In YDSE with `D = 1 m, d = 1 mm`, light of wavelength `500 nm` is incident at an angle of `0.57^(@)` w.r.t. the axis of symmetry of the experimental set-up. If center of symmetry of screen is O as shown figure.
a. find the position of centrla maxima,
b. find intensity at point O in terms of intensity of central maxima `I_(0)`, and
c. find number of maixma lying central maxima.
.

Explanation

(a) `theta = theta_(0) = 0.57^(@)`
`implies y = D tan theta = D theta`
`= -1 m xx ((0.57)/(57) rad) implies y = -1 cm`
(b) For point O, `theta = 0`
Hence, `Delta x = d sin theta_(0) = d theta_(0)`
`= 1 mm xx (10^(-2) rad)`
`= 10,000 nm = 20 xx (500 nm)`
`implies Delta x = 20 lambda`
Hence,point O corresponds to `20^(th)` maxima
= Intensity at `O = I_(0)`
(c) 19 maxima lie between central maxima and O, excluding maixma at O and central maxima.