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NVAMediumJEE 2025Interference (Young's Experiment)

JEE Physics 2025 Question with Solution

A thin transparent film with refractive index 1.41.4 is held on a circular ring of radius 1.8 cm1.8 \, \text{cm}. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm560 \, \text{nm} goes to a minimum every 12 s12 \, \text{s}. Assuming that the film is flat on its two sides, the rate of evaporation is:

Answer

Correct answer:1.67

Step-by-step solution

Standard Method

Given: refractive index of film μ=1.4\mu = 1.4, wavelength λ=560×10−9 m\lambda = 560 \times 10^{-9} \, \text{m}, and successive transmission minima occur every 12 s12 \, \text{s}.

Find: the rate of evaporation of the film.

For a thin film, successive minima in transmitted light occur when the thickness changes by Δt\Delta t such that

2μΔt=λ2\mu \Delta t = \lambda

Therefore,

Δt=λ2μ\Delta t = \frac{\lambda}{2\mu}

Detailed Calculation

Substitute the values:

Δt=560×10−92×1.4=200×10−9 m\Delta t = \frac{560 \times 10^{-9}}{2 \times 1.4} = 200 \times 10^{-9} \, \text{m}

The film reaches the next minimum in 12 s12 \, \text{s}, so the evaporation rate is

Rate=Δt12=200×10−912 m/s\text{Rate} = \frac{\Delta t}{12} = \frac{200 \times 10^{-9}}{12} \, \text{m/s} Rate=16.67×10−9 m/s=1.67×10−8 m/s\text{Rate} = 16.67 \times 10^{-9} \, \text{m/s} = 1.67 \times 10^{-8} \, \text{m/s}

Therefore, the numerical answer is 1.671.67.

The solution states the final answer as 1.671.67. The question text also shows a conflicting printed value in volume-per-time units, but the solution working supports the numerical answer 1.671.67 for the asked format.

Common mistakes

  • Using the condition for reflected-light minima instead of transmitted-light minima. That changes the thickness difference incorrectly. Use the successive transmitted minima relation 2μΔt=λ2\mu \Delta t = \lambda.

  • Ignoring the refractive index μ\mu of the film. The optical path difference depends on 2μt2\mu t, not just 2t2t. Always include the film refractive index in thin film interference.

  • Treating the asked rate as direct thickness change without dividing by time. The film thickness changes by Δt\Delta t every 12 s12 \, \text{s}, so the rate is Δt/12\Delta t/12.

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