Question

With the help of a diagram obtain an expression for finding the distance between two consecutive bright or dark fringes in the interference pattern produced by double slits.

Answer 1

We already know that the path difference

$△x=n\lambda$ for bright fringes and

$△x=(n+1/2)\times \lambda$ for dark fringes

Now let us study the given figure.

Suppose $S_{1}A$ is the perpendiculr from $S_1$ to $S_{2}P$.

Suppose,

$D=OB$ = seperation between slits and the screen,

d = seperation between th slits

and $D>>d$.

Under the above approximation $S_{1}P$ and $S_{2}P$ are nearly parallel and hence $S_{1}A$ is very nearly perpendicular to $S_{1}P$, $S_{2}P$

and $OP$. AS $S_1{S}_2$ is perpendicular to $OB$ and $S_{1}A$ is perpendicular to $OP$, we have

$\angle S_2{S}_{1}A=\angle POB=\theta$

This is a small angle as $D>>d$.

The path difference is
$△x=P{S}_2-P{S}_1\approx P{S}_2-PA$

= $S_{2}A$ = $dsin\theta \approx dtan\theta$

= $d\times \frac{y}{D}$

The centers of the bright fringes are obtained atdistances y from the point B, where

$△x=d\times \frac{y}{D}$ = n$\lambda$

or, $y=\frac{nD\lambda }{d}$ i.e.,

at y = 0, $\pm \frac{D\lambda }{d}$, $\pm \frac{2D\lambda }{d}$, $\pm \frac{3D\lambda }{d}$, ....etcSimilarly we can show for dark fringes $y=\pm \frac{D\lambda }{2d}$, $y=\pm \frac{3D\lambda }{2d}$, $y=\pm \frac{5D\lambda }{2d}$,.......etc

the width fringe is therefore ,

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