The slits in a Young's double-slit experiment have equal widths, and the source is placed symmetrically relative to the slits. The intensity at the central fringe is $I_0$. If one of the slits is closed, the intensity at this point will be

What happens to the fringe width in Young's double-slit experiment if it is performed in glycerine instead of air?

In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be:

In Young's experiment,the ratio of maximum to minimum intensities of the fringe system is $4:1$. The amplitudes of the coherent sources are in the ratio: (in $:1$)

In a Young's double slit experiment,a student observes $8$ fringes in a certain segment of the screen when a monochromatic light of $600 \ nm$ wavelength is used. If the wavelength of light is changed to $400 \ nm$,then the number of fringes he would observe in the same region of the screen 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