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Question

In a Young's double slit experiment, the slits S1 and S2 are illuminated by a parallel beam of light of wavelength 4000 ˙A from the medium of refractive index μ1=1.2. A thin film of thickness 1.2 μm and refractive index μ=1.5 is placed in front of S1 perpendicular to the path of light. The refractive index of the medium between the plane of slits and screen is μ2=1.4. If the light coming from S1 and S2 have equal intensity I, then the intensity at the geometrical centre of screen O is:


A
0
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B
2I
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C
4I
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D
I
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Solution

The correct option is B 2I
Path difference between interfering waves at O is,

Δx=[S1O+(μrel1)t]S2O

S1O=S2O

Δx=(μrel1)t

Δx=(μμ21)t

Therefore, phase difference at O is given by:
Δϕ=(2πλ)Δx

Δϕ=(2πλ2)(μμ21)t

Where, λ2 is the wavelength of light in medium between slits and screen.

Using Snell's law,
μ1λ1=μ2λ2

λ2=μ1λ1μ2

Δϕ=2πμ2μ1λ1(μμ21)t

Δϕ=2π(1.4)1.2(4000×1010)(1.51.41)1.2×106

Δϕ=2π4=π2

Therefore, resultant intensity at O is given by,

IR=I1+I2+2I1I2cosΔϕ

IR=I+I+2IIcosπ2

IR=2I

Hence, option (B) is correct.

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