Suppose that one of the slit in Young's experiment is wider than the order, so the amplitude of the light reaching the central part of the screen from one slit, acting alone, is twice that from the other slit, acting alone. Obtain an expression for the light intensity I at the screen as a function or `theta` the angular position.

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]`