Solution
Concept: Thin Film Interference
The phenomenon observed is thin film interference in reflected light. A broad beam of light incident at an angle $\theta$ is reflected from the top surface (Air-Liquid) and the bottom surface (Liquid-Glass) of the film.
Step 1: Optical Path Difference
The optical path difference ($\Delta x$) between the two interfering rays for a film of refractive index $\mu$ and thickness $t$ is given by the standard cosine law for thin films: $$\Delta x = 2\mu t \cos r$$ where $r$ is the angle of refraction inside the liquid.
Note: While there may be phase shifts due to reflection at the boundaries (Air-Liquid and Liquid-Glass), these shifts add a constant value (like $\lambda/2$) to the path difference. Since we are analyzing the change in thickness required to go from one maximum to the next, the constant phase shifts cancel out. We focus purely on the change in the term $2\mu t \cos r$.
The optical path difference ($\Delta x$) between the two interfering rays for a film of refractive index $\mu$ and thickness $t$ is given by the standard cosine law for thin films: $$\Delta x = 2\mu t \cos r$$ where $r$ is the angle of refraction inside the liquid.
Note: While there may be phase shifts due to reflection at the boundaries (Air-Liquid and Liquid-Glass), these shifts add a constant value (like $\lambda/2$) to the path difference. Since we are analyzing the change in thickness required to go from one maximum to the next, the constant phase shifts cancel out. We focus purely on the change in the term $2\mu t \cos r$.
Step 2: Condition for One Full Cycle
The problem states that the intensity changes from Maxima $\rightarrow$ Minima $\rightarrow$ Maxima.
This sequence represents exactly one full cycle of interference fringe shift. For the intensity to return to the same state (Maxima to Maxima), the optical path difference must change by exactly one wavelength ($\lambda$).
Let the change in thickness be $\Delta t_{thickness}$. The corresponding change in optical path difference is: $$|\Delta (\Delta x)| = 2\mu |\Delta t_{thickness}| \cos r$$ For one full cycle: $$2\mu |\Delta t_{thickness}| \cos r = \lambda$$ $$|\Delta t_{thickness}| = \frac{\lambda}{2\mu \cos r}$$
The problem states that the intensity changes from Maxima $\rightarrow$ Minima $\rightarrow$ Maxima.
This sequence represents exactly one full cycle of interference fringe shift. For the intensity to return to the same state (Maxima to Maxima), the optical path difference must change by exactly one wavelength ($\lambda$).
Let the change in thickness be $\Delta t_{thickness}$. The corresponding change in optical path difference is: $$|\Delta (\Delta x)| = 2\mu |\Delta t_{thickness}| \cos r$$ For one full cycle: $$2\mu |\Delta t_{thickness}| \cos r = \lambda$$ $$|\Delta t_{thickness}| = \frac{\lambda}{2\mu \cos r}$$
Step 3: Expressing $\cos r$ in terms of $\theta$
We use Snell’s Law at the air-liquid interface to find $\cos r$. $$1 \cdot \sin \theta = \mu \sin r$$ $$\sin r = \frac{\sin \theta}{\mu}$$ Using the identity $\cos r = \sqrt{1 – \sin^2 r}$: $$\cos r = \sqrt{1 – \left(\frac{\sin \theta}{\mu}\right)^2}$$ $$\cos r = \sqrt{\frac{\mu^2 – \sin^2 \theta}{\mu^2}} = \frac{\sqrt{\mu^2 – \sin^2 \theta}}{\mu}$$
We use Snell’s Law at the air-liquid interface to find $\cos r$. $$1 \cdot \sin \theta = \mu \sin r$$ $$\sin r = \frac{\sin \theta}{\mu}$$ Using the identity $\cos r = \sqrt{1 – \sin^2 r}$: $$\cos r = \sqrt{1 – \left(\frac{\sin \theta}{\mu}\right)^2}$$ $$\cos r = \sqrt{\frac{\mu^2 – \sin^2 \theta}{\mu^2}} = \frac{\sqrt{\mu^2 – \sin^2 \theta}}{\mu}$$
Step 4: Determine the Rate of Evaporation
Substituting $\cos r$ back into the thickness equation from Step 2: $$|\Delta t_{thickness}| = \frac{\lambda}{2\mu \left( \frac{\sqrt{\mu^2 – \sin^2 \theta}}{\mu} \right)}$$ $$|\Delta t_{thickness}| = \frac{\lambda}{2\sqrt{\mu^2 – \sin^2 \theta}}$$
The rate of decrease of thickness, $R$, is the change in thickness divided by the time interval $\Delta t$. $$R = \frac{|\Delta t_{thickness}|}{\Delta t}$$ $$R = \frac{1}{\Delta t} \left( \frac{\lambda}{2\sqrt{\mu^2 – \sin^2 \theta}} \right)$$
Substituting $\cos r$ back into the thickness equation from Step 2: $$|\Delta t_{thickness}| = \frac{\lambda}{2\mu \left( \frac{\sqrt{\mu^2 – \sin^2 \theta}}{\mu} \right)}$$ $$|\Delta t_{thickness}| = \frac{\lambda}{2\sqrt{\mu^2 – \sin^2 \theta}}$$
The rate of decrease of thickness, $R$, is the change in thickness divided by the time interval $\Delta t$. $$R = \frac{|\Delta t_{thickness}|}{\Delta t}$$ $$R = \frac{1}{\Delta t} \left( \frac{\lambda}{2\sqrt{\mu^2 – \sin^2 \theta}} \right)$$
Rate of decrease of thickness = $\frac{\lambda}{2 \Delta t \sqrt{\mu^2 – \sin^2 \theta}}$