Scientists are testing a transparent material whose index of refraction for visible light varies with wavelength as n = 30.0 nm^(1/2)/λ^(1/2), where λ is in nm.
A 310-nm-thick coating is placed on glass (n = 1.50) what visible wavelengths will the reflected light have maximum constructive interference?
λ_constructive = (2nd)/m where m = 1,2,3,4
Δφ = 2π*(2nd)/λ
d is thickness of film[/B]
The Attempt at a Solution
I am really confused by the index of refraction notation n = 30.0 nm^(1/2)/λ^(1/2), but what I got out was that in the visible spectrum, n < 1. So you have air where n = 1. the film where n < 1, and glass where n = 1.5 . So light will reflect off the film with no phase difference, and light reflects off the glass with a phase shift of π. So the light has to travel 2d and have a phase shift of mπ (m = 1,3,5...) to have constructive interference.
π = 2π*(2 * (sqrt(30nm)/(sqrt(λ))) * 310nm) / λ
using a calculator I got λ = 358.6nm
using 3π, I got λ = 191.4nm which isn't in the visible spectrum
This answer is sadly wrong because the online answer box says there are two wavelengths that will have constructive interference.