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

The intensities of sources A and B are
` I_(A) = ((10)/(pi)) xx pi r_(a)^(2) = 10 x (0.001)^(2) = 10^(5) W`
`I_(B) = ((10)/(pi)) xx pi r _(B)^(2) = 10 xx (0.002)^(2) = 4 xx 10^(-5) W`
Intensities of sources A and B recevied along F are
`I_(A) = (10)/(100) (1 xx 10^(-5)) = 10(-6) W`
`I_(B) = (10)/(100) (4 xx 10^(-5)) = 4 xx 10^(-6)`
Path difference,
`Delta = (mu - 1) t = (1.5 - ) xx 2000 Å = 1000 Å`
Phase difference,
`delta = (2pi)/(lambda) xx Delta = (2 pi)/(6000) xx 1000 = (pi)/(3)`
Intensity at Point F is given by
`I_(F) = a_(A)^(2) + a_(A)^(2) + 2a_(A) a_(B) cos delta`
`= I_(A) + I_(B) + 2sqrt I_(A) I_(B) cos delta`
`10^(-6) + 4 xx 10^(6) + 2 sqrt (10^(-6) xx 4 xx 10^(-6)) cos ((pi)/(3))`
`10^(-6) + 4 xx 10^(-6) + 2 xx 2 xx 10^(-6) xx (1)/(2)`
`= 10^(-6) + 4 xx 10^(-6) + 2 xx 10^(-6) = 7 xx 10^(-6) W`.