JEE Main 2025 Jan 28 Shift 2, Physics Q25: Interference by Thin Film

JEE Main2025Jan 28, Shift 2Physics

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 _______ $π \times 1 0^{- 13}$ m $^{3}$ /s.

Your answer

Solution

For transmission through a thin film, successive minima occur when the optical thickness changes by $λ / (2 μ)$ , since transmission minima of the same kind correspond to $2 μ t = (2 n - 1) λ /2$ with $Δ n = 1$ :

$Δ t = \frac{λ}{2 μ} = \frac{560 \times 1 0 ^{- 9}}{2 \times 1.4} = 2 \times 1 0^{- 7}$ m.

This thinning occurs over $Δ T = 12$ s, so the rate of thickness change is $\dot{t} = 2 \times 1 0^{- 7} /12$ m/s.

The film has area $A = π r^{2} = π (1.8 \times 1 0^{- 2})^{2} = 3.24 π \times 1 0^{- 4}$ m $^{2}$ .

Volume evaporation rate $= A \dot{t} = 3.24 π \times 1 0^{- 4} \times \frac{2 \times 1 0 ^{- 7}}{12} = 54 π \times 1 0^{- 13}$ m $^{3}$ /s.

$∴$ Answer (the integer multiplying $π \times 1 0^{- 13}$ ) is $54$ .

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JEE Main 2025 Jan 28 Shift 2, Physics Q25: Interference by Thin Film

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