Maximum intensity in YDSE is `I_0`. Find the intensity at a
point on the screen where
(a) The phase difference between the two interfering beams is `pi/3.`
(b) the path difference between them is `lambda/4`.

Explanation

(a) We have `I = I_(max) cos^2 (phi/2) `. Here , `I_(max)` is `I_0`
(i.e., intensity due to independent sources is `I_0//4` )
(b) Phase difference corresponding to the given path difference `Delta x = lamda/4` is
`phi = ((2pi)/(lamda)) (lamda/4) ( because phi = (2pi)/(lamda)Delta x) = pi/2 " or " phi/2 =pi/4`
`I = I_0 cos^2 (pi/4) = (I_0)/(2)`