The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

In a Young's double-slit experiment,the intensity at a point where the path difference is $\lambda$ is $K$. What will be the intensity at a point where the path difference is $\lambda / 4$?

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 an interference experiment,the distance between a maximum and the adjacent minimum is .......

Assertion: In Young's double slit experiment,if the wavelength of incident monochromatic light is doubled,the number of bright fringes on the screen will increase.
Reason: The maximum number of bright fringes on the screen is directly proportional to the wavelength of light used.

Assertion: In Young's double slit experiment,the two slits are at a distance $d$ apart. An interference pattern is observed on a screen at a distance $D$ from the slits. At a point on the screen directly opposite to one of the slits,a dark fringe is observed. Then,the wavelength of the wave is proportional to the square of the distance between the two slits.
Reason: For a dark fringe,the intensity is zero.