A soap film made with material of index of refraction n = 1.9 is illuminated by white light.If the reflected light is maximum at $\lambda$ = 350 nm,what is the minimum thickness the film could have had?

Question

Quizplus · Accepted Answer

For maximum reflection at a specific wavelength, we use the equation for constructive interference:

2nt = (m + 1/2)λ

where n is the refractive index of the medium (in this case, the soap film), t is the thickness of the film, λ is the wavelength of the reflected light, and m is an integer corresponding to the order of the interference maximum.

To minimize the thickness of the film, we want the lowest non-zero value of m, which is m = 1. We can plug in the values given:

2(1.9)t = (1 + 1/2)350 nm

Solving for t, we get:

t = (3/4)(350 nm)/(2.2) = 46.8 nm

Therefore, the minimum thickness the film could have had is approximately 46 nm, which is closest to choice D (46 nm).

A Soap Film Made with Material of Index of Refraction λ\lambdaλ

A Soap Film Made with Material of Index of Refraction $\lambda$