First slide

Ray optics

Question

Refractive index of the material of a prism is $\sqrt{3}$ and refracting angle of the prism is $60^{o}$ . Then the minimum deviation is

Easy

A

$15^{o}$
B

$45^{o}$
C

$30^{o}$
D

$60^{o}$

Solution

$\begin{array}{l} \sin (\frac{δ_{m} + A}{2}) = μ \sin (\frac{A}{2}) \\ \Rightarrow \sin (\frac{δ_{m} + 60^{o}}{2}) = \sqrt{3} \sin 30^{o} = \frac{\sqrt{3}}{2} \\ ∴ \frac{δ_{m} + 60^{o}}{2} = 60^{o} \to δ_{m} = 60^{o} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App