Two slits are made one millimetre apart and the screen is placed one metre away from the slits. The fringe width when light of wavelength $500 \,nm$ is used is

In Young's double-slit experiment,$I_0$ is the intensity of the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the center is:

The fringe width in a $YDSE$ pattern is $2.4 \times 10^{-4} \, m$ when red light of wavelength $6400 \, \mathring{A}$ is used. How much will it change if blue light of wavelength $4000 \, \mathring{A}$ is used?

In $YDSE$, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If $W$ represents the original fringe width, the new fringe width is (in $\,W$)

In a Young's double slit experiment,$I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the central maximum will be

In Young's experiment,the separation between the $5^{\text{th}}$ maxima and the $3^{\text{rd}}$ minima is how many times the fringe width?