Two slits separated by a distance of $1 \,mm$ are illuminated with light of wavelength $6.5 \times 10^{-7} \,m$. The interference fringes are observed on a screen placed at $1 \,m$ from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to (in $\,mm$)

In a double slit experiment,the distance between slits is increased $10$ times whereas their distance from the screen is halved,then what is the fringe width?

Two light waves of wavelengths $800 \, nm$ and $600 \, nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \, m$ away from the plane of the slits. If the two slits are separated by $0.35 \, mm$,then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be $............. \, mm$.

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 a Young's double slit experiment,the distance between the two identical slits is $6.1$ times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single slit diffraction pattern is

In a Young's Double slit experiment,the first maxima is observed at a fixed point $P$ on the screen. Now,the screen is continuously moved away from the plane of the slits. The ratio of the intensity at point $P$ to the intensity at point $O$ (centre of the screen) is: