Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A prism of refractive index $\sqrt 2$has a refracting angle of 60$^\circ$. At what angle must a ray incident on it so that it suffers minimum deviation?(A) 30$^\circ$(B) 45$^\circ$(C) 60$^\circ$(D) 75$^\circ$

Last updated date: 20th Jun 2024
Total views: 54.9k
Views today: 0.54k
Verified
54.9k+ views
Hint:The angle of incidence is equal to the angle of emergence for minimum deviation. For a prism, the refractive angle is given by $\mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$ where A is the angle of prism and ${\delta _m}$ is the angle of minimum deviation.

Complete step-by-step solution
The refractive index of the prism is given as $\mu = \sqrt 2$and the refracting angle of the prism is 60$^\circ$, which is also the angle of the prism which is denoted as A.
For a prism, the sum of angle of incidence and the angle of emergence is equal to the sum of angle of prism and angle of minimum deviation.
i + e = A + ${\delta _m}$
For the case of minimum deviation, the angle of incidence is the same as angle of emergence, i = e, therefore,
i = $\dfrac{{A + {\delta _m}}}{2}$ …equation (1)
Now that we have obtained the relation between angle of incidence and angle of prism and angle of minimum deviation, we can consider the relation of the refractive index to further solve the problem.
$\mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$ …equation (2)
We now substitute equation (1) in equation (2) and substitute the values of refractive index as well as angle of prism. We obtain,
$\sqrt 2 = \dfrac{{\sin \left( i \right)}}{{\sin \left( {\dfrac{{60^\circ }}{2}} \right)}}$
On simplifying, we get,
sin (i) = $\sqrt 2 \times \sin \left( {30^\circ } \right)$
The value of sin(30$^\circ$) is $\dfrac{1}{2}$. Therefore, the value of sin (i) becomes,
sin (i) = $\dfrac{1}{{\sqrt 2 }}$
Since sin (i) = $\dfrac{1}{{\sqrt 2 }}$, this means that i = 45$^\circ$
Hence, the angle of incidence for minimum deviation in the given prism is 45$^\circ$.

Therefore, option B is the correct answer.

Note: The equation for the refractive index of the prism and the equivalence relation between the sum of angle of incidence and the angle of emergence and the sum of angle of prism and the angle of minimum deviation are very important. In this question, if the angle of minimum deviation is given, then we do not need to use the relation for the refractive index.