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Question


In YDSE experiment, half of the arrangement is submerged in fluid of refractive index μ as shown in figure. Value of D so that we get maximum intensity at O is
  1. nλd
  2. [n2λ2(μ1)2d24]12
  3. (n2λ2(μ1)2d22)
  4. (nλ(μ1)d2)


Solution

The correct option is B [n2λ2(μ1)2d24]12

Δx=(S2OS1O)optical=μS1OS1O

Δx=(μ1)d24+D2

For maximum intensity at O, the path difference should be nλ. Therefore,

n2λ2=(μ1)2(d24+D2)
So, 
D=[n2λ2(μ1)2d24]12

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