In a Young's double slit experiment, the fringe width will remain same, if ( D = distance between screen and plane of slits, d = separation between two slits and `lambda` = wavelength of light used)

What is interference ? In Young's double-slit experiment, show that the fringe width beta for interference fringes is beta = (lambda D)/d Where D is the distance of the screen from the slit, d is the distance between two slits and lambda is the wavelength of light used.

In Young's double slit experiment, the distance between the two slits is 0.1 mm, the distance between the screen and the slit is 100 cm. if the width of the fringe is 5 mm, then the wavelength of monochromatic light used is

In Young's double slit experiment , the distance between two slits is 0.1 mm . The distance between the screen and slits is 20 cm . If the slit is illuminated by a light of wavelength 5460 overset (@)(A) then the angular position of the first dark fringe is

In Young's double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

In Young's double-slit experiment, the separation between the slits is halved and the distance between the slits and the screen in doubled. The fringe width is

In Young's double-slit experiment, the separation between the slits is halved and the distance between the slits and the screen in doubled. The fringe width is

In Young's double slit experiment, the sepcaration between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is