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=lambda,` the screen will contain only one maxima
B. If `lambda lt d lt 2lambda,` at least one more maximum (besides the central maximum) will be observed on the screen
C. If the intensity of light 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=lambda,` then maximum path difference will be less than `lambda.` So there will be only central maximum on the screen.
If `lambdaltdlt2lambda,` then the maximum path difference will be less than `2lambda`.
So there will be two more maximum on screen corresponding to path difference `Deltax=lambda`
So `(A)` and `(B)` are correct.
Intensity of dark fringe becomes zero when intensities of two slits are equal. Initial intensity at both the slits are unequal so there will some brightness at dark fringe. Hence when intensity of both slits is made same the intensity at dark on screen shall decrease to zero. So both `(C)` and `(D)` are false.