$(a)\;y = D \sqrt {\frac{2n \lambda}{d}} \\ (b)\;y=D \sqrt { 2 \bigg(1- \frac{n \lambda}{d} \bigg)} \\ (c)\;y=D \sqrt {1- \frac{n \lambda}{d}} \\ (d)\;None $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

The optical path difference at P is

$P= S_1 P -S_2 P = d\cos \theta$

Since $\cos \theta=1 - \large\frac{\theta^2}{2}$

When $\theta$ is small

$\therefore P=d \bigg[1- \large\frac{\theta^2}{2}\bigg]$

$\qquad= d \bigg[ 1- \large\frac{y^2}{2D^2}\bigg]$

Where $D +d =D$

For nth maxima , $P=n \lambda$

$\therefore y=D \sqrt { 2 \bigg(1- \frac{n \lambda}{d} \bigg)}$

Hence b is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...