Question

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?

Answer 1

In this situtaion, Eq. (i) gives the interference maxima. Solving for `lambda` and inserting the given data, we obtain
`lambda = (2 mu d)/(n + 1//1) = ((2)(1.33)(320 km))/(n + 1//2) = (851 nm)/(n + 1//2)`
For `m = 0`, this give us `lambda = 1700 nm` , which is in the infrared region. For `n = 1`, we find `lambda = 567 nm`, which is yellow-green light, near the middle of the visible spectrum. For `n = 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`.