A ring of radius 3 cm has a soap film which is getting evaporated. Light of wavelength \(\lambda=580 \mathrm{~nm}\) give…

To find the rate of evaporation of the soap film, we need to determine the rate at which its thickness decreases over time.

• Wavelength of light: \(\lambda =580\) = \(580\times 1{0}^{−9}\)
• Refractive index of soap film: \(\mu =1.45\)
• Minimum transmission occurs every \(T=12\)
• Radius of the ring: \(r=3\) = \(0.03\)

A thin-film interference pattern changes when the optical path difference changes by half a wavelength in the medium:

\(∆t=\frac{\lambda }{2\mu }\)​

This means that the thickness decreases by:

\(\Delta t=\frac{\lambda }{2\mu }\)​

Substituting values:

\(\Delta t=\frac{580\times 1{0}^{−9}}{2\times 1.45}\)

\(\Delta t=\frac{580\times 1{0}^{−9}}{2.9}\)

​ \(\Delta t=200\times 1{0}^{−9}\text{ m}=2\times 1{0}^{−7}\text{ m}\)

This thickness decreases every 12 seconds, so the rate of evaporation per second:

\(\frac{∆t}{dt}=\frac{2\times 1{0}^{−7}\text{ m}}{12}\)​ \(=1.67\times 1{0}^{−8}\text{ m/s}\)

The evaporating volume per second is:

\(\text{Rate of evaporation}=\text{Surface area}\times \frac{∆t}{dt}\)\(=\pi {r}^{2}\times \frac{∆t}{dt}\)​

Substituting values:

\(=\pi (0.03{)}^{2}\times (1.67\times 1{0}^{−8})\)

\(=\pi (9\times 1{0}^{−4})\times (1.67\times 1{0}^{−8})\)

\(=15\pi \times 1{0}^{−12}{\text{ m}}^{3}\mathrm{/}\text{s}\)

From the given options, the correct answer : \(15\pi \times 1{0}^{−12}\)