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 of sources A and B are
`P_(A) = ((10)/(pi)) xx pi r_(A)^(2) = 10 xx (0.001) 2 = 10^(-5) W`
`P_(B) = ((10)/(pi)) xx pi r_(B)^(2) = 10 xx (0.002) 2 = 4 xx 10^(-5) W`
Power of sources A and B received along F are
`P_(A) = (10)/(100) (4 xx 10^(-5)) = 10^(-6) W`
`P_(B) = (10)/(100) (4 xx 10^(-5)) = 4 xx 10^(-6) W`
Path difference,
`Delta = (mu - 1) t = (1.5 - 1) xx 2000 Å = 1000 Å`
Phase difference,
`delta = (2pi)/(lambda) xx Delta = (2 pi)/(6000) xx 1000 = (pi)/(3)`
Power at point F is given by
`P_(F) = P_(A) + P_(B) + 2sqrt((P_A) P_(B)) cos delta`
` = 10^(-6) + 4 xx 10^(-6) + 2 sqrt(10^(-6) xx 4 10^(-6)) cos ((pi)/(3))`
`= 7 xx 10^(-6) W`