    Question

# ′n′ transparent slabs of refractive index 1.5 each having thickness 1 cm,2 cm,...n cm are arranged one over the other. A point object is seen through this combination from top with perpendicular light. If the shift of the object by combination is 5 cm. Then the value of ′n′ is

A
5
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B
5.00
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C
5.0
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Solution

## For a small angle of incidence the shift S is S=(1−1μ)Δt where μ is the refractive index, and Δt is the thickness. So the total shift due to all the slabs will be given as, S=S1+S2+S3+ . . . +Sn S=Δt1[1−1μ]+Δt2[1−1μ]+...+Δtn[1−1μ] As the thickness of the slabs is given as 1 cm,2 cm,...n cm we have S=[1−1μ](1+2+...+n) Now the refractive index is given as μ=1.5=32 and the total shift is given as S=5 cm 5=(1−23)(n(n+1)2) 30=n(n+1) 5×6=n(n+1) ⇒n=5  Suggest Corrections  0      Related Videos   Playing with Glass Slabs
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