In a modified YDSE a monochromatic uniform and parallel beam of light of wavelength 6000A∘ and intensity (10/π) W/m2 is incident normally on two circular apertures A and B of radii 0.001m and 0.002m respectively. A perfectly transparent film of thickness 2000A∘ and refractive index 1.5 for wavelength 6000A∘ is placed in front of the aperture A(see Fig.). Calculate the power in watts received at the focal spot of the lens. The lens is symmetrically placed w.r.t. the apertures. Assume that 10% of the power received to each aperture goes in the original direction and is brought to the focal spot.

The power transmitted by apertures A and B arePA=10ππ0.0012=10-5W PB=10ππ0.0022=4×10-5WOnly 10% of transmitted power reaches the focus.PA'=10-5×10100=10-6W PB'=4×10-5×10100=4×10-6WThe resultant power at the focus after superposition of two waves isP=PA'+PB'+2PA'PB' cos ϕwhere 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)×2000A∘=1000A∘Phase difference ∆ϕ=2πλ(μ-1)t=2π6000×1000=π3Therefore, P=10-6+4×10-6+210-64×10-6 cosπ3=7×10-6W.

In a modified YDSE a monochromatic uniform and parallel beam of light of wavelength 6000A∘ and intensity (10/π) W/m2 is incident normally on two circular apertures A and B of radii 0.001m and 0.002m respectively. A perfectly transparent film of thickness 2000A∘ and refractive index 1.5 for wavelength 6000A∘ is placed in front of the aperture A(see Fig.). Calculate the power in watts received at the focal spot of the lens. The lens is symmetrically placed w.r.t. the apertures. Assume that 10% of the power received to each aperture goes in the original direction and is brought to the focal spot.

The power transmitted by apertures A and B arePA=10ππ0.0012=10-5W PB=10ππ0.0022=4×10-5WOnly 10% of transmitted power reaches the focus.PA'=10-5×10100=10-6W PB'=4×10-5×10100=4×10-6WThe resultant power at the focus after superposition of two waves isP=PA'+PB'+2PA'PB' cos ϕwhere 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)×2000A∘=1000A∘Phase difference ∆ϕ=2πλ(μ-1)t=2π6000×1000=π3Therefore, P=10-6+4×10-6+210-64×10-6 cosπ3=7×10-6W.