Final answer:
The wavelength of light reflected by a water film of 300 nm thickness and an index of refraction of 1.33 that is brightest to an observer is 532 nm, which corresponds to green light in the visible spectrum.
Step-by-step explanation:
To determine at what wavelength light reflected from a water film is brightest, we need to consider thin film interference. The index of refraction of the water is given as 1.33, and the thickness of the film is 300 nm. When light hits the film, some of it is reflected at the top surface, and some is transmitted, then reflected off the bottom surface. The constructive interference happens when the phase shift between these two reflected waves leads to them being in phase. The phase shift due to reflection at a medium of higher refractive index is π, and the additional phase shift after traveling through the film and back is given by 2nt/λ, where n is the refractive index, t is the thickness of the film, and λ is the wavelength of light in the water film.
The total phase difference must be an integer multiple of 2π for constructive interference, so our condition is 2nt/λ' + π = m2π, where λ' is the wavelength of light in the film (related to the wavelength in air λ by λ' = λ/n), and m is an integer (the order of interference). Rearranging for λ, the reflected wavelength in air, we get:
λ = 2nt/(m-1/2)
For the first-order interference (m=1), we find:
λ = 2*1.33*300 nm/(1-1/2) = 4*1.33*300 nm = 1596 nm
Since 1596 nm is outside the visible spectrum, we look for the next order (m=2), which yields:
λ = 2*1.33*300 nm/(2-1/2) = 8*1.33*300 nm/3 = 1064 nm
Again, this is outside the visible spectrum. Continuing to higher orders until we find a wavelength within the range 400 nm - 690 nm:
For m=3, λ = 532 nm, which falls within the visible spectrum and corresponds to green light. Thus, for the given film thickness and refractive index, the wavelength of light that is most constructively reflected and visible is 532 nm.