A $360 - \mathrm { nm }$ thick oil film floats on the surface of a pool of water. The indices of refraction of the oil and the water are $1.50$ and $1.33$, respectively. When the surface of the oil is illuminated from above at normal incidence with white light, what is the wavelength of light between $400 \mathrm {~nm}$ to $800 \mathrm {~nm}$ wavelength that is most weakly reflected?

Question

Quizplus · Accepted Answer

The wavelength of light that is most weakly reflected corresponds to destructive interference. For normal incidence, the condition for destructive interference in a thin film is given by $2nt = (m + \frac{1}{2})\lambda_n$, where $n$ is the refractive index of the film (oil in this case, $n = 1.50$), $t$ is the thickness of the film ($360 \, \mathrm{nm}$), $m$ is an integer (starting from 0), and $\lambda_n$ is the wavelength of light in the film. The wavelength in the film is related to the wavelength in vacuum ($\lambda$) by $\lambda_n = \lambda / n$. Solving for $\lambda$ gives $\lambda = 2nt / (m + \frac{1}{2})$. For the first order of interference ($m = 0$), this simplifies to $\lambda = 4nt$. Substituting the given values, $\lambda = 4 \times 1.50 \times 360 \, \mathrm{nm} / (0.5) = 540 \, \mathrm{nm}$, which is the wavelength most weakly reflected.

A 360−nm360 - \mathrm { nm }360−nm Thick Oil Film Floats on the Surface of a Pool

A $360 - \mathrm { nm }$ Thick Oil Film Floats on the Surface of a Pool