Question

Derive an expression for the bandwidth of interference fringes in Young's double slit experiment.

Answer 1

Expression for bandwidth of interference fringes in youngs double experience : Let $d$ be the distance between two coherent sources $A$ and $B$ of wavelength $\lambda$. A screen $XY$ is placed parallel to $AB$ at a distance $D$ from the coherent sources. $C$ is the mid point of $AB$. $O$ is a point on the screen equidistant from $A$ and $B.P$ is a point at a distance $x$ from $O$, as shown in figure. Waves from $A$ and $B$ meet at $P$ in phase or out of phase depending upon the path difference between two waves.
Interference bandwidth
Draw $AM$ perpendicular to $BP$.
The path difference $S=BP-AP$
$AP=MP$
$\therefore \delta =BP-AP=BP-MP=BM$
In right angled $△ABM,BM=d\sin \theta$
If $Q$ is small, $\sin \theta =\theta$
$\therefore$ the path difference $\delta =\theta d$
In right angled triangle $COP$,
$\tan \theta =\frac{OP}{CP}=\frac{x}{D}$
For small values of $\theta ,\tan \theta =\theta$
$\therefore$ the path difference $\delta =\frac{xd}{D}$
Bright fringes : By the principle of interference, condition for constructive interference is the path difference $=n\lambda$
$\therefore \mu \frac{xd}{D}=\lambda$
where $n=0,1,2,.....$ indicates the order of the bright fringes.
$\therefore x=\frac{D}{d}\lambda$
This equation gives the distance of the $n^{th}$ bright fringe from the point $O$.
Dark fringes : By the principle of interference, condition for destructive interference is the path difference $=(2n-1)\frac{\lambda }{2}$.
where $n=1,2,3,....$ indicate the order of the dark fringes.
$\therefore x=\frac{D}{d}(2n-1)\frac{\lambda }{2}$
This equation gives the distance of the nth dark fringe from the point $O$. Thus, on the screen alternate dark and bright bands are seen on either side of the central bright band.
Bandwidth $\left(\beta \right)$ : The distance between any two consecutive bright or dark bands is called bandwidth. The distance between $(n+1{)}^{th}$ and $n^{th}$ order consecutive bright fringes from $O$ is given by
$x_{(n+1)}-x_n=\frac{D}{d}(n+1)\lambda -\frac{D}{d}n\lambda =\frac{D}{d}\lambda$
Bandwidth, $\beta =\frac{D}{d}\lambda$.