Final answer:
The smallest thickness of a soap film (n = 1.36) that would appear black under illumination with 540-nm light is approximately 99.265 nm, which results from the first instance of complete destructive interference.
Step-by-step explanation:
The smallest thickness of a soap film (with n = 1.36) that would appear black when illuminated with 540-nm light can be found using the principles of thin film interference. A black appearance is due to destructive interference, which occurs when the path length difference between light reflecting off the top and bottom surfaces of the film causes them to be out of phase by half a wavelength (or a multiple thereof).
For the first instance of complete destructive interference (the darkest black), we look for the condition where the path difference is half the wavelength of the light in the film medium. This means the thickness must be one-quarter of the wavelength of the light in air, as the light travels a round trip through the film. The wavelength within the film is the wavelength in air, λ, divided by the refractive index, n.
λ_{film} = λ / n
With λ = 540 nm and n = 1.36, we calculate:
λ_{film} = 540 nm / 1.36
λ_{film} = 397.06 nm
Since we want the thickness t for the film to be one-quarter of λ_{film} for destructive interference, we get:
t = λ_{film} / 4
t = 397.06 nm / 4
t = 99.265 nm
So the smallest thickness is approximately 99.265 nm.