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JEE Main 2022 · Physics · Waves and Sound · medium

A thin soap film of refractive index 4/3 in air is illuminated at normal incidence by light of wavelength 640 nm. The minimum non-zero thickness of…

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A thin soap film of refractive index 43\dfrac43 in air is illuminated at normal incidence by light of wavelength 640 nm640\ \mathrm{nm}. The minimum non-zero thickness of the film for which the reflected light appears brightest is
  1. A. 80 nm80\ \mathrm{nm}
  2. B. 120 nm120\ \mathrm{nm}
  3. C. 160 nm160\ \mathrm{nm}
  4. D. 240 nm240\ \mathrm{nm}
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Answer: (B) 120 nm120\ \mathrm{nm}

Reflection from the top surface (air to film) introduces a phase change of π\pi; reflection at the lower surface (film to air) does not. So the condition for constructive interference in reflection is 2μt=(m+12)λ2\mu t=\left(m+\tfrac12\right)\lambda. The smallest non-zero thickness corresponds to m=0m=0: t=λ4μ=6404×4/3=120 nmt=\dfrac{\lambda}{4\mu}=\dfrac{640}{4\times4/3}=120\ \mathrm{nm}. Using 2μt=λ2\mu t=\lambda (as for transmission) would give 240 nm240\ \mathrm{nm}, the thickness that appears dark.
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