In a double-slit experiment,instead of taking slits of equal widths,one slit is made twice as wide as the other. Then,in the interference pattern:

$A$ 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 $D = 100 \, cm$.

In a double slit experiment,when light of wavelength $400 \; nm$ was used,the angular width of the first minima formed on a screen placed $1 \; m$ away was found to be $0.2^{\circ}$. What will be the angular width of the first minima if the entire experimental apparatus is immersed in water (in $^{\circ}$)? $(\mu_{water} = 4/3)$

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.

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 ......... $\mu m$.

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