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

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?

In a Young's double-slit experiment,let $A$ and $B$ be the two slits. Thin films of thicknesses $t_A$ and $t_B$ and refractive indices $\mu_A$ and $\mu_B$ are placed in front of slits $A$ and $B$ respectively. If $\mu_A t_A = \mu_B t_B$,the central maximum will:

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 $OP$.

$6300 \mathring A$ wavelength light shines on two narrow slits separated by a distance of $1.0 \ mm$ and illuminates a screen at a distance of $1.5 \ m$ away. When one slit is covered by a thin glass plate of refractive index $1.8$ and the other slit by a thin glass plate of refractive index $\mu$,the central maxima shifts by $6^o$. Both plates have the same thickness of $0.5 \ mm$. The value of the refractive index $\mu$ of the plate is:

The central fringe of the interference pattern produced by light of wavelength $6000 \mathring A$ is found to shift to the position of the $4^{\text {th}}$ bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit in Young's experiment. The thickness of the glass plate will be ......... $\mu m$.