In a modified Young's double-slit experiment, a monochromatic uniform and parallel beam of light of wavelength `6000 Å` and intensity `(10//pi)` W `m^(-2)` is incident normally on two circular apertures A and B of radii 0.001 m and 0.002 m, respectively. A perfectly transparent film of thickness `2000 Å` and refractive index 1.5 for the wavelength of `6000 Å` is placed in front of aperture A (see the figure). Calculate the power (in mW) received at the focal spot F of the lens. Then lens is symmetrically placed with respect to the aperture. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.

Explanation

`I=10/pi W//m^(2)`
`P_(A)=IA_(A)=10/pixxpi(0.001)^(2)=10^(-5) W`
`P_(B)=IA_(B)=10/pixxpi(0.002)^(2)=4xx10^(-5) W`
The aperture allows only `10%` of power, hence
`P_(A)^(')=0.1xx10^(-5)=10^(-6)W`
`P_(B)^(')=0.1xx4xx10^(-5)=4xx10^(-6)W`
Film: `t=200 nm, mu=3//2, lambda=600 nm`
Path difference due ti film,
`Deltax=(mu-1)t=(3//2-1)xx200=100 nm`
Phase difference `phi=(2pi)/lambdaDeltax=(2pi)/600xx100=pi/3`
Power received at `F`
`P=P_(A)^(')+P_(B)^(')+2sqrt(P_(A)^(')-P_(B)^(')) cos phi`
`=10^(-6)+4xx10^(-6)+2sqrt(10^(-6)xx4xx10^(-6)) cos(pi//3)`
`=10^(-6)+4xx10^(-6)+2xx10^(-6)`
`=7xx10^(-6) W`