White light, with a uniform intensity across the visible wavelength range 430 - 690 nm, is perpendicularly incident on a wate film, of index of refraction `mu = 1.33` and thickness `d = 320`nm, that is suspended in air. At what wavelength `lambda` is the light reflected by the film brightest to an observer?

Explanation

This situation is like that of Figure shown, for which equation written below gives the interference maxima.
Solving for `lambda` and inserting the given data, we obtain
`lambda = (2mu d)/(m+1//2)`
`=((2)(1.33)(320 nm))/(m+1//2)`
`=(851 nm)/(m+1//2)`
for m = 0, this give us `lambda = 1700 nm`, which is in the infrared region. For m = 1, we find I = 567 nm, which is yellow-green light, near the middle of the visible spectrum. For `m = 2, lambda = 340 nm`, which is the ultraviolet region. So the wavelength at which the light seen by the observer is brightest is
`lambda = 567 nm`. Ans.