A thin transparent film with refractive index $1.4$ is held on a circular ring of radius $1.8\ \text{cm}$. The fluid in the film evaporates such that transmission through the film at wavelength $560\ \text{nm}$ goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is ______ $\pi\times10^{-13}\ \text{m}^3/\text{s}$.
Numerical value type. Enter your answer.
Answer: 54
Successive transmission minima correspond to a change in thickness of
$$\Delta t = \frac{\lambda}{2\mu} = \frac{560}{2\times1.4} = 200\ \text{nm}$$
This happens every $12\ \text{s}$, so the volume lost per second is
$$\frac{\pi r^2\,\Delta t}{12} = \frac{\pi(1.8\times10^{-2})^2\times200\times10^{-9}}{12} = \pi\times\frac{3.24\times10^{-4}\times2\times10^{-7}}{12} = 54\pi\times10^{-13}\ \text{m}^3/\text{s}$$
Solution by Sreeraj P, M.Sc Physics