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) In Young's double slit experiment, the wavefronts from the two illuminated slits superpose on the screen. This results in formation of alternate bright and dark fringes because of constructive and destructive interference, respectively. The intensity of light it maximum at the centre C of the screen and it is called central maxima.



Let $S_1$ and $S_2$ be two slits separated by a distance d. GG' is the screen at a distance D from the slits $S_1$ and $S_2$. Both the slits are equidistant from point C. The intensity of light will be maximum at this point due to the path difference of the waves reaching this point will be zero.

At point P, the path difference between the rays coming from the slits $S_1$ and $S_2$ is $S_{2}P-S_{1}P$.

Now, $S_1{S}_2=d,EF=d$, and $S_{2}F=D$

$\therefore$ In $\Delta S_{2}PF$,

$S_{2}P={[S_2{F}^2+P{F}^2]}^{\frac{1}{2}}$

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

= $D{[1+\frac{{(x+\frac{d}{2})}^2}{D^2}]}^{\frac{1}{2}}$

Similarly, in $\Delta S_{1}PE$

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

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

On expanding it binomially,

$S_{2}P-S_{1}P=\frac{1}{2D}[4x\times \frac{d}{2}]=\frac{xd}{D}$

For bright fringes (constructive interference), the path difference is an integral multiple of wavelengths, i.e., path difference is $n\lambda$.

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

$x=\frac{n\lambda D}{d}$, where $n=0,1,2,3,4,...........$

For $n=0,x_0=0$

$n=1,x_1=\frac{\lambda D}{d}$

$n=2,x_2=\frac{2\lambda D}{d}$

$n=3,x_3=\frac{3\lambda D}{d}$

$⋮$

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

The separation between the centres of two consecutive bright interference fringes is the width of a dark fringe.

$\therefore \begin{array}{c}{\beta }_1=x_n-x_{n-1}=\frac{\lambda D}{d}\end{array}$

Similarly, for dark fringes,

$x_n=(2n-1)\frac{\lambda D}{2d}$

For $n=1,x_1=\frac{\lambda D}{2d}$

For $n=2,x_2=\frac{3\lambda D}{2d}$

The separation between the centres of two consecutive dark interference fringes is the width of a bright fringe.

$\therefore \begin{array}{c}{\beta }_2=x_n-x_{n-1}=\frac{\lambda D}{d}\end{array}$

$\therefore {\beta }_1={\beta }_2$

All the bright and dark fringes are of equal width as ${\beta }_1={\beta }_2$

(b) Let $\omega$, a and I represent the slit width, amplitude and intensity respectively.

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

$\frac{(a_1-a_2)}{(a_1+a_2)}=\frac{3}{5}$

$\frac{(a_1-a_2)+(a_1+a_2)}{(a_1+a_2)-(a_1-a_2)}=\frac{3+5}{5-3}$

$\frac{2a_1}{2a_2}=\frac{8}{2}$

or $\frac{a_1}{a_2}=\frac{4}{1}$

and $\frac{{\omega }_1}{{\omega }_2}=\frac{(a_1{)}^2}{(a_2{)}^2}=\frac{16}{1}$