In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength `lamda`. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.

Options

A. `(lambda D)/(3d)`
B. `(lambda D)/(2d)`
C. `(lambda D)/(d)`
D. `(lambda D)/(4d)`

Answer

D

Explanation

`I=I_(0)cos^(2)((theta)/(2)), y = ((n lambda)D)/(d)`
`y = x. (D)/(d) = (lambda)/(4) (D)/(d)`