Question

With the help of a labelled diagram, show that fringe separation $\beta$ in Young's double slit experiment is given by:
$\beta =\frac{\lambda D}{d}$
where the terms have their usual meaning

Answer 1

Let ${}^{\prime }{S}^{\prime }$ be the monochromatic source of light and let $S_1,S_2$ be the two parallel narrow slits close to each other. According to Huygen's principle $S_1$ and $S_2$ act as new wave source from which secondary wave lets originate. Let $S_{1}A$ is the perpendicular drawn on $S_{2}P$ from $S_1$. Path difference is given by
$S_{2}A=S_{2}P-S_{1}P$
In right angled $△S_1{S}_{2}A\sin \theta =\frac{S_{2}A}{S_1{S}_2}$
In right angled $△PCO\tan \theta =\frac{PO}{CO}$
If $\theta$ is small; then $\frac{S_{2}A}{S_1{S}_2}=\frac{OP}{CO}$
$\frac{S_{2}A}{d}=\frac{x}{D}$
So, Path difference $=S_{2}A=\frac{xd}{D}$
The intensity of light will be maximum at those places where the path difference between the interfering waves is zero or integral multiple of $\lambda$.
So, $\frac{xd}{D}=m\lambda$
or $x=\frac{mD\lambda }{d}$