Final answer:
The smallest thickness of a soap film (n = 1.37) that would appear black when illuminated with 540-nm light is approximately 98.54 nm.
Step-by-step explanation:
The color of a soap film is determined by the interference of light waves as they reflect off the top and bottom surfaces of the film. When the thickness of the film satisfies the condition for constructive interference for a particular wavelength, that color will be most strongly reflected.
To determine the smallest thickness of a soap film that appears black when illuminated with 540-nm light, we need to find the minimum thickness that corresponds to destructive interference for this wavelength. When light waves undergo destructive interference, they cancel each other out, resulting in no reflection or a black appearance.
The condition for destructive interference is given by the equation:
2nt = (2k + 1)*(λ/2)
Where t is the thickness of the film, n is the refractive index of the film, k is an integer representing the order of the interference, and λ is the wavelength of light. In this case, we have n = 1.37 and λ = 540 nm. To find the smallest thickness, we set k = 0 and solve for t.
Plugging in the values:
2*(1.37)*t = (2*0 + 1)*(540 nm/2)
Simplifying:
2.74*t = 270 nm
t = 98.54 nm
Therefore, the smallest thickness of a soap film (n = 1.37) that would appear black when illuminated with 540-nm light is approximately 98.54 nm.