JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Wave Optics

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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

Answer: 54

Solution

For a thin film of refractive index

μ

and thickness

t

, transmitted light is minimum (equivalently, reflected light is maximum) when

2 μ t = (2 n - 1) \frac{λ}{2}

Successive minima occur at

t = \frac{λ}{4 μ}, \frac{3 λ}{4 μ}, \frac{5 λ}{4 μ}, \dots

, so the thickness change between consecutive minima is

Δ t = \frac{2 λ}{4 μ} = \frac{λ}{2 μ}

Substituting

λ = 560 nm

μ = 1.4

Δ t = \frac{560 \times 1 0 ^{- 9}}{2 \times 1.4} = 200 \times 1 0^{- 9} m

The corresponding change in volume of the flat circular film of radius

r = 1.8 \times 1 0^{- 2} m

Δ V = π r^{2} Δ t = π \times (1.8 \times 1 0^{- 2})^{2} \times 200 \times 1 0^{- 9} = π \times 648 \times 1 0^{- 13} m^{3}

This volume evaporates every 12 s, so the rate of evaporation is

Rate = \frac{Δ V}{12} = \frac{648}{12} π \times 1 0^{- 13} = 54 π \times 1 0^{- 13} m^{3} / s

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Integer answer

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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Wave Optics

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