Question

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 ______ $\pi \times 10^{-13} m^3/s$.

Answer 1

54

Answer 2

The condition for successive minima (destructive interference in transmission) is that the optical path difference changes by one wavelength. For a film with two equal flat sides the change in thickness Δt between two minima is given by

$2 n\,\Delta t = \lambda \quad \Longrightarrow \quad \Delta t = \frac{\lambda}{2n}$

Substitute $\lambda = 560\,\text{nm} = 560\times10^{-9}\,\text{m}$ and $n = 1.4$:

$\Delta t = \frac{560\times10^{-9}}{2\times 1.4} = \frac{560\times10^{-9}}{2.8} = 200\times10^{-9}\,\text{m}$

The area of the circular ring is

$A = \pi r^2\quad \text{with} \quad r=1.8\,\text{cm}=0.018\,\text{m}$,

$A = \pi (0.018)^2 = \pi(0.000324)\,\text{m}^2$

The loss in volume in one cycle (when the thickness drops by $\Delta t$) is

$\Delta V = A \times \Delta t = \pi (0.000324)\times (200\times10^{-9})\,\text{m}^3$

Calculate the product:

$0.000324 \times 200 \times10^{-9} = 0.0648\times10^{-9} = 6.48\times10^{-11}$

Thus,

$\Delta V = 6.48\times10^{-11}\,\pi\,\text{m}^3$

This change happens every 12 seconds, so the rate of evaporation is

$\text{Rate} = \frac{\Delta V}{12} = \frac{6.48\times10^{-11}\,\pi}{12} \, \text{m}^3/s = 0.54\times10^{-11}\,\pi\,\text{m}^3/s$

Expressing $0.54\times10^{-11}$ in the form $\_ \times10^{-13}$:

$0.54\times10^{-11} = 54\times10^{-13}$

Thus, the rate of evaporation is $54\pi \times10^{-13}\,\text{m}^3/s$.