Verified Solution • 2025–26 Session moderate

QJEEM25Ph9c75: JEE Main 2025 Physics - Class 12 JEEM25Ph9c75 Solution: A thin transparent film with refractive index 1.4 , is held on…

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} \mathrm{~m}^3 / \mathrm{s} .

Step-by-Step Verified Solution

Maxima condition

2μt=nλ⇒t=nλ2μ⇒t=λ2μ,2λ2μ,……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 \ldots

Minima condition 2μt=(2n−1)λ/22 \mu \mathrm{t}=(2 \mathrm{n}-1) \lambda / 2

amp;⇒t=(2n−1)λ4μ⇒t=λ4μ,3λ4μ,…..amp;Δt=2λ4μ\begin{aligned}& \Rightarrow \mathrm{t}=\frac{(2 \mathrm{n}-1) \lambda}{4 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{4 \mu}, \frac{3 \lambda}{4 \mu}, \ldots . . \\& \Delta \mathrm{t}=\frac{2 \lambda}{4 \mu}\end{aligned}

Rate of evaporation =A(Δt) time =54×10−13 m3/s=\frac{\mathrm{A}(\Delta \mathrm{t})}{\text { time }}=54 \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}

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