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

(i) `theta=theta_0=57^@`
`impliesy=-D tanthetaapprox`
`-D theta=-1 meter times (0.57/57 rad)`
`implies y=-1 cm`
(ii) for point 0, `theta=0`
Hence, `Delta p= d sin theta_0`,
`d theta_0=1mmtimes(10^-2 rad)`
`=10,000nm`
`=20times(500nm)`
`impliesDeltap=20 lamda`
Hence point O corresponds to 20 th maxima
`implies` intensity at `O=I_0`
(iii) 19 maxima lie between central maxima and O, excluding maxima at O and central maxima