In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

Young's double-slit experiment is performed in air and then performed in water. How does the fringe width change?

$A$ beam of electrons is used in a $YDSE$ experiment. The slit width is $d$. When the velocity of the electrons is increased,then

In Young's double slit experiment,if the widths of the slits are in the ratio $4 : 9$,the ratio of the intensity at maxima to the intensity at minima will be

In a $YDSE$ with two identical slits,when the upper slit is covered with a thin,perfectly transparent sheet of mica,the intensity at the centre of the screen reduces to $75\%$ of the initial value. The second minimum is observed to be above this point and the third maximum below it. Which of the following cannot be a possible value of the phase difference caused by the mica sheet?

In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to