Q.

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=2πλδ​Let 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
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