In Young's double slit experiment, one of the slit is wider than other, so that the amplitude of the light from one slit is double of that from other slit. If `I_(m)` be the maximum intensity, the resultant intensity I when they interfere at phase difference `phi` is given by

Explanation

The expression for resultant intensity at any point is
`I = k (A_(1)^(2) + A_(1)^(2) + 2A_(1) A_(2) cos phi)`
where k is constant of proportinality, `phi` is the difference between interfering waves, `A_(1)` and `A_(2)` are amplitudes of superposing waves. As `A_(1) = 2 A_(2) = 2A`, we have the intensity at the central position.
`I_(0) = k (A_(1) A_(2))^(2) = k (2 A + A)^(2) = 9 k A^(2)`
So, `k = (I_(0))/(9 A ^(2))`
The path difference at any point P is `Delta x = d sin theta`
The phase difference `(2 pi)/(lambda) d sin theta`
Thus, Eq. (i) becomes
`I = (I_(0))/(9 A^(2)) [ (2A)^(2) + A^(2) + 2(2A) (A) cos ((2 pi d sin theta)/(lambda))]`
`= (I_(0))/(9A^(2)) [(5A^(2) +4A^(2) cos ((2 pi sin theta)/(lambda))]`
`= (I_(0))/(9) [(1 + 8 cos^(2)) (pi d sin theta)/(lambda)]` as `cos 2 theta = [ cos^(2) .(theta)/(2) - 1]`