Final answer:
When white light falls perpendicular to the surface of the film, certain wavelengths are absent from the reflected light due to interference. Using the equation for constructive/destructive interference in thin films, we can determine these wavelengths. For a film of magnesium fluoride with a thickness of 105 nm, the wavelengths that result in destructive interference are around 172.43 nm.
Step-by-step explanation:
When white light (which contains all the visible wavelengths) falls perpendicular to the surface of the film, some of the light reflects and some of it passes through the film. The reflected light experiences interference, resulting in certain wavelengths being absent in the reflected light. This interference occurs when the thickness of the film is such that there is a path difference between the reflected rays that is equal to a multiple of the wavelength.
To determine the wavelengths that are absent from the reflected light, we can use the equation for constructive/destructive interference in thin films:
2nt = (m + 0.5)λ,
where n is the index of refraction of the film, t is the thickness of the film, m is an integer representing the order of the interference, and λ is the wavelength.
In this case, the film is made of magnesium fluoride (n = 1.38) and has a thickness of 105 nm. We want to find the wavelengths that result in destructive interference, which corresponds to the first-order interference (m = 1).
Substituting the given values into the equation, we have:
2(1.38)(105 nm) = (1 + 0.5)λ,
simplifying gives:
λ = (2(1.38)(105 nm))/(1 + 0.5).
Calculating the value of λ gives:
λ ≈ 172.43 nm.
Since we only want the wavelengths within the range of visible light (between 400-700 nm), we can conclude that the reflected light will be missing the wavelengths around 172.43 nm.