Question

In YDSE arrangement as shown in the figure, fringes are seen on screen using monochromatic source S having wavelength $3000\stackrel{{}^{\circ }}{A}$ (in the air). $S_{1}and{S}_2$ are two slits separated by $d=1mmandD=1m$ . Left of slits ${\text{S}}_1\text{a}\text{n}\text{d}{\text{S}}_2$ medium of refractive index ${\text{n}}_1=2$ is present and to the right of $S_{1}and{S}_2$ medium of ${\text{n}}_2=\frac{3}{2}$ , is present. A thin slab of thickness 't' is placed in front of ${\text{S}}_1$ . The refractive index of ${\text{n}}_3$ of the slab varies with distance from it's starting face as shown in figure.





In order to get central maxima at the centre of the screen, the thickness of the slab (in $\mu m$ ) required is:

Answer 1

Answer: 2

Path difference,
$\Delta x=n_{1}S{S}_2+n_2{S}_{2}P-\left[\left(n_{1}S{S}_1+n_2{S}_{1}P\right)-{\int }_0^t\left(n_3-n_2\right)dx\right]$
$=n_1\left(S{S}_2-S{S}_1\right)+n_2\left(S_{2}P-S_{1}P\right)-{\int }_0^t{n}_{3}dx-n_{2}t$
In order to get central maxima at the centre of the screen

$o=\frac{2\times {\left(1\times {\left(10\right)}^{-3}\right)}^2}{2\times 1}+0-2t+\frac{3t}{2}$
$0.5t=1\mu m$
$t=2\mu m$