A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film?
A) 47.0 nm
B) 74.0 nm
C) 94.0 nm
D) 152 nm
E) 268 nm

Question

Quizplus · Accepted Answer

For constructive interference (i.e., bright spots), the thickness of the soap film, d, should be:

d = (m + 1/2)λ/n

where λ is the wavelength of the incident light, n is the index of refraction of the soap film, and m is an integer representing the order of the interference (0, 1, 2, ...).

Since the problem asks for the minimum thickness of the film (i.e., the first order of interference), we can simplify the equation to:

d = λ/2n

Plugging in the values given, we get:

d = (500 nm) / (2 x 1.33) = 187.97 nm

Therefore, the closest choice is C, 94.0 nm.

A 500 Nm Wavelength Light Illuminates a Soap Film with an Index