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.

Answer

A

Explanation

The power transmitted through A
`=[10%of(10/(pi))]xxpi(0.001)^(2)=10^(-6)W`
The power transmitted through B
`=[10%of((10)/(pi))xxpixx(0.002)^(2)=4xx10^(-6)W`
let `trianglephi` be the phase difference introduced by film `:'``trianglephi=(2pi)/(lamda)` (path difference introduced by the film) `=(2pi)/(lamda)xx(mu-1)t=(2pi)/(6000xx10^(-10))[1.5-1]xx2000xx10^(-10)=(pi)/(3)` radian
The power received at F
`P=P_(1)+P_(2)+2sqrt(P_(1)P_(2))costrianglephi`
`=10^(-6)+4xx10^(-6)+2sqrt(10^(-6)4xx10^(-6))cos((pi)/(3))`
``7xx10^(-6)W`.