Question

A prism of refractive index $\mu $ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of $\mu $ is:
A.${\sin ^{ - 1}}\left( {\dfrac{\mu }{2}} \right)$
B.${\sin ^{ - 1}}\sqrt {\dfrac{{\mu - 1}}{2}} $
C.$2{\cos ^{ - 1}}\left( {\dfrac{\mu }{2}} \right)$
D.$2{\cos ^{ - 1}}\left( {\dfrac{\mu }{8}} \right)$

Answer 1

Hint: A prism is an optically transparent material which is used to refract the light which falls on the surface of the prism,it has well-polished surfaces. Refractive index is defined as the ratio of speed of light in vacuum to the speed of light in the second medium.
Formula used:
The formula of the refractive index is given by,
$\mu = \dfrac{{\sin \left( {\dfrac{{A + D}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Where D is the angle of deviation, A is the angle of prism and $\mu $ is the coefficient of
refractive index.

Complete answer:
It is given in this problem that there is a prism having refractive index $\mu $ and the angle of incidence is placed such that there is minimum deviation of position and we need to find the value of A in terms of $\mu $ which is a refractive index.
Since, the formula of the refractive index given by,
$\mu = \dfrac{{\sin \left( {\dfrac{{A + D}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Where D is the angle of deviation, A is the angle of prism and $\mu $ is the coefficient of refractive index.
$ \Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{A + D}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
It is mentioned in the problem that the prism is placed in the minimum deviation position therefore,
$ \Rightarrow A = D$
$ \Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{A + D}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Replacing angle D with A in the formula of the refractive index we get,
$ \Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{A + A}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
$ \Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{2A}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Since, $\sin A = 2\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{A}{2}} \right)$.
$ \Rightarrow \mu = \dfrac{{\sin \left( A \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
Replacing $\sin A$ in the above relation we get,
$ \Rightarrow \mu = \dfrac{{2\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{A}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
$ \Rightarrow \mu = 2\cos \left( {\dfrac{A}{2}} \right)$
The refractive index is given by$\mu = 2\cos \left( {\dfrac{A}{2}} \right)$.

The correct answer for this problem is option C.

Note:
It is given in the problem that the angle of prism equal to the angle of deviation. The refractive index of the material decides to which extent the light which is entering the prism will deviate from the normal, the more the denser material the light will bend far away from the normal.