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Old Feb 1st 2017, 04:25 AM   #1
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Layer thickness reflection

Hi,

For my exam-preparation I want to solve this question:
Three layers of material, with n1<n2<n3.
Light enters layer 1 perpendicular to the surface.
What should be the layer thickness of layer 2 to obtain reduced reflection (reflected wave at interface 12 is cancelled out by the reflected wave at interface 23)?

The solution should be: (wavelength)/4.
When I draw the picture, I do not understand this, because it seems to be increased reflection instead of reduced reflection.

My elaboration: https://postimg.org/image/qanujf30p/

Can anyone please help me?

Thanks!

Bas
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Old Feb 1st 2017, 09:47 AM   #2
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Originally Posted by BasWage View Post
Hi,

For my exam-preparation I want to solve this question:
Three layers of material, with n1<n2<n3.
Light enters layer 1 perpendicular to the surface.
What should be the layer thickness of layer 2 to obtain reduced reflection (reflected wave at interface 12 is cancelled out by the reflected wave at interface 23)?

The solution should be: (wavelength)/4.
When I draw the picture, I do not understand this, because it seems to be increased reflection instead of reduced reflection.

My elaboration: https://postimg.org/image/qanujf30p/

Can anyone please help me?

Thanks!

Bas
The main idea behind the cancellation we need for the wave at the interface to be at a half wavelength, so the reflected wave completely cancels...ie. the incoming wave has a node at the interface and the reflected wave is pi out of phase from the incident wave.

-Dan
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Old Feb 2nd 2017, 04:13 AM   #3
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Hi Dan,

Thank you for your response!
Yeah, I totally agree with you, I know that that is the intention of this exercise. However, as I worked it out in a drawing, it seemed that this phase shift of pi is not going to happen.

But, I think I found my fault.
Assuming that at the interface a phase shift of pi happens.
Apparently, if a wave strikes this interface at its node (coming from a top), the reflected wave travels back in the exact same way as it came in (so going back from the node to a top).
This is very counter intuitive (for me), because I thought a phase shift of pi implies that the wave gets mirrored about its neutral axis.

Anyhow, I think I understand it now!

Thanks again!
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Old Feb 6th 2017, 12:44 AM   #4
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Since there is a phase change of pi / 2 at both surfaces, we can neglect the effect of this phase change. It is as though it didn't happen at all.
For minimum intensity, the path difference has to be lambda / 2, of which
lambda / 4 is covered on the first leg and lambda / 4 on the return trip. Hence the thickness should be lambda / 4.
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