In a double slit experiment,the distance between slits is increased $10$ times whereas their distance from the screen is halved,then what is the fringe width?

In a Young's double slit experiment,the ratio of the slit widths is $4 : 1$. The ratio of the intensity of maxima to minima,close to the central fringe on the screen,will be

The ratio of the distance of the $n^{\text{th}}$ bright band and the $m^{\text{th}}$ dark band from the central bright band in an interference pattern is

Electrons accelerated from rest by an electrostatic potential are collimated and sent through a Young's double slit experiment. The fringe width is $\omega$. If the accelerating potential is doubled,then the width is now close to ............. $\omega$.

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 $YDSE$,$a=2 \, mm$,$D=2 \, m$,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 $\mu m$).