A triple slit experiment (with basic conditions similar to a YDSE) is performed as shown in the figure with monochromatic light of amplitude a0. Consider light waves from S1, S2, S3 reaching a point P (shown by straight lines) on screen. S2M, S1N are perpendicular to S3P, S2P respectively. S3M≈S2N=δ . The graph between resultant amplitude of light (A) versus δ is plotted. Choose the CORRECT graph between options (A) and (B) and appropriate relation between A10 and A20 between options (C) and (D).
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a
b
c
A10+A20A10−A20=2 , where A10 and A20 are the values of local maxima of A in the graph (A10>A20)
d
A10+A20A10−A20=5 , where A10 and A20 are the values of local maxima of A in the graph (A10>A20)
answer is A.
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Detailed Solution
ϕ=2πλδLet say y2=a0sin(ωt−kx)⇒ y1=a0sin(ωt−kx−ϕ) and y3=a0sin(ωt−kx+ϕ)Resultant Amplitude=A=|a0+2a0cosϕ|=a0|(1+2cosϕ)|⇒ A=a0|(1+2cos(2πδλ))|For δ=0 , A10=a0|(1+2cos(0))|=3a0For δ=λ3 , A=a0|(1+2cos(2π3))|=0For δ=λ2 , A20=a0|(1+2cos(π))|=a0For δ=2λ3 , A=a0|(1+2cos(4π3))|=0∴A10+A20A10−A20=3a0+a03a0−a0=2