Question
Describe Young’s double slit interference experiment and derive conditions for occurrence of dark and bright fringes on the screen. Define fringe width and derive a formula for it.
✓
Answer
Description of Young’s double-slit interference experiment:
1. A plane wavefront is obtained by placing a linear source $S$ of monochromatic light at the focus of a convex lens. It is then made to pass through an opaque screen $A B$ having two narrow and similar slits $S_1$ and $S_2 \cdot S_1$ and $S_2$ are equidistant from $S$ so that the wavefronts starting simultaneously from $S$ and reaching $S_1$ and $S_2$ at the same time are in phase. A screen PQ is placed at some distance from screen $A B$ as shown in below figure
- $S_1$ and $S_2$ act as secondary sources. The crests/-troughs of the secondary wavelets superpose and interfere constructively along straight lines joining the black dots shown in above figure. The point where these lines meet the screen have high intensity and are bright.
- Similarly, there are points shown with red dots where the crest of one wave coincides with the trough of the other. The corresponding points on the screen are dark due to destructive interference.
- These dark and bright regions are called fringes or bands and the whole pattern is called interference pattern.
Consider Young's double-slit experimental set up. Two narrow coherent light sources are obtained by wavefront splitting as monochromatic light of wavelength $\lambda$ emerges out of two narrow and 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 and at considerable distance $D(D \gg d)$ from the slits. $O O$ ' is the perpendicular bisector of segment $S_1 S_2$.
Consider, a point $P$ on the screen at a distance $y$ from $O^{\prime}(y \& D)$. The two light waves from $S_1$ and $S_2$ reach $P$ along paths $S_1 P$ and $S_2 P$, respectively. If the path difference $(\Delta I)$ between $S_1 P$ and $S_2 P$ is an integral multiple of $\lambda$, the two waves arriving there will interfere constructively producing a bright fringe at $P$. On the contrary, if the path difference between $S_1 P$ and $S_2 P$ is half integral multiple of $\lambda$, there will be destructive interference and a dark fringe will be produced at $P$.
From above figure,
Expression for the fringe width (or band width) : The distance between consecutive bright (or dark) fringes is called the fringe width (or band width) W. Point $P$ will be bright (maximum intensity), if the path difference, $\Delta l = y _{ n } \frac{d}{D}=n \lambda$ where $n=0,1,2,3 \ldots$, Point $P$ will be dark (minimum intensity equal to zero), if $y_m \frac{d}{D}=(2 m-1) \frac{\lambda}{2}$, where, $m=1,2,3 \ldots$, Thus, for bright fringes (or bands),
$y_n=0, \lambda \frac{D}{d}, \frac{2 \lambda D}{d} \ldots \ldots$
and for dark fringes (or bands),
$y_m=\frac{\lambda D}{2}, 3 \frac{\lambda}{2} \frac{D}{d}, 5 \frac{\lambda}{2} \frac{D}{d} \cdots \cdots$
These conditions show that the bright and dark fringes (or bands) occur alternately and are equally - spaced. For Point $O ^{\prime}$, the path difference $\left( S _2 O ^{\prime}- S _1 O ^{\prime}\right)=0$. Hence, point $O ^{\prime}$ will be bright. It corresponds to the centre of the central bright fringe (or band). On both sides of $O^{\prime}$, the interference pattern consists of alternate dark and bright fringes (or band) parallel to the slit.
Let y$_n $and $y_{n + 1},$ be the distances of the nth and $(n + 1)^{th}$^ bright fringes from the central bright fringe.
$\therefore \frac{y_n d}{D}=n \lambda \quad \therefore y_n=\frac{n \lambda D}{d}$
and $\frac{y_{n+1} d}{D}=(n+1) \lambda \quad \therefore y_{n+1}=\frac{(n+1) \lambda D}{d} \ldots$
The distance between consecutive bright fringes $=y_{n+1}-y_n=\frac{\lambda D}{d}[(n+1)-n]=\frac{\lambda D}{d}$
Hence, the fringe width,
$\therefore W=\Delta y=y_{n+1}-y_n=\frac{\lambda D}{d} \text { (for bright fringes) } \ldots$
Alternately, let $y_m$ and $y_{m+1}$ be the distances of the $m$ th and $(m+1)^{\text {th }}$ dark fringes respectively from the central bright fringe.
$ \therefore \frac{y_m d}{D}=(2 m-1) \frac{\lambda}{2} \text { and }$
$\frac{y_{m+1} d}{D}=[2(m+1)-1] \frac{\lambda}{2}=(2 m+1) \frac{\lambda}{2}$
$\therefore y_m=(2 m-1) \frac{\lambda D}{2 d} \text { and }$
$y_{m+1}=(2 m+1) \frac{\lambda D}{2 d} $
$\therefore$ The distance between consecutive dark fringes,
$ y_{m+1}-y_m=\frac{\lambda D}{2 d}[(2 m+1)-(2 m-1)]=\frac{\lambda D}{d}$
$\therefore W=y_{m+1}-y_m$
$=\frac{\lambda D}{d} \text { (for dark fringes) }$
Eqs. $(7)$ and $(11)$ show that the fringe width is the same for bright and dark fringes.
[Note: In the first approximation, the path difference is $d \sin \theta$.]
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