Find the ratio of intensities at the two points `X and Y` on a screen in Young's double slit experiment, where waves from the two source `S_(1) and S_(2)` have path difference of zero, and `lambda//4` respectively.

Explanation

Data: `Delta_(1)=0, Delta_(2)= lambda/6`,
Phase difference `(phi) = (2pi)/lambda xx` Path difference `(Delta)`
`therefore phi_(1) = (2pi)/lambda xx Delta_(1)=0` and `phi_(2) = (2pi)/(lambda) xx Delta_(2) = (2pi)/(lambda) xx lambda/6 = pi/3` rad
Assuming equal intensities `I_(0)` of the two interfering waves, the intensity at a point on the screen is
`I=4I_(0)cos^(2)phi/2`
`therefore I_(P)/I_(Q) = cos(phi_(1)/2)/cos(phi_(1)/2) = ((cos 0)/(cospi//6))^(2) = (1/(sqrt(3)//2))^(2) = 4/3`
`therefore` the ratio of the intensities at the points P and Q is `4:3`.
`I= 2I_(0)(1 + cosphi) therefore I_(P)/I_(Q) = (1+cosphi_(1))/(1+cosphi_(2)) = (1+cos0)/(1+ cospi//3)`
`=2/(1+0.5) = 2/(3//2) = 4/3`