Chapter 26, Problem 58GP

Set Up: Interference occurs due to the path difference of light in the thin film. Originally the path difference was an odd number of half-wavelengths for cancellation to occur. If the path difference decreases by $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ wavelength, it will be a multiple of the wavelength, so constructive interference will occur.

Solve: Calling ΔT the thickness that must be removed, we have the path difference = 2ΔT = $\frac{1}{2}$λ/n and ΔT— λ/4n = (525 nm)/(4×1.40) = 93.75 nm. At 4.20 nm year, we have (4.20 nm/year)t = 93.75 nm and t = 22.3 years.

58. A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index 1.55. This film has the proper thickness to cancel normally incident light of wavelength 525 nm that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 nm per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?