In a Young's double slit experiment, the separaton between the two slits is d and the wavelength of the light is `lambda`. The intensity of light fallin 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 `lamdaltdlt2lamda`, 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

Given d, `lamda`, `I_(1)=4I_(2),I_(1)=4I&I_(2)=I`
if `d=lamda` then maximum path difference `(dsintheta)` will be less than `lamda`. So there will be only central maxima on the screen, because in the equation d `sintheta=nlamda`, n can take only one value if `lamdalt d lt 2lamda`, then the maximum path difference will be less tha `2lamda`. So there will be two more maximum on screen in additio to the central maximum intensity of dark fringes becomes zero if intensities at the two slits made are equal. [so C & D are not corret]