Shift of fringes when a slab/lens in introduced in front of a slit
Example: n transparent slabs of refractive index 1.5 each, having thickness 1cm,2cm,.......... to ncm are arranged one over another. A point object is seen through this combination with the perpendicular light. If the shift of object by the combination is 1cm then find the value of n.
Solution: As the refractive index of all the slabs are 1.5 so those may be a considered as a single slab with the thickness summed up. Summed up thickness of n slabs Δt=1+2+3+...+n=2n(n+1) Given, for small angle of incidence, shift of object: s=(1−μ1)Δt so, ⇒1cm=(1−1.51)2n(n+1) ⇒6=n(n+1) ⇒n2+n−6=0 ⇒n2+3n−2n−6=0 ⇒n(n+3)−2(n+3)=0 ⇒(n+3)(n−2)=0 So, n=−3 or n=2 As, n can not be negative. so n=2