Thin-Film Interference: Coating a Lens - Answers found, explanation needed

[FishStik]

Thin-Film Interference: Coating a Lens -- Answers found, explanation needed!

Homework Statement

Boxed answers are the correct ones.

Homework Equations

2L=(m+1/2)(λ/n) for constructive interference (maximum reflection, bright film)
2L=(m)(λ/n) for destructive interference (minimum reflection, dark film)

Where:
L = thickness of film
m = fringe count (for m=0,1,2,3,4...)
λ = wavelength of light
n = refractive index of coating

The Attempt at a Solution

I got the correct answer (a) for Fall2008 #20 by using the equation for constructive interference, plugging in 0 for m, 1000 nm for λ, and 1.8 for n.
Now for Sp2011 #11, I took it as essentially the same question except this one is asking for minimum reflection, so I should use the equation for destructive interference. Doing that gets me (e), which is incorrect. The correct answer is (b), which is half of (e), implying that I should use the equation for constructive interference.

Am I missing a fundamental difference between the two problems besides one asking for minimum and another asking for maximum reflection? Thanks for the help.

 

[technician]

The λ of the light in the coating is 550/n. The light must travel a distance of λ/2 in the coating to emerge from the coating out of phase with the reflected light byλ/2 and cause destructive interference. In the first example this means that the coating needs to be λ/4 thick.(106nm)
In the second example the logic is exactly the same but this time a phase change of λ/2 occurs at the junction of the coating and the visor of the helmet. A phase change of λ/2 occurs when there is reflection from a boundary when the light passes from high refractive index (1.8) into low refractive index (1.5). The wavelength of the light in the coating is 1000/n
A phase change of 0.5 of this wavelength occurs at the coating/visor junction. To MAXIMISE the reflection the 2 waves should emerge in phase therefore the path difference through the coating (2 x thickness) needs to be λ/2 giving a total phase difference of 1λ
In the first example there is no λ/2 phase difference at the junction between the coating and the lens because the refractive index of the lens is greater than the refractive index of the coating.

 

[ehild]

In the second example the logic is exactly the same but this time a phase change of λ/2 occurs at the junction of the coating and the visor of the helmet. A phase change of λ/2 occurs when there is reflection from a boundary when the light passes from high refractive index (1.8) into low refractive index (1.5).

The opposite is true: There is no phase change when the light reflects from a lower-index medium, and the phase of the reflected wave changes by pi when the light enters from a lower index medium to the boundary with a higher index one.

The pi phase change corresponds to half-wavelength optical path difference.

ehild

 

[technician]

ehild... you are correct, I have checked my notes and got it the wrong way round!
I think that in the first question there is a phase change of pi at both surfaces but in the second question there is only 1 phase change of pi at the air/coating surface.
I think it makes no difference to the answers but what I wrote was confusing.
Apologies to FishStik and to you... and thanks.

 

[ehild]

You are welcome

ehild