In a `YDSE` experiment the two slits are covered with a transparent membrane of negligible thickness which allows light to pass through it but does not allow water. A glass slab of thickness `t=0.41 mm` and refractive index `mu_(g)=1.5` is placed infront of one of the slits as shown in the figure. The separation between the slits is `d=0.30 mm`. The entire space to the left of the slits is filled with water of refractive index `mu_(w)=4//3`.
A coherent 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=1m` from the slits, find
(`a`) the position of central maxima.
(`b`) the intensity at point `O`.

Explanation

(`a`) Let central maxima lies at point `P` at a distance `x` from the central line. Then optical path difference `Deltap` at point `P` is
`(S_(2)P-t)+tmu_(g)-mu_(w)dsintheta-S_(1)P`
`rArr S_(2)P-S_(1)P+(mu_(g)-1)t-mu_(q)dsintheta`
` rArr (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)xx3xx10^(-4)xx(1)/(2)-0.5xx0.41xx10^(-3)]`
`=(1)/(3xx10^(-4))[2xx10^(-4)-2.05xx10^(-4)]=-(5xx10^(-6))/(3xx10^(-4))`
`= -1.66xx10^(-2)m=-1.66 cm`
Central maxima lies at a distance `1.66 cm` below the central line
(`b`) At point `O`, optical path difference is
`(mu_(g)-1)t-mu_(w)dsintheta=-5xx10^(-6)m`
So intensity at `O`
`I_(0)=I+I+2sqrt(II) cos"(2pi)/(5xx10^(-7))(-5xx10^(-6))`
`I_(0)=4I`