Optics question regarding rays.

  • #1
Ishida52134
139
0

Homework Statement


Given a cellphone layered with an anti reflection film with width L. A ray denoted with lambda hits the screen and some of the ray passes through while some of it reflects off from both inside the film and where the ray hits the film. The index of reflection from top to bottom is denoted by n1, n2, and n3 and n1 < n2 < n3. At what thickness L would the two rays interfere.
If you are confused here is a picture of what I'm talking about. Just the diagram, the labels are different. http://upload.wikimedia.org/wikipedia/commons/f/f7/Optical-coating-1.png
It's the bottom picture, just flip it then rotate it -90 degrees.


Homework Equations


Snell's Law n1sinx = n2siny


The Attempt at a Solution


I just tried applying Snell's Law and using concepts of terminal refraction. I'm just confused with the whole question in general...
thanks.
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,435
62

Homework Statement


Given a cellphone layered with an anti reflection film with width L. A ray denoted with lambda hits the screen and some of the ray passes through while some of it reflects off from both inside the film and where the ray hits the film. The index of reflection from top to bottom is denoted by n1, n2, and n3 and n1 < n2 < n3. At what thickness L would the two rays interfere.
If you are confused here is a picture of what I'm talking about. Just the diagram, the labels are different. http://upload.wikimedia.org/wikipedia/commons/f/f7/Optical-coating-1.png
It's the bottom picture, just flip it then rotate it -90 degrees.


Homework Equations


Snell's Law n1sinx = n2siny


The Attempt at a Solution


I just tried applying Snell's Law and using concepts of terminal refraction. I'm just confused with the whole question in general...
thanks.

I assume you mean the two rays interfere destructively [the reflection is reduced]

For destructive interference, the two reflected rays [surface and bottom interface] need to be out of phase when they eventually get to you.
That usually means a path difference in terms of wavelengths.
You also need to account for whether the light will undergo a phase shift as it reflects - depends on the refractive index each side of the interface.
You also need to consider the medium in which the extra path difference is to occur, as you want to use the wavelength in that medium.
 

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