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

KEY IDEA
The reflected light from the film is brightest at the wavelengths `lambda` for which the reflected rays are in phase with one another. The equation relating these wavelengths `lambda` to the given film thickness L and film index of refraction `n_(2)` is either Eq. 35-50 or Eq. 35-51, depending on the reflection phase shifts for this particular film.
Calculations: To determine which equation is needed, we should fill out an organizing table. However, because there is air on both sides of the water film, the situation here is exactly like that, and thus the table would be exactly. Then, we see that the reflected rays are in phase (and thus the film is brightest) when
`2L=("odd number")/(2) xx (lambda)/(n_(2))`
which leads to Eq. 35-50:
`2L=(m+(1)/(2))(lambda)/(n_(2))`
Solving for `lambda` and substituting for L and `n_(2)`, we find
`lambda=(2n_(2)L)/(m+(1)/(2))=((2)(1.33)(320 nm))/(m+(1)/(2))=(851 nm)/(m+(1)/(2)`
For m = 0, this gives us `lambda = 1700 nm`, which is in the infrared region. For m = 1, we find `lambda= 567 nm`, which is yellow-green light, near the middle of the visible spectrum. For m = 2, `lambda = 340 nm`, which is in the ultraviolet region. Thus, the wavelength at which the light seen by the observer is brightest is
`lambda=567 nm." "("Answer")`