Question

The refracting angle of a prism is $A$ and the refractive index of the material of the prism is $\cot \left( \dfrac{A}{2} \right)$​. The angle of minimum deviation of the prism can be given as,
$\begin{aligned} & A.180{}^\circ -A \\\ & B.180{}^\circ -2A \\\ & C.180{}^\circ -3A \\\ & D.180{}^\circ -4A \\\ \end{aligned}$

Answer 1

The refractive index of the material is given. The refractive can be found out using the general formula in terms of angle of prism and angle of minimum deviation. Substitute the given values in this equation. Then rearrange the terms in the equation in order to find the angle of minimum deviation. These all will help you to solve this question.

Formula used:
$\mu =\dfrac{\sin \left( \dfrac{\delta m+A}{2} \right)}{\sin \dfrac{A}{2}}$
Where $\delta m$ is the angle of minimum deviation and $A$ be the angle of the given prism.

Complete step by step answer:
In the question, it is already mentioned that the refractive index of the material of the prism is given as
$\mu =\cot \left( \dfrac{A}{2} \right)$
As we all know, the refractive index of a material can be found out using the general equation,
$\mu =\dfrac{\sin \left( \dfrac{\delta m+A}{2} \right)}{\sin \dfrac{A}{2}}$
Where $\delta m$ is the angle of minimum deviation and $A$ be the angle of the given prism.
Substituting these values in the equation will give,
$\cot \left( \dfrac{A}{2} \right)=\dfrac{\sin \left( \dfrac{\delta m+A}{2} \right)}{\sin \dfrac{A}{2}}$
As we all know it is mentioned in the question that the angle of minimum deviation is to be calculated. For that, we have to re-arrange the given equation in terms of angle of minimum deviation.
For that first of all convert the $\cot \left( \dfrac{A}{2} \right)$.
Generally the $\cot \left( \dfrac{A}{2} \right)$ can be written as,
$\cot \left( \dfrac{A}{2} \right)=\dfrac{\cos \left( \dfrac{A}{2} \right)}{\sin \left( \dfrac{A}{2} \right)}$
Substitute this in the given equation,
$\dfrac{\cos \left( \dfrac{A}{2} \right)}{\sin \left( \dfrac{A}{2} \right)}=\dfrac{\sin \left( \dfrac{\delta m+A}{2} \right)}{\sin \dfrac{A}{2}}$
The denominator can be cancelled here as they are equal.
$\cos \left( \dfrac{A}{2} \right)=\sin \left( \dfrac{\delta m+A}{2} \right)$
We can convert the $\cos \left( \dfrac{A}{2} \right)$ into $\sin \left( \dfrac{A}{2} \right)$.
This conversion is done using the formula,
$\sin \left( \dfrac{\pi }{2}-\dfrac{A}{2} \right)=\cos \left( \dfrac{A}{2} \right)$
Substitute this in the equation will give,
$\sin \left( \dfrac{\pi }{2}-\dfrac{A}{2} \right)=\sin \left( \dfrac{\delta m+A}{2} \right)$
Therefore,
$\left( \dfrac{\pi }{2}-\dfrac{A}{2} \right)=\left( \dfrac{\delta m+A}{2} \right)$
From this we can calculate the angle of minimum deviation.
$\pi -2A=\delta m$

Therefore the correct answer is given as option B.

Note:
The angle of deviation is equivalent to the change in the angle of incidence and the angle of refraction of a ray of light travelling through the surface between one medium and another medium. Both this medium will have different refractive indices in the angle of minimum deviation, the angle of incidence will be equal to angle of refraction.