Question

A thin film of alcohol (n=1.36) lies on a flat glass plate (n=1.51). When monochromatic light, whose wavelength can be changed, is incident normally, the reflected light is a minimum for λ=512 nm and a maximum for λ= 650 nm. What is the minimum thickness of the film? (Answer: 471 nm)

Answer 1

2n_at = mλ₁, and 2n_at = (n+1/2) λ₂

with λ₁= 650 nm, and λ1= 512 nm and n >= m,

which tells that: λ₁/λ₂ = (2n + 1)/2m ⟹ 4n + 2 = 5m, with n >= m

=====> (m, n) = (2, 2), (10, 12), (18, 22) and so on...

The thickness of the film is t = mλ₁/2n_a = 471 nm, etc

or

Both of the reflections for interference occur from rarer to denser medium–one reflection from air(n0 = 1.0) to alcohol(n1 = 1.36) and the other one from alcohol (n1 = 1.36) to glass (n2 = 1.51). So, in both of the reflected waves, additional π phase difference will be introduced. Hence, condition for maximum and minimum are

2n₁d cos θ_t = pλ₁ (for maximum)

and    (q + 1/2)λ₂ (for minimum)

For normal incidence cos θ_t = 1. Given that λ₁ = 650 nm, λ₂ = 512 nm. Hence we equating the condition

we get 5p = 2(2q + 1). For the smallest integer solutions p = 2, q = 2. Then the minimum thickness is d = 0.47 µm.