The width of the fringe is $2\,mm$ on the screen in a double-slit experiment for light of wavelength $400\,nm$. The width of the fringe for light of wavelength $600\,nm$ will be $..............\,mm$.

Consider a Young's double slit experiment as shown in the figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit $S_1$?

In a double-slit experiment,the distance between the slits is $d$. The screen is at a distance $D$ from the slits. If a bright fringe is formed opposite to one of the slits,its order $n$ is:

In a Young's double slit experiment,$15$ fringes are observed on a small portion of the screen when light of wavelength $500 \; nm$ is used. Ten fringes are observed on the same section of the screen when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is (in $nm$):

In Young's double slit experiment,if the distance between the two slits is halved,then the fringe width will become

In a Young's double slit experiment,slits are separated by $0.5 \ mm$,and the screen is placed $150 \ cm$ away. $A$ beam of light consisting of two wavelengths,$650 \ nm$ and $520 \ nm$,is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is...... $mm$.