A piece of glass has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavelength of light incident on the film is 560 nm and the indices of refraction of
Gasoline and glass are 1.40 and 1.50, respectively, what is the minimum nonzero thickness of the
Film to see a bright reflection?

Question

Quizplus · Accepted Answer

For constructive interference (bright reflection) in a thin film, the path difference must be a multiple of the wavelength in the medium. The minimum nonzero thickness for a bright reflection occurs when the path difference is half the wavelength in the medium. The effective path difference is $2t = \frac{\lambda}{2n}$, where $t$ is the thickness of the film, $\lambda$ is the wavelength of light in vacuum, and $n$ is the refractive index of the film. Rearranging for $t$, we get $t = \frac{\lambda}{4n}$. Substituting $\lambda = 560 \, \text{nm}$ and $n = 1.40$ for gasoline, we find $t = \frac{560}{4 \times 1.40} = 100 \, \text{nm}$. However, this calculation assumes a single pass through the film, but for constructive interference in reflection, there's an additional half-wavelength phase shift upon reflection at the higher refractive index boundary (gasoline-glass interface), effectively doubling the minimum thickness for the first bright fringe to $2 \times 100 \, \text{nm} = 200 \, \text{nm}$.

A Piece of Glass Has a Thin Film of Gasoline