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 isnλd[n2λ2(μ−1)2−d24]12(n2λ2(μ−1)2−d22)(nλ(μ−1)−d2)

Solution

## The correct option is B [n2λ2(μ−1)2−d24]12 Δx=(S2O−S1O)optical=μS1O−S1O Δx=(μ−1)√d24+D2 For maximum intensity at O, the path difference should be nλ. Therefore, n2λ2=(μ−1)2(d24+D2) So,  D=[n2λ2(μ−1)2−d24]12

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