In a modified YDSE a monochromatic uniform and parallel beam of light of wavelength $6000 Å$ and intensity $(10 / π) W / m^{2}$ 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 figure). Calculate the power in watts received at the focal spot F of the lens. The lens is symmetrically placed w.r.t. the apertures. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.

Difficult

Solution

The power transmitted by apertures A and B are
$P_{A} = (\frac{10}{π}) [π (0.001)^{2}] = 10^{- 5} W$
$P_{B} = (\frac{10}{π}) [π \times (0.002)^{2}] = 4 \times 10^{- 5} W$
Only 10% of transmitted power reaches the focus.
$P_{A}^{'} = 10^{- 5} \times \frac{10}{100} = 10^{- 6} W$
$P_{B}^{'} = 4 \times 10^{- 5} \times \frac{10}{100} = 4 \times 10^{- 6} W$
The resultant power at the focus after superposition of two waves is
$P = P_{A}^{'} + P_{B}^{'} + 2 \sqrt{P_{A}^{'} P_{B}^{'}} \cos ϕ$
where $ϕ$ is phase difference
The introduction of mica sheet in the path of A creates a path difference $(μ - 1) t .$
$(μ - 1) t = (1.5 - 1) \times 2000 Å = 1000 Å$
Phase difference
$Δ ϕ = \frac{2 π}{λ} (μ - 1) t = \frac{2 π}{6000} \times 1000 = \frac{π}{3}$
Therefore
$P = 10^{- 6} + 4 \times 10^{- 6} + 2 \sqrt{(10^{- 6}) (4 \times 10^{- 6})} \cos \frac{π}{3}$
$= 7 \times 10^{- 6} W$

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