In Young's double-slit experiment,if the separation between the slit and the screen increases,the fringe width:

In Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ bright fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is:

Two ideal slits $S_1$ and $S_2$ are at a distance $d$ apart and are illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_2$ as shown. The distance between the plane of the slits and the source slit is $D$. $A$ screen is held at a distance $D$ from the plane of the slits. The minimum value of $d$ for which there is darkness at $O$ is

In Young's double slit interference experiment,the slit widths are in the ratio $1 : 25$. Then the ratio of intensity at the maxima and minima in the interference pattern is

In Young's double slit experiment, two slits $S_1$ and $S_2$ are $d$ distance apart and the separation from slits to screen is $D$. If two transparent slabs of equal thickness $t = 0.1 \text{ mm}$ but refractive indices $\mu_1 = 1.51$ and $\mu_2 = 1.55$ are introduced in the path of the beam $(\lambda = 4000 \text{ Å})$ from $S_1$ and $S_2$ respectively, the central bright fringe will shift by how many fringes?