Question

In Young's double slit experiment, two slits $S_1$ and $S_2$ are ' $d$ ' distance apart and the separation from slits to screen is $D$ (as shown in figure). Now if two transparent slabs of equal thickness $0.1 \,mm$ but refractive index $1.51$ and $1.55$ are introduced in the path of beam $(\lambda=4000 \mathring {A})$ from $S_1$ and $S_2$ respectively, The central bright fringe spot will shift by ______number of fringes.

Answer 1

Path difference at $P$ be $\Delta x$
$\Delta x =\left(\mu_2-\mu_1\right) t $
$ =(1.55-1.51) 0.1 \,mm$
$ =0.04 \times 10^{-4} $
$\Delta x =4 \times 10^{-6}=4 \,\mu m$
$y =\frac{\Delta xD }{ d }=4 \times 10^{-6} \frac{ D }{ d }$
$\{ y$ is the distance of central maxima from geometric center $\}$
$\underset{(\beta)}{\text { fringe width }}=\frac{\lambda D}{d}=4 \times 10^{-6} m \frac{D}{ d }=4 \mu m \frac{D}{ d }$
$\therefore $ Central bright fringe spot will shift by ' $x$ '
Number of shift $=\frac{y}{\beta}$
$=\frac{4 \times 10^{-6} D / d }{4 \times 10^{-7} D / d }=10 $