In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

The figure shows a double slit experiment where $P$ and $Q$ are the slits. The path lengths $PX$ and $QX$ are $n\lambda$ and $(n + 2)\lambda$ respectively,where $n$ is a whole number and $\lambda$ is the wavelength. Taking the central fringe as zero,what is formed at $X$?

Light of wavelength $6000 \, Å$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. Find the angular position of the $10^{th}$ maximum in radians.

In $YDSE$ setup,light of wavelength $640 \, nm$ is used with $d = 0.8 \, mm$ and $D = 1 \, m$. If intensity at central maximum is $I_0$ and its position is $y = 0$,then:

In the Young's double slit experiment,the distance between the slits varies in time as $d(t) = d_{0} + a_{0} \sin \omega t$; where $d_{0}$,$\omega$,and $a_{0}$ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as: