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.

Explanation

`I_(R)=4I^(') cos^(2) phi//2=I_(0) cos^(2) (phi//2)`
where `I_(1)=I_(2)=I^('), phi=(2pi)/lambdaDeltax, I_(0)`: intensity at centre
(a) `I_(R)=I_(0)/2=I_(0) cos^(2) (phi//2)implies cos(phi//2)=1/sqrt(2)= cos(pi//4)`
`phi=pi/2`
`(2pi)/lambda Deltax=pi/2`
`Deltax=lambda//4`
`(yd)/D=lambda/4 implies y=(Dlambda)/(4d)`
`(b) I_(R)=I_(0)/4=I_(0) cos^(2) (phi//2) implies cos(phi//2)=1/2=cos (pi//3)`
`phi=(2pi)/3`
`(2pi)/lambda Deltax=(2pi)/3`
`Deltax=lambda//3`
`(yd)/D=lambda/3 impliesy=(Dlambda)/(3d)`