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Consider the arrangement shown in figure...

Consider the arrangement shown in figure. By some mechanism,
the separation between the slits `S_3 and S_4` can be changed. The intensity is
measured at the point P which is at the common perpendicular bisector of `S_1 S_2`
and `S_3 S_4`. When `z=(Dlambda)/(2d)`, the intensity measured at P is I. Find the intensity when
z is equal to

`(a)(Dlambda)/d (b)(3Dlambda)/(2d) (c) (2Dlambda)/d ` .

Text Solution

Verified by Experts

The correct Answer is:
2
Where `Z= (Dlambda)/(2d)=(beta)/(2) rArr OS_(4) = (beta)/(4)` as shown.
If intensity at `'p'` is `I` then intensity of light at `S_(3)` and `S_(4)` is `I//4 & I//4`
`therefore` Path difference `S_(4) P-S_(3)P=0`
So, intensity of slits `S_(1)` and `S_(2)`
`Deltaphi` at `S_(4)`
`Deltap=(yd)/(D)=(d)/(D)((z)/(2))=(lambda)/(4)`
`Deltaphi=(2pi)/(lambda).(lambda)/(4)=(pi)/(2)`
`(I)/(4)-I_(R)=I_(1)+I_(1)+2sqrt(I_(1)I_(1)) cos' (pi)/(2)`
`I_(1)=I_(2)=(I)/(8)` intensity of `S_(1)` and `S_(2).`
If `z=4.(lambdaD)/(d)=4beta`
`Deltap=(yd)/(D)=(z)/(2).(d)/(D)=((4lambdaD)/(d))xx(1)/(2)xx(d)/(D)=2lambda`
`Deltaphi=(2pi)/(lambda).2d=4pi`
`I_(3)=I_(4)=(I)/(8)+(I)/(8)+(2I)/(8) cos 4pi=(1)/(2)`
at `'p'I_(p)=(I)/(2)+(I)/(2)+2sqrt(I/(2).(I)/(2)) cos 0^(@)`
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