Physics · Wave Optics

JEE Main 2025 — 28 January, Evening Shift — Question 68

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 $\qquad$ $\pi \times 10^{-13} \mathrm{~m}^{3} / \mathrm{s}$ .

Answer: 54

Numerical answer — enter this value.

Step-by-step solution

Maxima condition

$2 \mu \mathrm{t}=\mathrm{n} \lambda \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{2 \mu}, \frac{2 \lambda}{2 \mu}, \ldots$ .

Minima condition $2 \mu \mathrm{t}=(2 \mathrm{n}-1) \lambda / 2$

$\Rightarrow \mathrm{t}=\frac{(2 \mathrm{n}-1) \lambda}{4 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{4 \mu}, \frac{3 \lambda}{4 \mu}, \ldots \ldots$

$\Delta t=\frac{2 \lambda}{4 \mu}$

Rate of evaporation $=\frac{\mathrm{A}(\Delta \mathrm{t})}{\text { time }}=54 \times 10^{-13} \mathrm{~m}^{3} / \mathrm{s}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Wave Optics
Topic: Interference of Light Waves with Thin Films