In a double slit experiment, the distance between the slits is $0.1 \ cm$ and the screen is placed at $50 \ cm$ from the slits plane. When one slit is covered with a transparent sheet having thickness $t$ and refractive index $n = 1.5$, the central fringe shifts by $0.2 \ cm$. The value of $t$ is . . . . . . $cm$.

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,a glass plate is placed before one slit which absorbs half the intensity of light. Under this case:

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 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:

$A$ plate of refractive index $1.6$ is introduced in the path of light from one of the slits in Young's double slit experiment,then: