Question

(a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the bubble can be and appear dark at all visible wavelengths? Assume the same index of refraction as water. (b) Discuss the fragility of the film considering the thickness found

Answer 1

Answer:

t< 75 nm

Explanation:

A soap bubble is a thin film where when the beam enters the film it has a 180º phase change due to the refractive index and the wavelength changes between

              λ = λ₀ / n

In the case of constructive interference in the curve of the spherical film it is

              2 nt = (m + ½) λ₀

Where t is the thickness of the film and n the refractive index that does not indicate that we use that of water n = 1.33, m is an integer. The thickness of the film for the first interference (m = 0) is

              t = λ₀ / 4 n

A thickness less than this gives destructive interference.

Let's look for the thickness for the visible spectrum

Violet light λ₀ = 400 nm = 400 10⁻⁹ m

      t₁ = 400 10⁻⁹ / 4 1.33

      t₁ = 75.2 10-9 m

Red light λ₀ = 700 nm = 700 10⁻⁹ m

       t₂ = 700 10⁻⁹ / 4 1.33

       t₂ = 131.6 10⁻⁹ m

Therefore, for all wavelengths to have destructive interference, the thickness must be less than 75 10⁻⁹ m = 75 nm

b) a film like eta is very thin, it is achieved when gravity thins the pomp, but any movement or burst of air breaks it,