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Question

k transparent slabs are arranged one over another. The refractive indices of the slabs are μ1,μ2,μ3,μk and the thickness are t1,t2,t3,tk. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.

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Solution

Total no. of slab = k,

Thickness =t1,t2,t3,tk

Refractive index =μ1,μ2,μ3,μk

The shift Dt=[11μ1]t1+[11μ2]t2+[11μk]tk(i)

If μ refractive index of combination of slabs and image is formed at same place.

Δt=[1(1μ)](t1+t2)+tk(ii)

Equating (i) and (ii), we get (error).


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