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

We can multiply intensity of light with area of circular aperture of slits to get power received by slits but it is given that only 10% of the received energy goes in original direction, hence we can write power of the sources as follows:
`P_1 = 10/100 xx I pi r_1^2 = 0.1 xx 10/pi xx pi(0.001)^2 = 10^(-6)W`
`P_2 = 10/100 xx I pi r_2^2 = 0.1 xx 10/pi xx pi (0.002)^(2) = 4 xx 10^(-6)W`
Path difference produced by the film can be written as follows:
`delta = (2pi)/lambda Delta x = (2pi)/6000 xx 1000 = pi/3`
Resultant intensity is given by the following formula:
`I = I_1 + I_2 + 2sqrt(I_1 I_2) cos delta`
We can rewrite the same relation for power due to proportionality.
`P = P_1 + P_2 + 2 sqrt(P_1P_2) cos delta`
On substituting values, we get the following :
`P = (1 + 4 + 2 sqrt(1 xx 4) cos pi/3) xx 10^(-6) = 7 xx 10^(-6) W`.