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Question

A slab of transparent material is made as shown in the figure. The thickness of C is twice the thickness of B. If the number of waves in A is equal to the number of waves in combination of B and C, then the refractive index of B is


A
1.33
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B
1.8
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C
1.6
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D
1.4
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Solution

The correct option is B 1.8
Let the thickness of B=t
thickness of C=2t
thus, thickness of A=t+2t=3t
if λ0 is wavelength in air, then as number of waves in A is equal to number of waves in B + number of waves in C
3tλ0μ0=tλ0μ1+2tλ0μ2
3μ0=μ1+2μ2
μ1=3μ02μ2=3×1.42×1.2
μ1=1.8
Hence, refractive index of B is μ1=1.8

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