In Young's double-slit experiment,the maximum intensity is $I_0$. If one slit is closed,the new maximum intensity is:

Two light waves of wavelengths $800 \, nm$ and $600 \, nm$ are used in Young's double slit experiment to obtain interference fringes on a screen placed $7 \, m$ away from the plane of the slits. If the two slits are separated by $0.35 \, mm$,then the shortest distance from the central bright maximum to the point where the bright fringes of the two wavelengths coincide will be $............. \, mm$.

In a Young's double-slit experiment,the distance between the two sources is $0.1 / \pi \, mm$. The distance between the source and the screen is $25 \, cm$. The wavelength of light is $5000 \, \mathring{A}$. The angular position of the first dark fringe is ........$^o$.

In Young's double-slit experiment,the width of one slit is twice that of the other. What is the ratio of the maximum to the minimum intensity?

The ratio of intensities at two points $P$ and $Q$ on the screen in a Young's double-slit experiment,where the phase differences between two waves of the same amplitude are $\pi / 3$ and $\pi / 2$ respectively,is:

In $YDSE$,the $Y$-coordinates of the central maxima and the $10^{th}$ maxima are $2 \, cm$ and $5 \, cm$ respectively. When the $YDSE$ apparatus is immersed in a liquid of refractive index $1.5$,what will be the corresponding $Y$-coordinates?