$A$ 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 Young's double slit experiment,if the width (aperture) of the source slit $S$ is increased while keeping other parameters constant,then the interference fringes will:

In a double-slit experiment,green light $(5303\,\mathring{A})$ falls on a double slit having a separation of $19.44\,\mu m$ and a width of $4.05\,\mu m$. The number of bright fringes between the first and the second diffraction minima is

If a thin mica sheet of thickness $t$ and refractive index $\mu = 5/3$ is placed in the path of one of the interfering beams as shown in the figure,then the displacement of the fringe system is:

In a Young's double slit experimental arrangement shown here,if a mica sheet of thickness $t$ and refractive index $\mu$ is placed in front of the slit $S_1$,then the path difference $(S_1P - S_2P)$