The figure shows a double-slit experiment. Each slit has a width $W$. $A$ thin glass slab of thickness $\delta$ and refractive index $\mu$ is placed in front of one of the slits. The intensity at the central point $C$ is measured as a function of the thickness $\delta$. At what thickness $\delta$ will the intensity at $C$ be minimum?

If a transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{-5} \, m$ is inserted in front of one of the slits of Young's Double Slit experiment,how much will be the shift in the interference pattern? The distance between the slits is $0.5 \, mm$ and that between slits and screen is $100 \, cm$. (Answer in $cm$)

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.

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 $YDSE$,the source $S$ placed symmetrically with respect to the slits $S_1$ and $S_2$ is now moved parallel to the plane of the slits so that it is closer to the upper slit $S_1$,as shown. Then,

When a glass plate of refractive index $1.44$ is introduced in the path of one of the interfering beams, the fringes are displaced by a distance '$y$'. If this plate is replaced by another plate of same thickness but of refractive index $1.66$, the fringes will be displaced by a distance