In Young's double slit experiment,if the widths of the slits are in the ratio $4 : 9$,the ratio of the intensity at maxima to the intensity at minima will be

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

In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.

In Young's double slit experiment,the intensity at a point is $1/4$ of the maximum intensity. The angular position of this point is:

Light of wavelength $520\, nm$ passing through a double slit produces an interference pattern of relative intensity versus deflection angle $\theta$ as shown in the figure. The separation $d$ between the slits is

Two wavelengths of light $\lambda_1$ and $\lambda_2$ are sent through a Young's double-slit experiment simultaneously. If the third-order bright fringe of $\lambda_1$ coincides with the fourth-order bright fringe of $\lambda_2$,then