JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2025
- Hard
- Numerical
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 ____ \(\pi \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 54
Step-by-step Solution
Detailed explanation
For a thin film interference, a fringe for transmission is formed.
When
\(2 \mu x=n \lambda\)
\(\begin{aligned} \Rightarrow & \frac{d x}{d t}=\left(\frac{d n}{d t}\right) \frac{\lambda}{2 \mu} \\ & =\left(\frac{1}{12}\right) \frac{560 \times 10^{-9}}{2 \times 1.4}=\frac{5}{3} \times 10^{-8} \mathrm{~m} / \mathrm{s}\end{aligned}\)
\(\begin{aligned} & V=\text { Volume of film }=\pi R^2 x \\ & \frac{d V}{d t}=\pi R^2 \frac{d x}{d t} \\ & =\pi\left(1.8 \times 10^{-2}\right)^2 \frac{5}{3} \times 10^{-8} \mathrm{~m}^3 / \mathrm{s} \\ & =54 \pi \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}\end{aligned}\)
When
\(2 \mu x=n \lambda\)
\(\begin{aligned} \Rightarrow & \frac{d x}{d t}=\left(\frac{d n}{d t}\right) \frac{\lambda}{2 \mu} \\ & =\left(\frac{1}{12}\right) \frac{560 \times 10^{-9}}{2 \times 1.4}=\frac{5}{3} \times 10^{-8} \mathrm{~m} / \mathrm{s}\end{aligned}\)
\(\begin{aligned} & V=\text { Volume of film }=\pi R^2 x \\ & \frac{d V}{d t}=\pi R^2 \frac{d x}{d t} \\ & =\pi\left(1.8 \times 10^{-2}\right)^2 \frac{5}{3} \times 10^{-8} \mathrm{~m}^3 / \mathrm{s} \\ & =54 \pi \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}\end{aligned}\)
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