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JEE Main 2021 · Physics · Waves and Sound · medium

In a Young's double slit experiment d=0.5 mm, D=1 m and λ=600 nm. A thin transparent film of refractive index 1.5 and thickness 6 mum is placed in…

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In a Young's double slit experiment d=0.5 mmd=0.5\ \mathrm{mm}, D=1 mD=1\ \mathrm{m} and λ=600 nm\lambda=600\ \mathrm{nm}. A thin transparent film of refractive index 1.51.5 and thickness 6 μm6\ \mu\mathrm{m} is placed in front of one slit. The central bright fringe shifts by
  1. A. 3 mm3\ \mathrm{mm}
  2. B. 5 mm5\ \mathrm{mm}
  3. C. 12 mm12\ \mathrm{mm}
  4. D. 6 mm6\ \mathrm{mm}
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Answer: (D) 6 mm6\ \mathrm{mm}

Extra optical path =(μ−1)t=0.5×6 μm=3 μm=(\mu-1)t=0.5\times6\ \mu\mathrm{m}=3\ \mu\mathrm{m}, which equals 3000600=5\dfrac{3000}{600}=5 wavelengths, so the pattern shifts by 55 fringe widths. β=λDd=600×10−9×10.5×10−3=1.2 mm\beta=\dfrac{\lambda D}{d}=\dfrac{600\times10^{-9}\times1}{0.5\times10^{-3}}=1.2\ \mathrm{mm}. Shift =5×1.2=6 mm=5\times1.2=6\ \mathrm{mm} towards the slit covered by the film.
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