In Young's double slit experiment,the distance between two sources is $0.1 \, mm$. The distance of the screen from the sources is $20 \, cm$. The wavelength of light used is $5460 \, \mathring{A}$. The angular position of the first dark fringe is: (in $^\circ$)

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width is

In $YDSE$,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. Then,the fringe width:

$A$ 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 . . . . . . $\times 10^{-5} \ m$.

In a $YDSE$,bi-chromatic light of wavelengths $400 \, nm$ and $560 \, nm$ are used. The distance between the slits is $0.1 \, mm$ and the distance between the plane of the slits and the screen is $1 \, m$. The minimum distance between two successive regions of complete darkness is......$mm$.

In Young's double slit experiment shown in the figure,$S_1$ and $S_2$ are coherent sources and $S$ is the screen having a hole at a point $1.0 \, mm$ away from the central line. White light ($400 \, nm$ to $700 \, nm$) is sent through the slits. Which wavelength passing through the hole has strong intensity (in $, nm$)? (Given: $d = 0.5 \, mm$,$D = 50 \, cm$)