In Young.s double slit experiment, using monochromatic light of wavelength `lamda `, the intensity of light at a point on the screen where path difference is `lamda ` is K units. The intensity of light at a point where path difference is `lamda//3` is.

Options

A. 4K
B. K
C. 2K
D. `(K)/(4)`

Answer

D

Explanation

(d): Intensity at any point on the screen is
`I = 4I_(0) cos^(2)(phi)/2`
where `I_(0)` is the intensity of either wave and `phi` is the phase difference between two waves.
Phase difference, `phi = (2pi)/(lamda) xx` Path difference
When path difference is `lamda` then `phi = (2pi)/(lamda) xx lamda = 2pi`
`:. I = 4I_(0) cos^(2) ((2pi)/2) = 4I_(0) cos^(2) (pi) = 4I_(0) = K` ...(1)
When path difference is `(lamda)/3` , then
`phi = (2pi)/(lamda) xx (lamda)/3 = (2pi)/3`
`:. I = 4I_(0) cos^(2) ((2pi)/(3)) = K/4` [Using (i)]