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Question

In the arrangement shown in figure, slits S1 and S4 are having a variable separation Z. Point O on the screen is at the common perpendicular bisector of S1S2 and S3S4.
When Z=λD2d the intensity measured at O is I0. The intensity at O when Z=2λDd is

A
I0
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B
2I0
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C
3I0
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D
4I0
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Solution

The correct option is B 2I0

Intensity at IS3=I+I+2IIcosθ ...(1)
here,θ=phase difference=k×Δx
wherek=2πλ,Δx=λD4d

then,
θ=2πλ×dD×λD4dθ=π2
so, IS3=I+I+2IIcosπ2=2I

in the given confriguration S3=S4, because of coherent sources.
so, IS4=2I

resultant intensity from S3 and S4
I0=IS3+IS4+2IS3IS3 =2I+2I+22I2I =8I.....(2)

Case-2:
Zinitial=λd2D
Zfinal=2λdD
so,by the factor 4 the intensity changes.

hence, new phase difference,θ=4×π2=2π
Intensity from source IS3=I+I+2IIcos2πIS3=4I
so, as IS4=4I
Noe the resultant intensity:
I0=IS3+IS4+2IS3IS3 =4I+4I+24I4I =16I ....(3)

On comparing eq. (2) and eq.(3),
Iinitial=8I=I0
Iinitial=16I=I
so, I=2I0


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