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A thin film of soap solution (μs=1.4) lies on the top of a glass plate (μg=1.5). When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420nm and 630nm. The minimum thickness of the soap solution is


  1. 420nm

  2. 500nm

  3. 450nm

  4. 490nm

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Solution

The correct option is C

450nm


Step 1: Given Data:

Refractive index of soap solution, μs=1.4

Refractive index of glass, μg=1.5

Two adjacent reflection maxima are observed at two wavelengths,

λ1=420nmλ2=630nm

Let the thickness of the soap solution is t.

Step 2: We have to find the minimum thickness of the soap solution is,

For reflection at the air-soap solution interface, the phase difference isπ.

For reflection at the interface of soap solution to glass also, there will be a phase difference of π.

The condition for maximum intensity:

2μt=nλ (where, μis the refractive index and n is order)

Step 3: For order n:

As we know, the formula of refractive index is n1λ1=n2λ2

For n,

nλ1=(n1)λ2

n×420=(n1)630

n×420=n×630-630

n×420-n×630=-630

n(420-630)=-630

n(-210)=-630

210n=630

n=630210

n=3

Step 4: Maximum order, where they coincide:

2×1.4×t=3×420

t=2×1.43×420

t=450nm

Hence, the correct option is (C).


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