A thin uniform film of oil that can be varied in thickness covers a sheet of glass of refractive index 1.52. The refractive index of the oil is 1.64. Light of wavelength 555 nm is shone from air at normal
Incidence on the film. Starting with no oil on the glass, you gradually increase the thickness of the
Oil film until the first interference maximum in the reflected light occurs. What is the thickness of
The oil film at that instant?

Question

Quizplus · Accepted Answer

For the first interference maximum in the reflected light to occur at normal incidence, the optical path difference must be equal to the wavelength of the light in the medium divided by 2 (since it's the first maximum). The optical path difference is given by $2nd$, where $n$ is the refractive index of the oil and $d$ is the thickness of the oil film. This path difference must equal half the wavelength in the oil, so $2nd = \lambda/2$, where $\lambda$ is the wavelength of light in vacuum. The wavelength of the light in the oil is given by $\lambda/n_{oil}$, where $n_{oil}$ is the refractive index of the oil. Therefore, $2d = \frac{\lambda}{2n_{oil}}$. Substituting the given values, $2d = \frac{555}{2 \times 1.64}$ nm, solving for $d$ gives approximately 84.6 nm.

A Thin Uniform Film of Oil That Can Be Varied