A thin film of liquid is held in a horizontal circular ring, with air on both sides of the film. A beam of light at wavelength 550 nm is directed perpendicularly onto the film, and the intensity I of its reflection is monitored. Figure 35-47 gives intensity I as a function of time the horizontal scale is set by ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$. The intensity changes because of evaporation from the two sides of the film. Assume that the film is flat and has parallel sides, a radius of $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{cm}$, and an index of refraction of 1.40. Also assume that the film’s volume decreases at a constant rate. Find that rate.

Radius of circular film $r=1.8\text{cm}$

Index of refraction of film ${\text{n}}_2\text{=1.4}$

Wavelength of light $\lambda =550nm$

The thin liquid film is a phase of small thickness, in which the two interfacial layers overlap to form a unified non-homogeneous structure of specific properties.

In the figure at $t=0$, intensity is minimum and again at $t=\left(6\right)\left(2\text{s}\right)\text{= 12s}$

it is minimum.

The change in time from one minimum to next minimum

$\Delta t=12\text{s - 0}\text{s}$

But we have condition for minima

$\begin{array}{rcl}2L& =& m\frac{\lambda }{n_2}\\ L& =& m\frac{\lambda }{2n_2}\end{array}$

Change in thickness from one minimum to next minimum is

$\Delta L=\Delta m\frac{\lambda }{2n_2}$

Here, $\Delta m=1$

Therefore $\Delta L=\frac{\lambda }{2n_2}$

But change in volume $\Delta v=\left(\pi r^2\right)\Delta L$

(Since the film is circular)

$\Delta L=\frac{\lambda }{2n_2}$

Rate of change of volume $\frac{dv}{dt}=\frac{\pi r^2\lambda }{2n_2\Delta t}$

Given radius of circular film $r=1.8\text{cm}$

$$\begin{gathered} =\left(1.8\right)\left({10}^{-2}\text{m/cm}\right) \\ =0.018\text{m} \end{gathered}$$

Index of refraction of film $n_2=1.4$

Wavelength of light

$$\begin{gathered} =\text{550}\text{nm} \\ =\left(\text{550}\text{nm}\right)\left({10}^{-9}\text{m/nm}\right) \\ =550\times {10}^{-9} \end{gathered}$$

Rate of change of volume:

$\begin{array}{rcl}\frac{dv}{dt}& =& \frac{\left(\pi \right){\left(0.018\text{m}\right)}^2\left(\text{550}\times {\text{10}}^{-9}\text{m}\right)}{\left(2\right)\left(1.4\right)\left(12s\right)}\\ & =& 0.0166684\times {10}^{-9}{\text{m}}^3/\text{s}\\ & =& \text{1.67}\times {\text{10}}^{-11}{\text{m}}^3/\text{s}\end{array}$

Therefore, the rate is $\text{1.67}\times {\text{10}}^{-11}{\text{m}}^3/\text{s}$.