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 $............. \, mm$.

In a Young's double slit experiment, a small detector measures an intensity of illumination of $I$ units at the centre of the fringe pattern. If one of the two (identical) slits is now covered, the measured intensity will be

$A$ double slit interference experiment performed with a light of wavelength $600 \ nm$ forms an interference fringe pattern on a screen with $10^{\text{th}}$ bright fringe having its centre at a distance of $10 \ mm$ from the central maximum. The distance of the centre of the same $10^{\text{th}}$ bright fringe from the central maximum when the source of light is replaced by another source of wavelength $660 \ nm$ would be . . . . . . $mm$.

In a $Young's$ double-slit experiment,light of wavelength $6000 \, \mathring{A}$ is used. If the third dark fringe is formed at point $P$ on the screen,then the path difference $(S_1P - S_2P)$ is ......... $\mu m$.

In a Young's double-slit experiment, the distance between two coherent sources is $0.90 \, mm$ and the distance from the sources to the screen is $1 \, m$. If the distance of the second bright fringe from the center of the central bright fringe is $1 \, mm$, find the wavelength of the light used.

In Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?