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 2 lambda`, 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 is 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 silt 1, the intensitites of the observed dark and bright fringes will increases

Answer

A, B

Explanation

Maximum possible path difference obtained in YDSE is equal to separation between sources but such point is obtained at infinite separation from the centre, hence we can say that path difference will always be less than separation between sources.Hence when `d = lambda` then path difference will be less than `lambda` and therefore only maximum obtained will be at the centre . Hence option (a) is correct.
If `d < 2 lambda` then on either side of centre we shall get one maxima where path difference becomes `lambda`. In option (b) it is said that at least one more maxima is obtained and hence this option is correct. Precisely we shall get total 3 bright points including centre. Option (b) is correct.
When intensities of two sources are unequal, then we get non-zero intensity at dark point, but when intensities are made equal, then intensity at dark point becomes zero. Hence on making intensity at dark point, but when intensities are made equal, then intensity at dark point becomes zero. Hence on making intensity of two sources equal intensity of dark will decrease to zero and hence both the options (c ) and (d) are not correct.