Question

An equilateral prism has u =√3. its angle of minimum deviation will be : :
(A) 30°
(B) 60°
(C) 120°
(D) 45°​

Answer 1

An equilateral prism has u = √3. Its angle of minimum deviation will be 60°

Answer 2

Given :

Refractive index of equilateral prism = √3

To Find :

Angle of minimum deviation.

Solution :

★ The refractive index of the material of the prism is given by

$:\implies\sf\:n=\dfrac{\bigg[\dfrac{(A+\delta_m)}{2}\bigg]}{sin\bigg(\dfrac{A}{2}\bigg)}$

• n denotes refractive index
• A denotes angle of prism
• $\sf{\delta_m}$ denotes angle of minimum deviation

For an equilateral prism, A = 60°

By substituting the given values;

$:\implies\sf\:\sqrt{3}=\dfrac{sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]}{sin\bigg(\dfrac{60^{\circ}}{2}\bigg)}$

$:\implies\sf\:\sqrt{3}=\dfrac{sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]}{sin\:30^{\circ}}$

$\sf:\implies\:\sqrt{3}\times sin\:30^{\circ}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$\sf:\implies\:\sqrt{3}\times\dfrac{1}{2}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$:\implies\sf\:\dfrac{\sqrt3}{2}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

• sin 60° = √3/2

$:\implies\sf\:sin\:60^{\circ}=sin\:\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$:\implies\sf\:60^{\circ}=\bigg[\dfrac{(60^{\circ}+\delta_m)}{2}\bigg]$

$:\implies\sf\:60^{\circ}+\delta_m=120^{\circ}$

$:\implies\:\underline{\boxed{\bf{\purple{\delta_m=60^{\circ}}}}}$

∴ (B) is the correct answer!