JEE MAIN - Physics (2025 - 28th January Evening Shift - No. 21)
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 ____________ $\pi\times 10^{-13} \mathrm{~m}^3 / \mathrm{s}$.
Answer
54
Explanation
Maxima condition
$$2 \mu \mathrm{t}=\mathrm{n} \lambda \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{2 \mu}, \frac{2 \lambda}{2 \mu}, \ldots \ldots$$
Minima condition $2 \mu \mathrm{t}=(2 \mathrm{n}-1) \lambda / 2$
$$\begin{aligned} & \Rightarrow \mathrm{t}=\frac{(2 \mathrm{n}-1) \lambda}{4 \mu} \Rightarrow \mathrm{t}=\frac{\lambda}{4 \mu}, \frac{3 \lambda}{4 \mu}, \ldots . . \\ & \Delta \mathrm{t}=\frac{2 \lambda}{4 \mu} \end{aligned}$$
Rate of evaporation $=\frac{\mathrm{A}(\Delta \mathrm{t})}{\text { time }}=54 \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}$
Comments (0)
