
The change in path difference due to the two slabs at P is Δx=(μ2−μ1)t
=(1.55−1.51)×0.1=0.04×10−4
⇒Δx=4×10−6=4μm
Now, the distance of central maxima from geometrical center is y=dΔxD=4×10−6dD
Now, fringe width, nβ=4×10−6dD
Or, dnλD=4×10−6dD
Number of shift, n=4×10−74×10−6=10.