Question

Give analytical treatment of YDSE interference bands and hence obtain the expression for fringe width.

Answer 1

Let us consider young's double slit experimental setup. Two narrow coherent light sources are obtained by wavefront splitting as mono chromatic light of wavelength $\lambda$ emerges out of two closely spaced, parallel slits $S_1$ and $S_2$ of equal widths. The separation $S_1{S}_2=d$ is very small. The interference pattern is observed on a screen placed parallel to the plane of $S_1{S}_2$ and some considerable distance.
$D(D>>d)$ from the slits. OP is the perpendicular bisector of segment $S_1{S}_2$. Consider point Q on the screen at a distance x from $P(x<<D)$ the two light waves from $S_1,S_2$ reach Q along paths $S_1$Q and $S_2$Q respectively.
From the figures
$(S_{2}Q{)}^2=(S_{2}N{)}^2+(QN{)}^2=D^2+{(x+\frac{d}{2})}^2$ .....$\left(1\right)$
and $(S_{1}Q{)}^2=(S_{1}M{)}^2+(QM{)}^2=D^2+{(x-\frac{d}{2})}^2$ .....$\left(2\right)$
$\therefore (S_{2}Q{)}^2-(S_{1}Q{)}^2=D^2+{(x+\frac{d}{2})}^2-D^2-{(x-\frac{d}{2})}^2$
$\therefore (S_{2}Q+S_{1}Q)(S_{2}Q-S_{1}Q)=2xd$
$\therefore S_{2}Q-S_{1}Q=\frac{2xd}{S_{2}Q+S_{1}Q}$
In partice, $D>>x$ and $D>>d$, so that
$S_{2}Q-S_{1}Q\simeq 2D$
$\therefore$ Path difference, $(S_{2}Q-S_{1}Q)=\frac{2xd}{2D}=\frac{xd}{D}$ .....$\left(3\right)$
Expression for the fringe width or band width: The distance between consecutive bright(or dark) fringes is called the fringe width or band width X.
Point Q will be bright(maximum intensity), if the path difference $\frac{xd}{D}=n\lambda$, where $n=0,1,2,3,...$
point Q will be dark(minimum intensity equal to zero) if $\frac{xd}{D}=(2m-1)\frac{\lambda }{2}$, where $m=1,2,3,...$
These conditions show that the bright and dark fringes occur alternately and are equally speed.
For point P, the path difference $(S_{2}P-S_{1}P)=0$. Hence, point P will be bright, the central bright fringe or band. On both sides of point P, the interference pattern consists of alternate bright and dark bands or fringes parallel to the slit.
Let $x_n$ and $x_{n+1}$ be the distances of the $n^{th}$ and $(n+1{)}^{th}$ bright fringes from the central bright fringe.
$\therefore \frac{x_{n}d}{D}=n\lambda$ or $x_n=\frac{n\lambda D}{d}$ ....$\left(4\right)$
and $\frac{x_{n+1}d}{D}=(n+1)\lambda$ or $x_{n+1}=\frac{(n+1)\lambda D}{d}$ ....$\left(5\right)$
$\therefore$ The distance between consecutive bright fringes
$=x_{n+1}-x_n=\frac{\lambda D}{d}\left[\right(n+1)-n]=\frac{\lambda D}{d}$ ....$\left(6\right)$