Will Reflected Light Be Eliminated by Interference in These Thin Film Scenarios?

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Homework Help Overview

The discussion revolves around the conditions under which reflected light can be eliminated by interference in thin films, specifically examining various configurations of materials with different indices of refraction. The problem involves determining the validity of statements regarding the thickness of the film relative to the wavelength of light within the film.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are analyzing the conditions for destructive interference and the role of phase reversals at material interfaces. There are attempts to validate specific cases based on the number of phase reversals and the relationship between the indices of refraction of the materials involved.

Discussion Status

Some participants are questioning the accuracy of the phase reversal counts in the original poster's analysis. Clarifications are being sought regarding the criteria for determining phase reversals when light reflects off different materials. There is acknowledgment of correct interpretations in some cases, while others remain under scrutiny.

Contextual Notes

Participants are operating under the assumption that the thickness of the film and the wavelength of light are critical to understanding interference patterns, and they are exploring how these factors interact with the indices of refraction of the materials involved.

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Homework Statement


The diagram shows light incident from above on a film of thickness d. Each of the three materials in the figure can be chosen to be air, with index of refraction n=1.00, water, with index n=1.33, or glass, with index n=1.50. Under which of the following conditions will the reflected light be completely or nearly eliminated by interference?
(lambda2 refers to the wavelength of the light inside the thin film.)
(Note: In the case that d<<lambda2, you can assume the thickness is so small that the travel distance in material 2 has negligible affect on the phase of the wave. Treat this as if the two interfaces are almost right on top of each other.)
http://s3.amazonaws.com/answer-board-image/390a9f5e-a2a9-4fbe-a6f6-c33689d8251c.gif

1) T/F d=lambda2/4, material 1 is air, 2 is water, 3 is glass.
2) T/F d=lambda2/2, material 1 is air, 2 is water, 3 is glass.
3) T/F d=lambda2/4, material 1 is glass, 2 is air, 3 is glass.
4) T/F d=lambda2/4, material 1 is water, 2 is glass, 3 is air.
5) T/F d<<lambda2, material 1 is water, 2 is glass, 3 is air.

Homework Equations



2*n*d*sinθ = (m-(1/2))*λ
m=integer
λ=wavelength
n=index of refraction = speed of light in vacuum/speed of light in medium
d= thickness

The Attempt at a Solution


We are looking for places where the refracted light is eliminated by interference, so we are looking for destructive interference.
Destructive if Path-length Difference(PLD) 2nd= (m + (1/2))λ when the sources are exactly in phase.
Destructive if Path-length Difference(PLD) 2nd = mλ when the sources are exactly out of phase.
There are phase reversals when it is reflecting of a higher n.

1) d=lambda2/4, material 1 is air(n=1.00), 2 is water, 3 is glass.
True because
There would be 2 phase reversals [1 between air and water and the other between water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ
d=2λ/4
(2λ/4) = (1/2)λ
(2/4)λ= (1/2)λ
(1/2)λ= (1/2)λ

2) d=lambda2/2, material 1 is air(n=1.00), 2 is water(n=1.33), 3 is glass(n=1.50).
False because
There would be 2 phase reversals [1 between air and water and the other between water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ
d=2λ/2
2λ/2 = (1/2)λ
(2/2)λ = (1/2)λ​

3) d=lambda2/4, material 1 is glass(n=1.50), 2 is air(n=1.00), 3 is glass(n=1.50).
True because
There would be 2 phase reversals [1 between air and glass and the other between air and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ.
d=2λ/4
(2λ/4) = (1/2)λ
(2/4)λ= (1/2)λ
(1/2)λ= (1/2)λ

4) d=lambda2/4, material 1 is water(n=1.33), 2 is glass(n=1.50), 3 is air(n=1.00).
True because
There would be 2 phase reversals [1 between air and water and the other between the water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ.
d=2λ/4
(2λ/4) = (1/2)λ
(2/4)λ= (1/2)λ
(1/2)λ= (1/2)λ

5) d<<lambda2, material 1 is water(n=1.33), 2 is glass(n=1.50), 3 is air(n=1.00).
False because
There would be 2 phase reversals [1 between air and water and the other between the water and glass] So the wavelength would be exactly in phase which means it needs to have PLD of (1/2)λ.
d<<lambda2
λ2= (1/2)λ

I thought i was doing right but the answer is incorrect.
Any help would be greatly appreciated.
 

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Hello.

You are not getting the number of phase reversals correct for some of the cases. How do you decide if a phase reversal occurs?
 
TSny said:
Hello.

You are not getting the number of phase reversals correct for some of the cases. How do you decide if a phase reversal occurs?
A phase reversal happens when the n it is "hitting" is larger then what the "n" was before.

1) d=lambda2/4, material 1 is air(n=1.00), 2 is water(n=1.3), 3 is glass(n=1.5).
So for this one the wave is traveling in air and then it hits water, which has a higher "n" value. this means that part of the wave is reflected back and part continues through the water. The wave traveling in the water "hits" the glass and glass has a higher "n" value then water so part of the wave reflects back and some continues through the object.

I thought this was how you are supposed to tell. Is it incorrect?
 
Yes. You got this one right. But some of the other cases are incorrect. For example, check case(3).
 

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