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 wavelength of 6000 Å is placed in front of aperture A (see Fig.). 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.
Text Solution
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Sol. The power transmitted by apertures A and B are
P A = 
[ π (0.001) 2 ] = 10 –5 W
P B = 
[ π (0.002) 2 ] = 4 × 10 –5 W
Only 10% of transmitted power reaches the focus.
= 10 –5 × 
= 10 –6 W
= 4 × 10 –5 × 
= 4 × 10 –6 W
The resultant power at the focus after superposition of two waves is
P = 
+ 
+ 2 
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) × 2000 Å = 1000 Å
Phase difference Δφ = 
(µ – 1)t = 
× 1000 = 
Therefore, P = 10 –6 + 4 × 10 –6 + 2 
cos 
= 7 × 10 –6 W.
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