In Young's double slit arrangement,water is filled in the space between the screen and the slits. Then:

If one of the slits of a standard $YDSE$ apparatus is covered by a thin parallel-sided glass slab so that it transmits only one-half of the light intensity of the other, then:

In Young's double slit experiment, the aperture screen distance is $2 \, m$. The fringe width is $1 \, mm$. Light of $600 \, nm$ is used. If a thin plate of glass $(\mu = 1.5)$ of thickness $0.06 \, mm$ is placed over one of the slits, then there will be a lateral displacement of the fringes by $... \, cm$.

Two coherent narrow slits emitting light of wavelength $\lambda$ in the same phase are placed parallel to each other at a small separation of $3 \lambda$. The light is collected on a screen $S$ which is placed at a distance $D (>> \lambda)$ from the slits. Find the smallest distance $x$ from the center $O$ such that the point $P$ is a maxima.

$6300 \mathring A$ 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:

$A$ transparent medium of refractive index $\mu = 1.5$ and thickness $t = 2.5 \times 10^{-5} \, m$ is placed in front of one of the slits in a Young's double-slit experiment. By what distance (in $cm$) will the interference pattern shift? The distance between the two slits is $d = 0.5 \, mm$ and the distance between the screen and the slits is $D = 100 \, cm$.