In $YDSE$, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If $W$ represents the original fringe width, the new fringe width is (in $\,W$)

In a $YDSE$,light of two different wavelengths $(\lambda_1, \lambda_2)$ is incident normal to the plane of slits. The $n^{th}$ maxima of $\lambda_1$ coincides with the $m^{th}$ maxima of $\lambda_2$ exactly in front of one of the slits. Given $D = 1.5 \ m$,$d = 3 \ mm$,and $4500 \ \mathring{A} < \lambda_1, \lambda_2 < 7000 \ \mathring{A}$,then $n, m$ and $\lambda_1$ are:

In a $YDSE$,if the slits are of unequal width:

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

In $YDSE$,$a=2 \, mm$,$D=2 \, m$,and $\lambda=500 \, nm$. Find the distance of the point on the screen from the central maxima where the intensity becomes $50 \%$ of the central maxima (in $\mu m$).

In a $YDSE$ apparatus,if we use white light,then: