Question

The angle of incidence for a ray of light at a refracting surface of a prism is $45^\circ $. The angle of the prism is $60^\circ $. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are
(A) $45^\circ ;\dfrac{1}{{\sqrt 2 }}$
(B) $30^\circ ;\sqrt 2 $
(C) $45^\circ ;\sqrt 2 $
(D) $30^\circ ;\dfrac{1}{{\sqrt 2 }}$

Answer 1

Hint:
The angle of minimum deviation can be found out from the given angle of prism and the angle of incidence by the formula ${\delta _m} = 2i - A$. Then by using the values of the angle of prism and the previously derived angle of minimum deviation we can find the refractive index of the material of the prism.
Formula used: In this solution, we will be using the following formula
$\Rightarrow {\delta _m} = 2i - A$
where ${\delta _m}$ is the angle of minimum deviation, $i$ is the angle of incidence, $A$ is the prism.
$\Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
where $\mu $ is the refractive index of the prism.

Complete step by step answer:
In the question, we are given the values of the angle of prism $A = 60^\circ $ and the angle of incidence $i = 45^\circ $. By using those we can calculate the angle of minimum deviation by the formula,
$\Rightarrow {\delta _m} = 2i - A$
Therefore by substituting the values we get,
$\Rightarrow {\delta _m} = \left( {2 \times 45 - 60} \right)^\circ $
By calculating we get,
$\Rightarrow {\delta _m} = \left( {90 - 60} \right)^\circ $
$\Rightarrow {\delta _m} = 30^\circ $
Hence the value of the angle of minimum deviation is ${\delta _m} = 30^\circ $.
Now to find the value of the refractive index of the material, we use the value of the angle of minimum deviation and the angle of the prism
$\Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
by substituting the values of $A = 60^\circ $ and ${\delta _m} = 30^\circ $ in the above equation, we get
$\Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{60 + 30}}{2}} \right)}}{{\sin \left( {\dfrac{{60}}{2}} \right)}}$
Now we can calculate the angle in the numerator and denominator as,
$\Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{90}}{2}} \right)}}{{\sin \left( {\dfrac{{60}}{2}} \right)}}$
$\Rightarrow \mu = \dfrac{{\sin \left( {45} \right)}}{{\sin \left( {30} \right)}}$
The values of the sine in the above equation are given by,
$\Rightarrow \sin \left( {45} \right) = \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {30} \right) = \dfrac{1}{2}$
So by using these values in the above equation we get,
$\Rightarrow \mu = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{2}}}$
We can write this as,
$\Rightarrow \mu = \dfrac{1}{{\sqrt 2 }} \times 2$
So by cancelling $\sqrt 2 $ from the denominator, we get
$\Rightarrow \mu = \sqrt 2 $
So by solving we get the values of the angle of minimum deviation ${\delta _m} = 30^\circ $ and refractive index of the material as, $\mu = \sqrt 2 $
So the correct answer is (B); $30^\circ ;\sqrt 2 $.

Note:
In the question, the ray suffers refraction through the prism at an angle of minimum deviation. The angle of minimum deviation is the specific position where due to the angle of incidence the angle of deviation is the minimum. During this position of minimum deviation, the refracted ray is parallel to the base of the prism.