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=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, C, D

Explanation

The condition to obtain maxima in the phenomenon observed in young's double slit experiment is
`dsintheta=nlamda` where n is an integer
`lamdasintheta=nlamda`
`implies` `sintheta=n`
When `n=0`,`theta=0`
When `n=1`,`theta=90degree` (This will be a point on the screen which will be at infinity and therefore not practical) Other values of n are invalid as `-1lesinthetale1`.
`implies` The screen will have only one maxima.
When `lamdaltdlt2lamda`
`implies` `lamdalt(nlamda)/(sintheta)lt2lamda` `[d=(nlamda)/(sintheta)]`
`implies` `1lt(n)/(sintheta)lt2`
The possible values of n are 0, +1, _1.
`implies` There is at least one more maxima (besides the central maxima, option[B]is correct.
We know that `I_(max)=(sqrtI_(1)+sqrtI_(2))^(2)`,`I_(min)=(sqrtI_(1)-sqrtI_(2))^(2)`
Initially `I_(2)=4I` and `I_(2)=I`
`:'` `I_(max)=9 I` and `I_(min)=I`
When `I_(1)=I_(2)=I` then `I_(max)=4I` and `I_(min)=0`
i.e., when the intensities become equal, `I_(min)` reduces to zero. Options [C] and [D] are incorrect.