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

For point 0, `theta = 0`
Hence, `Delta p = d sin theta_(0)`,
`d theta_(0)=1 mm xx (10^(-2)"rad")`
`=10,000 nm`
`= 20xx(500 nm)`
`rArr Delta p = 20 lambda`
Hence point O corresponds to 20th maxima
`rArr` intensity at `O=I_(0)`