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)/(2)lt d lt 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 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increases
D. if the intensity of light falling on slit 2 is reduced so that it becomes equal to that of slit 1, the intensikties of the observed dark and bright fringes will increase.

Answer

A

Explanation

Condition for maxima is `sin theta = n lambda`
If `d = lambda, sin theta = n` Possible value is `theta`
only one maxima will be obtained
(A) is correct. (B) is correct.
`I_(max) = (sqrt(I_(1)) + sqrt(I_(2)))^(2) & I_(min) = (sqrt(I_(1)) - sqrt(I_(2)))^(2)`
with `I_(1) = 4I_(2)`
`I_(max) = 9I_(2) , I_(min) = I_(2)`
but when `I_(1) = I_(2)`
`I_(max) = 4I_(2) and I_(min) = 0`
(C ) and (D ) are wrong.