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JEE Main 2025 (28 Jan, Shift 2) · Biprism, Thin Films & Other Devices

Biprism, Thin Films & Other Devices question from JEE Main 2025 (28 Jan, Shift 2)

Q12JEE Main 2025 · 28 JAN · SHIFT 2Difficult

Biprism, Thin Films & Other Devices

A thin transparent film with refractive index 1.4, is held on circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is ____________ π×10−13 m3/s\pi \times 10^{-13}\text{ m}^3/\text{s}.

Numerical answer type. Work it out, then open the answer.

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Answer: 54

  1. Successive transmission minima: thickness changes by Δt=λ2μ=5602×1.4=200 nm\Delta t = \frac{\lambda}{2\mu} = \frac{560}{2\times1.4} = 200\ \text{nm}.
  2. Volume lost per 12 s: πr2Δt=π(1.8×10−2)2×2×10−7=6.48π×10−11 m3\pi r^2 \Delta t = \pi(1.8\times10^{-2})^2 \times 2\times10^{-7} = 6.48\pi\times10^{-11}\ \text{m}^3.
  3. Rate =6.48π×10−1112=5.4π×10−12=54π×10−13 m3/s= \frac{6.48\pi\times10^{-11}}{12} = 5.4\pi\times10^{-12} = 54\pi\times10^{-13}\ \text{m}^3/\text{s}.
  4. Answer: 54.
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