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

7

Explanation

The power transmitted through A
`P_(A) = [10% of (10)/(lambda)]xxpi(0.001)^(2)`
`= (10)/(100)xx(10)/(pi)xxpixx10^(-6)= 10^(-6)watt`
The power transmitted through 'B' is
`P_(B) = [10%` of `(10)/(lambda)]xxpi(0.002)^(2)`
`= (10)/(100)xx(10)/(pi)xxpixx4xx10^(-6)watt`
Let `phi` be the phase difference t = introduced by film
`phi = (2pi)/(lambda)x = (2pi)/(lambda)xx(mu-1)t = (2pi)/(6000xx10^(-10))xx0.5xx2000xx10^(-10)`
`phi = (pi)/(3)` redian. The power recevied at F
`P = P_(A)+P_(B)+2sqrt(P_(A)P_(B))cos phi`
`= 10^(-6)+4xx10^(-6)+2sqrt(10^(-6)xx4xx10^(-6))xx(1)/(2)`
`= 5xx10^(-6)+2xx10^(-6)= 7xx10^(-6) = 7mu W`