Question

For glass prism $\left( {\mu = \sqrt 3 } \right)$, the angle of minimum deviation is equal to the angle of prism. The angle of prism is:
(A) ${30^ \circ }$
(B) ${45^ \circ }$
(C) ${60^ \circ }$
(D) ${90^ \circ }$

Answer 1

Hint:The angle of prison is determined by using the refractive index of the prism formula, when the angle of minimum deviation is equal to the angle prism. By using this given information in the refractive index of the prism formula, the angle of the prism is determined.

Useful formula:
The refractive index of the prism is given by,
$\mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \dfrac{A}{2}}}$
Where, $\mu $ is the refractive index, $A$ is the refracting angle of the prism and ${\delta _m}$ is the angle of the minimum deviation.

Complete step by step solution:
Given that,
The refractive index of the prism is, $\mu = \sqrt 3 $.
The angle of minimum deviation is equal to the angle of prism, ${\delta _m} = A$.
Now, by using the refractive index formula, then
The refractive index of the prism is given by,
$\mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \dfrac{A}{2}}}\,........................\left( 1 \right)$
By substituting the refractive index of the prism value and the angle of the minimum deviation in the equation (1), then the above equation is written as,
$\sqrt 3 = \dfrac{{\sin \left( {\dfrac{{A + A}}{2}} \right)}}{{\sin \dfrac{A}{2}}}$
By adding the terms in the above equation, then the above equation is written as,
$\sqrt 3 = \dfrac{{\sin \left( {\dfrac{{2A}}{2}} \right)}}{{\sin \dfrac{A}{2}}}$
Now by cancelling the same terms in the numerator and in denominator, then the above equation is written as,
$\sqrt 3 = \dfrac{{\sin A}}{{\sin \dfrac{A}{2}}}\,......................\left( 2 \right)$
From the trigonometry $\left( {\sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}} \right)$ by substituting this term in the numerator in RHS, then the above equation is written as,
$\sqrt 3 = \dfrac{{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}{{\sin \dfrac{A}{2}}}$
By cancelling the same terms in the numerator and in denominator, then the above equation is written as,
$\sqrt 3 = 2\cos \dfrac{A}{2}\,...................\left( 3 \right)$
By rearranging the terms, then the above equation is written as,
$\dfrac{{\sqrt 3 }}{2} = \cos \dfrac{A}{2}$
By taking the $\cos $ from RHS to LHS, then
$\dfrac{A}{2} = {\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2}$
From trigonometry the ${\cos ^{ - 1}}\dfrac{{\sqrt 3 }}{2} = {30^ \circ }$, then
$\dfrac{A}{2} = {30^ \circ }$
By rearranging the terms, then
$A = {60^ \circ }$
Thus, the above equation shows the angle of the prism.

Hence, the option (C) is the correct answer.

Note: After equation (2), some trigonometry equation is taken, it is taken for further calculation and also for easy calculation. In equation (3), the term $2$ will not get cancelled each other, in trigonometry there are no rules to cancel the terms which are like this.