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Q.

A thin glass slab G1 is held over a large glass slab G2, creating an air gap of uniform thickness  t=0.5μm between them. Electromagnetic wave having wavelengths ranging from  0.4 μm  to 1.15 μm is incident normally on the slab G1. When interference between waves reflected from boundaries of air gap (the two reflected waves are shown in fig as R1 and R2) was studied, it was found that only two wavelengths interfered constructively. One of these two wavelengths is  λ1=0.04μm. Find the other wavelength (λ2) that interferes constructively. (In micrometers)

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answer is 0.67.

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Detailed Solution

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Let thickness of air gap be t. 
On reflection R2 suffers a phase change of π.
  Condition for constructive interference is
2t=(2n1+1)λ12  where λ1=0.4μm                  2t=(2n1+1)0.2  ....(1)

For other wavelength

2t=(2n2+1)λ22 ....(2)

From (1),  2×0.5μm=(2n1+1)0.42μm
           n1=2
According to the question;  λ2>λ1
(Since λ1=0.4μm  is the smallest incident wavelength)

             n2<n1             n2=1            λ2=4t2n2+1=4×0.5μm3=0.67μm

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