Question

A soap film (index of refraction 1.33) is surrounded on both sides by air. When white light shines nearly perpendicularly on the film, you see bright interference colours of wavelength = 588 nm in the reflected light. What is the second smallest possible value for the thickness of the film?

Answer 1

When light reflects off a film, we can use the principles of constructive interference to determine the conditions for a bright reflection. The film in this case is surrounded by air, which has a lower index of refraction than the soap film. Thus, when light reflects off the top and bottom surfaces of the soap film, a phase shift occurs at the top surface.

For constructive interference in a thin film where light reflects from a medium of lower refractive index (air) to one of higher refractive index (soap film), the additional path difference that contributes to constructive interference is given by:

\[
2nt = m\lambda
\]

where:
- \( n \) is the index of refraction of the film,
- \( t \) is the thickness of the film,
- \( m \) is the order of interference (an integer),
- \( \lambda \) is the wavelength of light in vacuum or air.

Since the light in air is 588 nm, we need to use this wavelength in the equation.

Given:
- \( n = 1.33 \) (index of refraction of the soap film),
- \( \lambda = 588 \, \text{nm} \) (wavelength in air).

Substituting these values into the equation we get:

\[
2(1.33)t = m(588 \, \text{nm})
\]
\[
2.66t = m(588)
\]
\[
t = \frac{m(588)}{2.66}
\]

Now we need to find the thickness for different values of \( m \).

Let's calculate this for \( m = 1 \) and \( m = 2 \):

1. **For \( m = 1 \)**:
\[
t_1 = \frac{1 \times 588 \, \text{nm}}{2.66} \approx \frac{588}{2.66} \approx 220.3 \, \text{nm}
\]

2. **For \( m = 2 \)**:
\[
t_2 = \frac{2 \times 588 \, \text{nm}}{2.66} \approx \frac{1176}{2.66} \approx 442.2 \, \text{nm}
\]

The second smallest possible value for the thickness of the film corresponds to \( m = 2 \):

\[
\boxed{442.2 \, \text{nm}}
\]