## wave-optics Question 30

Question: Q. 9. (i) In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.

(ii) The ratio of the intensities at minima to the maxima in the Young’s double slit experiment is $9: 25$. Find the ratio of the width of the slits.

U [O.D. I, II, III 2014]

## Show Answer

Solution:

Ans. (i) Try yourself similar to Q. 2 (i) LAT Questions. 311/2 Now, fringe width,

$\beta=$ separation between two successive maxima (or two successive minima)

$$ \begin{aligned} & =x_{n}-x_{n-1} \ \therefore \quad \beta & =\frac{\lambda \mathrm{D}}{d} \end{aligned} $$

(ii) We have

$$ \frac{I_{\max }}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}=\frac{25}{9} $$

$$ \begin{array}{ll} \therefore & \frac{a_{1}+a_{2}}{a_{1}-a_{2}}=\frac{5}{3} \Rightarrow \frac{a_{1}}{a_{2}}=\frac{4}{1} \ \therefore & \frac{w_{1}}{w_{2}}=\frac{I_{1}}{I_{2}}=\frac{\left(a_{1}\right)^{2}}{\left(a_{2}\right)^{2}}=\frac{16}{1} \end{array} $$

[CBSE Marking Scheme 2014]