In the ideal double-slit experiment,when a glass plate (refractive index $\mu = 1.5$) of thickness $t$ is introduced in the path of one of the interfering beams (wavelength $\lambda$),the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

In Young's double slit experiment,we get $60$ fringes in the field of view using monochromatic light of wavelength $4000 \; \mathring{A}$. If we use monochromatic light of wavelength $6000 \; \mathring{A}$,then the number of fringes obtained in the same field of view is

In a Young's double slit experiment,slits are separated by $0.5 \ mm$,and the screen is placed $150 \ cm$ away. $A$ beam of light consisting of two wavelengths,$650 \ nm$ and $520 \ nm$,is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is...... $mm$.

In two separate set-ups of the Young's double slit experiment,fringes of equal width are observed when lights of wavelengths in the ratio $1:2$ are used. If the ratio of the slit separation in the two cases is $2:1$,the ratio of the distances between the plane of the slits and the screen in the two set-ups is

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

In Young's double-slit experiment,one of the slits is painted such that its intensity is half that of the other slit. Then: