In Young's double slit experiment,when two light waves form the third minimum,they have:

In a Young's double-slit experiment,interference fringes are obtained on a screen placed at a certain distance from the slits using monochromatic light. If the screen is moved towards the slits by $5 \times 10^{-2} \, m$,the fringe width changes by $3 \times 10^{-5} \, m$. If the distance between the slits is $10^{-3} \, m$,find the wavelength of the light.

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 a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

In the Young's double slit experiment,the ratio of intensities of bright and dark fringes is $9$. This means that

In Young's double-slit experiment,monochromatic light is used to illuminate the two slits $A$ and $B$. Interference fringes are observed on a screen placed in front of the slits. If a thin glass plate is placed normally in the path of the beam coming from slit $A$,then: