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 = lambda` , the screen will contain only one maximum
B. If `lambda lt d lt 2lambda`, at least one more maxcimum (besides the central masximum) will be obsetved on the screen
C. If the intensity of light falling on slit 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 intensitites of the observed dark and bright fringes will increases.

Answer

A, B

Explanation

If `d = lambda` , then maximum path difference (path difference is given by d sin `theta`) will be less than `lambda` . So there will be only central maximum on the screen, because in the equation d sin `theta = n lambda` , n can take only value. If `lambda lt d lt 2lambda` . 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 intiensity 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 fringe on screen will decreases to zero. The alternatives (c ) and (d) are not correct.