Light of wavelength λ is incident normally on a thin film. The reflected light shows a minima for reflection. The thickness of film is minimum possible. Now the same film is placed in front of upper slit of the YDSE apparatus. The fringe pattern shifts up by 7.5 mm, the refractive index of material is: (Take: D = 5 m, d = 0.1 mm, λ = 500 nm).
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Question

Light of wavelength λ is incident normally on a thin film. The reflected light shows a minima for reflection. The thickness of film is minimum possible. Now the same film is placed in front of upper slit of the YDSE apparatus. The fringe pattern shifts up by 7.5 mm, the refractive index of material is: (Take: D = 5 m, d = 0.1 mm, λ = 500 nm).

Options

G1 Physics · Accepted Answer

The thinnest film giving minimum reflected intensity at normal incidence satisfies 2μt=λ, so t=λ/(2μ). Placing it before one slit gives optical path difference (μ−1)t. The fringe shift is y=[(μ−1)t]D/d. Thus 7.5×10⁻³=[(μ−1)/μ](λD/2d). With λ=500 nm, D=5 m and d=0.1 mm, λD/(2d)=12.5 mm. Hence (μ−1)/μ=7.5/12.5=0.6, giving μ=2.5. Option B.