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 ____________ π× 10⁻¹³ m³ / s.

Solution

Physics sir · 4 steps

Each new transmission minimum means the film got thinner by λ/2μ. Turn that thickness into volume.

1. Thickness per minimum. The two interfering beams differ in path by 2μ t. To go from one minimum to the next, this path difference must change by exactly one wavelength. The phase changes at the surfaces stay the same as the film thins, so they do not affect the spacing.

   Δ t = λ/2μ = (560 × 10⁻⁹)/(2 × 1.4) = 2 × 10⁻⁷ m

2. Volume lost. The film is a flat disc of radius 1.8 cm, so the volume lost in 12 s is π r² Δ t = π × (1.8 × 10⁻²)² × 2 × 10⁻⁷ = π × 6.48 × 10⁻¹¹ m³.
3. Divide by time. Rate = (π × 6.48 × 10⁻¹¹)/12 = π × 5.4 × 10⁻¹² = 54π × 10⁻¹³ m³/s.
4. So. The rate of evaporation is 54 π × 10⁻¹³ m³/s.

The common slip: 108. Dropping the factor 2 in 2μ t gives Δ t = λ/μ = 400 nm and a rate of 108, not 54. The light crosses the film twice.

The idea to keep. In a thin film, the light travels through the thickness twice and its wavelength shrinks by μ. So one fringe shift or one extra minimum always means a thickness change of λ/2μ.

Answer from the NTA final answer key. Solution by PrepWiser.