In a modified YDSE a monochromatic uniform and parallel beam of light of wavelength $$6000$$ Å and intensity (10/$$π$$)W/m2$$ is incident normally on two circular apertures A and B of radii $$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.

Question

In a modified YDSE a monochromatic uniform and parallel beam of light of wavelength $$6000$$ Å and intensity (10/$$π$$)W/m2$$ is incident normally on two circular apertures A and B of radii $$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.

Infinity Learn · Accepted Answer

The power transmitted by apertures A and B $$arePA=10$$ππ(0.001)2=10$$−$$5WPB=10$$ππ×$$(0.002)2=4$$×$$10$$−$$5WOnly$$ $$10$$% of transmitted power reaches the focus.PA′$$=10$$−$$5$$×$$10100=10$$−$$6WPB$$′$$=4$$×$$10$$−$$5$$×$$10100=4$$×$$10$$−$$6WThe$$ resultant power at the focus after superposition of two waves $$isP=PA$$′$$+PB$$′$$+2PA$$′PB′$$cos$$⁡ϕwhere ϕ is phase differenceThe introduction of mica sheet in the path of A creates a path difference $$($$μ−$$1)t$$. $$($$μ−$$1)t=(1.5$$−$$1)×$$2000$$Å$$=1000$$ÅPhase differenceΔϕ$$=2$$πλ$$($$μ−$$1)t=2$$$$π$$$$6000$$×$$1000=$$π$$3ThereforeP=10$$−$$6+4$$×$$10$$−$$6+210$$−$$64$$×$$10$$−$$6cos$$⁡π$$3=7$$×$$10$$−$$6W$$

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