In Young's double-slit experiment,if the separation between the slit and the screen increases,the fringe width:

In Young's double slit experiment,if the two slits are illuminated with separate sources,no interference pattern is observed because

$A$ beam of light consisting of wavelengths $650 \ nm$ and $550 \ nm$ illuminates the Young's double slits with separation of $2 \ mm$ such that the interference fringes are formed on a screen, placed at a distance of $1.2 \ m$ from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is . . . . . . $\times 10^{-5} \ m$.

The path difference between two identical light waves at a point $Q$ on the screen is $3 \mu m$. If the wavelength of the waves is $5000 \text{ Å}$,then at point $Q$ there is:

In a $YDSE$,bichromatic light of wavelengths $\lambda_1 = 400\, nm$ and $\lambda_2 = 560\, nm$ is used. The distance between the slits is $d = 0.1\, mm$ and the distance between the plane of the slits and the screen is $D = 1\, m$. The minimum distance between two successive regions of complete darkness is.....$mm$.

$A$ Young's double slit experiment apparatus has slits separated by $0.2 \ mm$ and a screen $60 \ cm$ away from the slits. The whole apparatus is immersed in a liquid medium of refractive index $\mu = \frac{11}{9}$ and the slits are illuminated with green light $(\lambda = 550 \ nm$ in vacuum$)$. Find the fringe width of the pattern formed on the screen. (in $mm$)