In the $YDSE$ shown,the two slits are covered with thin sheets having thickness $t$ and $2t$,and refractive index $2\mu$ and $\mu$ respectively. Find the position $(y)$ of the central maxima.

$A$ plate of thickness $t$ made of a material of refractive index $\mu$ is placed in front of one of the slits in a double slit experiment. What should be the minimum thickness $t$ for which the intensity at the centre of the fringe pattern is zero?

Consider the figure (not drawn to scale) in which a converging lens of radius $R = 1 \ cm$ and focal length $f = 20 \ cm$ is cut in the middle. The upper part is lifted up by $d = 1 \ mm$ and the lower part is pulled down by the same distance. The gap between them is blocked by an opaque sheet. $A$ point light source with wavelength $\lambda = 500 \ nm$ is placed on the optical axis at a distance of $2f$ from the split lens. $A$ large screen is placed at $L = 1 \ m$ from the right focus of the lens. Find the approximate number of interference fringes on the screen.

In a standard $YDSE$ setup,a small transparent slab of thickness $t$ and refractive index $\mu = 1.5$ is placed along the path $AS_2$ (as shown in the figure). Given that the slab thickness $t = d/4$,where $d$ is the slit separation,and the distance from the source $A$ to the slits is not explicitly needed for the shift calculation,find the position of the central maxima on the screen relative to $O$. Assume the distance between the slits and the screen is $D$.

In the setup shown in the figure,the two slits,$S_1$ and $S_2$,are not equidistant from the slit $S$. The central fringe at $O$ is then

Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between them is $2 \lambda$ as shown in the figure. The first bright fringe is formed at $P$ due to interference on a screen placed at a distance $D$ from $S_1$ $(D >> \lambda)$. Find the distance $OP$.