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

For a thin film of refractive index $$n$$ on both sides by air (refractive index 1), the net phase difference between the two reflected rays is given by the formula:

$$ \Delta \phi = \frac{4\pi n t}{\lambda} + \pi \quad\text{-(1)} $$

The extra $$\pi$$ arises from the phase inversion at the top surface.

Transmission minima correspond to reflection maxima, which require constructive interference of the reflected waves, so the condition is

$$ \Delta \phi = 2\,m\,\pi,\quad m=1,2,3,\dots \quad\text{-(2)} $$

Equating (1) and (2) gives

$$ \frac{4\pi n t}{\lambda} + \pi = 2\,m\,\pi. $$

Simplifying yields the film thickness at the $$m^{\text{th}}$$ minimum:

$$ t_m = \frac{(2m - 1)\,\lambda}{4\,n} \quad\text{-(3)} $$

The change in thickness between successive minima ($$m$$ to $$m+1$$) is

$$ \Delta t = t_{m+1} - t_m = \frac{\lambda}{2\,n} \quad\text{-(4)} $$

Substituting $$\lambda = 560 \times 10^{-9}\,\text{m}$$ and $$n = 1.4$$ gives

$$ \Delta t = \frac{560\times10^{-9}}{2\times1.4} = 2.0\times10^{-7}\,\text{m}. $$

The film is supported on a circular ring of radius $$r = 1.8\,\text{cm} = 0.018\,\text{m}$$, so its area is

$$ A = \pi\,r^2 = \pi\,(0.018)^2 = 3.24\times10^{-4}\,\pi~\text{m}^2. $$

When the film thickness decreases by $$\Delta t$$, the volume of liquid evaporated is

$$ \Delta V = A\,\Delta t = (3.24\times10^{-4}\,\pi)\,(2.0\times10^{-7}) = 6.48\times10^{-11}\,\pi~\text{m}^3. $$

This occurs every $$\Delta t_{\text{time}} = 12\,\text{s}$$.

The evaporation rate is

$$ \frac{\Delta V}{\Delta t_{\text{time}}} = \frac{6.48\times10^{-11}\,\pi}{12} = 5.4\times10^{-12}\,\pi~\text{m}^3/\text{s}. $$

Expressing this in the form $$\pi\times10^{-13}$$ gives

$$ 5.4\times10^{-12} = 54\times10^{-13}. $$

Hence the evaporation rate is

$$ 54\,\pi\times10^{-13}~\text{m}^3/\text{s}. $$

$$54\,\pi\times10^{-13}\,\text{m}^3/\text{s}$$