{r}_{1}}={{r}_{2}}=r$ (say)…….(i)
Again, for a prism
$A={{r}_{1}}+{{r}_{2}}$
Using equation (i), we get
$A=2r$ …….(ii)
Also,
$i+e=A+d$
$\Rightarrow$ $i+i=A+{{d}_{m}}$ ……..(Since for the prism in minimum deviation $i=e$ and $d={{d}_{m}}$ )
$\Rightarrow$ $2i=A+{{d}_{m}}$
$\Rightarrow$ $i=\dfrac{A+{{d}_{m}}}{2}$
Given: ${{d}_{m}}={{40}^{\circ }}$ and $A={{60}^{\circ }}$
$\therefore i=\dfrac{{{60}^{\circ }}+{{40}^{\circ }}}{2}={{50}^{\circ }}$

So, the correct answer is “Option C”.

Note: In case of minimum deviation through a prism following conditions are satisfied,
(i) The angle of incidence is equal to the angle of emergence.
(ii) The angle made by the ray, traveling through the material of the prism with, the normal to the two surfaces of the prism are equal to each other.
(iii) the ray passing through the material of the prism is parallel to its base.