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.

Options

A. (a) `3xx10^-6W`
B. (b) `1xx10^-6W`
C. (c) `5xx10^-6W`
D. (d) `7xx10^-6W`

Answer

D

Explanation

The power transmitted by apertures A and B are
`P_A=(10/pi)[pi(0.001)^2]=10^-5W`
`P_B=(10/pi)[pixx(0.002)^2]=4xx10^-5W`
Only 10% of transmitted power reaches the focus.
`P_A^'=10^-5xx(10)/(100)=10^-6W`
`P_B^'=4xx10^-5xx(10)/(100)=4xx10^-6W`
The resultant power at the focus after superposition of two waves is
`P=P_A^'+P_B^'+2sqrt(P_A^'P_B^')cosphi`
where `varphi` is phase difference.
The introduction of mica sheet in the path of A creates a path difference `(mu-1)t`.
`(mu-1)t=(1.5-1)xx2000Å=1000Å`
Phase difference
`Deltaphi=(2pi)/(lambda)(mu-1)t=(2pi)/(6000)xx1000=pi/3`
Therefore
`P=10^-6+4xx10^-6+2sqrt((10^-6)(4xx10^-6))cospi/3`
`=7xx10^-6W`