A coherent beam light of intensity I and absolute wavelength `lambda= 5000Å` is being incident on the slits making an angle `30^@` with horizontal. If screen is placed at a distance D=1 m from the slits and the separation between- the slits is `3 xx 10^(-4) m`. A thin film of glass `mu_(g)`= 3/2 and thickness 0.41 mm is placed near one slit as shown in the figure, find
(a) the position of central maxima.
(b) the intensity at'point O.

Explanation

Let central amxima lies at a point P at a distance x from the central line. Then optical path diffenrece `Delta`p at a point P is:
`(S_(2)P - t) + tmu_(g) - mu_(w)dsintheta - S_(1)p`
`implies S_(2)P - S_(1)P + (mu_(g) -1) t - mu_(w)dsintheta`
`implies (xd)/D + (mu_(g) -1 )t - mu_(w)dsintheta`
For central maxima `Deltap=0`
`x=D/d[(mu_(w)dsintheta) - (mu_(g)-1)t]`
`1/(3xx10^(-4)) [4/3 xx3 xx10^(-4) xx 1/2 -0.5 xx 0.41 xx10^(-3)]`
`= (5xx10^(-6))/(3xx10^(-4)) = -1.66xx10^(-2)m=-1.66m`
So central maxima lies at a distance 1.66 cm below the central line
(b) At pointO , optical path difference is
`(mu_(g)-1)t - mu_(w)dsintheta=-5xx10^(-6)m`
So intensity at 0,
`I_(0) =I + I + 2sqrt(II) "cos" (2pi)/(5xx10^(-7))(-5xx10^(-6))=4I`