The minimum thickness of a mica sheet having refractive index $\mu = 3/2$,which should be placed in front of one of the slits in $YDSE$ to reduce the intensity at the center of the screen to half of the maximum intensity,is:

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 a Young's double-slit experiment,let $A$ and $B$ be the two slits. $A$ thin film of thickness $t$ and refractive index $\mu$ is placed in front of $A$. Let $\beta =$ fringe width. The central maximum will shift:

In a $YDSE$,light of wavelength $\lambda = 5000 \; \mathring{A}$ is used,which emerges in phase from two slits separated by a distance $d = 3 \times 10^{-7} \; m$. $A$ transparent sheet of thickness $t = 1.5 \times 10^{-7} \; m$ and refractive index $\mu = 1.17$ is placed over one of the slits. What is the new angular position of the central maxima of the interference pattern,and what is its linear position $y$ from the center of the screen?

In Young's double-slit experiment performed using a monochromatic light of wavelength $\lambda$,when a glass plate $(\mu=1.5)$ of thickness $t = x \lambda$ is introduced in the path of one of the interfering beams,the intensity at the position where the central maximum occurred previously remains unchanged. The value of $x$ will be..........

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 $(S_1P - S_2P)$