Question

calculate the angle of minimum deviation in an equilateral triangle of refractive index 1.414 is


and also tell me how did you got it​

Answer 1

▶ Solution :

Given :

▪ Angle of prism = 60°

▪ Refractive index of prism = 1.41 = √2

To Find :

▪ Angle of minimum deviation.

Formula :

▪ Formula of refractive index of prism in terms of angle of prism and angle of minimum deviation is given by

$\boxed{\bf{\pink{\mu=\dfrac{\sin{\huge{(}}\dfrac{A+\delta_m}{2}{\huge{)}}}{\sin{\huge{(}}\dfrac{A}{2}{\huge{)}}}}}}$

Calculation :

$\implies\sf\:\sqrt{2}=\dfrac{\sin{\huge{(}}\dfrac{60\degree+\delta_m}{2}{\huge{)}}}{\sin30\degree}\\ \\ \implies\sf\:\dfrac{1}{\sqrt{2}}=\sin{\huge{(}}\dfrac{60\degree+\delta_m}{2}{\huge{)}}\\ \\ \implies\sf\:\dfrac{\pi}{4}=\dfrac{\frac{\pi}{3}\times \delta_m}{2}\\ \\ \implies\sf\:\dfrac{\pi}{2}=\dfrac{\pi}{3}+\delta_m\\ \\ \implies\boxed{\bf{\purple{\delta_m=\dfrac{\pi}{6}=30\degree}}}$