In Young's double slit experiment,light from two identical sources is superimposing on a screen. The path difference between the two light waves reaching a point on the screen is $\frac{7 \lambda}{4}$. The ratio of the intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

In Young's double slit experiment,when the wavelength used is $6000 \ \mathring{A}$ and the screen is $40 \ cm$ from the slits,the fringes are $0.012 \ cm$ wide. What is the distance between the slits in $cm$?

Assertion: In Young's experiment,the fringe width for dark fringes is different from that for bright fringes.
Reason: In Young's double slit experiment,if the fringes are performed with a source of white light,then only black and bright fringes are observed.

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:

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 Young's double-slit experiment,monochromatic light is used to illuminate two slits $A$ and $B$. Interference fringes are obtained on a screen in front of the slits. Now,if a thick glass plate is placed in the path of light from one of the slits,then: