Question

(a) 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.
(b) 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 widths of the two slits.

Answer 1

(a)

When light waves from two illuminated slits is incident on the screen, the path traveled by each light wave is different. This path difference leads to a phase difference in the two light waves. The path difference is different for each point on the screen and hence, intensity is different for all the points. This leads to the formation and bright and dark fringes on the screen.

Consider point P on the screen as shown in the figure.

$S_2{P}^2=S_2{F}^2+P{F}^2$

$S_{2}P=\sqrt{D^2+(x+\frac{d}{2}{)}^2}$

Similarly,

$S_{1}P=\sqrt{D^2+(x-\frac{d}{2}{)}^2}$

Path difference is given by:

$S_{2}P-S_{1}P=\sqrt{D^2+(x+\frac{d}{2}{)}^2}-\sqrt{D^2+(x-\frac{d}{2}{)}^2}$

Using binomial expansion,

$S_{2}P-S_{1}P=D(1+\frac{1}{2}(\frac{x}{D}+\frac{d}{2D}{)}^2+...)-D(1+\frac{1}{2}(\frac{x}{D}-\frac{d}{2D}{)}^2+....)$

Ignoring higher order terms,

$\Delta x=S_{2}P-S_{1}P\approx \frac{xd}{D}$

For constructive interference i.e. bright fringes,

$n\lambda =\frac{xd}{D}$

$x_n=\frac{n\lambda D}{d}$

Fringe width is equal to the distance between two consecutive maxima.

$\beta =x_n-x_{n-1}=\frac{n\lambda D}{d}-\frac{(n-1)\lambda D}{d}$

$\beta =\frac{\lambda D}{d}$

(b)

$\frac{I_{max}}{I_{min}}=\frac{(a_1+a_2{)}^2}{(a_1-a_2{)}^2}=\frac{9}{25}$

Solving, $\frac{a_1}{a_2}=\frac{4}{1}$

Ratio of slit widths, $\frac{w_1}{w_2}=\frac{a_1^2}{a_2^2}=16$