Question

In a young's double slit experiment, the distance between slits $S_1$ and $S_2$ is $d$ and distance of slit plane from the screen is $D(>>d)$. The point source of light (s) is placed a distance $\frac{d}{2}$ below the principal axis and the slits $S_1$ and $S_2$ are located symmetrically with respect to the principal axis of the lens. Focal length of the lens is $f(>>d)$. The distance of the central maxima of the fringe pattern from the centre $\left(O\right)$ of the screen is found to be $\frac{Dd}{xf}$. The value of $4x$ is

Answer 1

All rays transmitted through the lens will be parallel since the source (s) is in focal plane
$\tan \varphi =\frac{\frac{d}{2}}{f}$
Path difference in waves reaching the slits $S_1$ and $S_2$ is,
$\Delta x_1=S_{1}M=d\sin \varphi \approx d\tan \varphi =\frac{d^2}{2f}$
Central maxima will be formed at a point $P$ where, $S_{2}P-S_{1}P=\Delta x_1$
So that net path difference is zero.
If angular position of point P is $\theta$, then,
$S_{2}P-S_{1}P=d\sin \theta$
$\Rightarrow d\sin \theta =\frac{d^2}{2f}$
$\Rightarrow \sin \theta =\frac{d}{2f}$
$\therefore \tan \theta \approx \frac{d}{2_f}$
$\frac{OP}{D}=\frac{d}{2f}$
$OP=\frac{Dd}{2f}$