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 situationis like that of Figure `(9.1),` for which equation `(9.1)` given the interference maxima. Solving for `lambda` and inserting the given date, we obtain
`lambda=(2mud)/(m+1//2)=((2)(1.33)(320nm))/(m+1//2)=(851nm)/(m+1//2)`
for `m=0,` this give us `lambda=1700 nm,` which is in the infrared region. For `m=1,` we find `lambda=567 nm,` which is yellow-green light, the middle of the visible spectrum. For `m=2,lambda=340 nm,` which is in the ultraviolet region. So the wavelength at which the light seen by the observer is brightest is
`lambda=567 nm.`