Red light having a wavelength of 690 nm (in air) is vertically incident on a thin film of oil that has an index of refraction of 1.19. The film of oil floats above water (having an index of refraction of 1.30). No red light is reflected. The minimum (nonzero) thickness of the film of oil is
A) 290 nm.
B) 345 nm.
C) 580 nm.
D) 145 nm.

Question

Quizplus · Accepted Answer

For destructive interference (no reflection) in a thin film, the condition is $2nt = (m + 0.5)\lambda/n_0$, where $n$ is the refractive index of the film, $t$ is the thickness, $\lambda$ is the wavelength in air, and $n_0$ is the refractive index of the surrounding medium. Here, $n = 1.19$, $\lambda = 690 \text{ nm}$, and $n_0 = 1$. For the minimum thickness, $m = 0$, so $2(1.19)t = (0.5)(690)$. Solving for $t$ gives approximately 145 nm.

Red Light Having a Wavelength of 690 Nm (In Air)