Note that the angles in the sketch are exaggerated for clarity. Interference is governed by phase differences caused by reflection and by differing path lengths.
First consider the phase difference caused by reflection. Ray 1 is reflected at an air/film interface. Since nair < nfilm, the phase of Ray 1 is flipped relative to the incident ray. Ray 2 is reflected at an film/glass interface. Since nfilm < nglass, the phase of Ray 2 is also flipped relative to the incident ray. Since both rays have flipped, Δreflection/2π = 0.
Next consider the phase difference caused by the fact that Ray 2 travels an extra distance 2t (remember that we assume that the angles involved are small). This corresponds to a phase difference (number of wavelengths) of Δtravelled/2π = 2t/λfilm, where λfilm is the wavelength of the light in the thin film. The wavelength in the film is related to the wavelength in air by nfilmλfilm = nairλair.
Since Δreflection/2π = 0, we get destructive interference when
Δtravelled/2π = ½n, where n = 1, 3, 5, … .
Substituting in our expression for Δtravelled we have
2t / (nairλair/nfilm) = ½n.
Solving for t yields,
t = ¼n(nairλair/nfilm) , where n = 1, 3, 5, … .
The thinnest film for this to occur is when n = 1, so
t = ¼(1)(1.00)(650nm)/(1.42) = 114.4nm
What is the thinnest film of a coating with n=1.42 on glass n=1.52 for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?
I am confused on the n's and I have no idea what equation to use, it seems like I don't have enough info! HELP please!
1 answer
0 answers