In Young's double-slit experiment `d//D = 10^(-4)` (d = distance between slits, D = distance of screen from the slits) At point P on the screen, resulting intensity is equal to the intensity due to the individual slit `I_(0)`. Then, the distance of point P from the central maximum is `(lambda = 6000 Å)`

In young's double slit experiment, if the distance between the slits is halved and the distance between the slits and the screen is doubled, the fringe width becomes

In Young's double slit experiment, if d, D and lambda represent the distance between the slits, the distance of the screen from the slits and wavelength of light used respectively, then the fringe width is inversely proportional to

In Young's double slit experiment if d, D and lamda represent the distance between the slits, the distance of the screen from the slits and wavelenfth of light used respectively, then the band width is width is inversely proportional to

In Young's slit experiment, if the distance between two slits is halved and the distance between the slits and the screen is doubled then the width of fringe will be.....

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 equal to

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 Young's double - slit experiment, the distance between slits is d = 0.25 cm and the distance of the screen D = 120 cm from the slits. If the wavelength of light used is lambda=6000Å and I_(0) is the intensity of central maximum, then the minimum distance of the point from the centre, where the intensity is (I_(0))/(2) is kxx10^(-5)m . What is the value of k?