In a Young's double slit experiment, the separation between the two slits is d and the wavelength of the light is `lamda`. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice (s).

Options

A. If `d=lamda` the screen will contain only one maximum
B. If `lamda lt d lt 2lamda` at least one more maximum (besides the central maximum) will be observed on the screen.
C. If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase
D. if the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase.

Answer

A, B

Explanation

If `d=lamda`, then maximum path difference (path difference is given by `dsintheta`) will be less than or equal `lamda`. So there will be only central maximum on the screen, because in the equal d`sintheta=nlamda`, n can have onlyone value.
If `lamda lt d lt 2lamda,` then the maximum path difference will be less than `2lamda`. so there will be two more maximum on screen in addition to the central maximum. intensity of the dark fringes become zero when intensities at the two slits are equal. initial intensity at both the slits are unequal so there will some brightness as dark fringe. hence when intensity of both slits is made samme the intensity at dark fringe on screen will decrease to zero. the alternatives (c) and (d) are not correct.