In a Young's double slit experiment with light of wavelength `lamda` the separation of slits is d and distance of screen is D such that `D gt gt d gt gt lamda`. If the fringe width is `bea`, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is

In a double slit experiment the separation between the slits is d and distance of screen from slits is D. If the wavelenth of light used is lamda and I is the intensity of central bright fringe, then intensity at distance x from central maximum is given by

In Young’s double slit experiment, the wavelength of the light used is doubled and distance between two slits is half of initial distance, the resultant fringe width becomes

In a double slit experiment, the separation between the slits is d and distance of the screen from slits is D. If the wavelength of light used is lambda and I is theintensity of central bright fringe, then intensity at distance from central maximum is

In a Young's double slit experiment, D equals the distance of screen and d is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is same as that due to a single slit is equla to

In a Young's double slit experiment, slits are seperated by a distance d and screen is at distance D, (D gt gt d) from slit plane. If light source of wavelength lambda(lambda ltlt d) is used. The minimum distance from central point on the screen where intensity is one fourth of the maximum is (D lambda)/(nd) . Find the value of n.

In Young's double-slit experiment , d is separation between the slits. Separation between plane of the slits and screen is D. Wavelength of light used is lambda . Number of fringes per unit distance on the screen is