Question

What is the minimum thickness of a soap film (excluding the case of zero thickness) that appears black when illuminated with light having a wavelength of 575 nm? The soap film has an index of refraction of 1.35, and air is present on both sides of the film. Express your answer in nanometers.

Answer 1

The minimum thickness of the soap film that appears black when illuminated with 575 nm light is 213.9 nm.

Step-by-step explanation:

To determine the minimum thickness of a soap film that appears black when illuminated with light of a specific wavelength, we can use the equation for constructive interference in thin films:

2nt = mλ

where n is the index of refraction of the soap film, t is the thickness of the film, m is an integer representing the order of interference, and λ is the wavelength of light.

Since we want the film to appear black, we are looking for destructive interference, which occurs when there is a phase shift of π radians. For destructive interference, the path length difference is equal to one-half the wavelength:

2nt = (m + 1/2)λ

Solving for the minimum thickness (t):

t = ((m + 1/2)λ) / (2n)

Substituting the given values:

t = ((1/2)λ) / (2n)

Plugging in the values λ = 575 nm and n = 1.35:

t = ((1/2)(575 nm)) / (2(1.35)) = 213.9 nm

Therefore, the minimum thickness of the soap film that appears black when illuminated with 575 nm light is 213.9 nm.