GREPhysics.NET: GR8677 #73

GR8677 #73

Problem

GREPhysics.NET Official Solution

Optics $\Rightarrow$ }Thin film

In order for the thin film layer to be non-reflecting, it must cancel the reflected wavelengths---as in a destructive interference. The change in wavelengths is $\Delta \lambda = \lambda - \lambda/2 = \lambda /2$ , since the wave changes phase by $\pi$ at the interface between air and the coating, and changes phase again at the second interface between coating and glass. (Assume that $n_{air} < n_{coating} < n_{glass}$ .) Destructive interference is thus given by a half-integer wavelength change $m \lambda /2$ , where the smallest change is $\lambda/2$ .

The change in wavelength occurs over twice the thickness $t$ of the coating, thus

$2t = \Delta \lambda = \lambda/2. <br />$

This implies that $t =\lambda/4$ , as in choice (A).

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Comments

vsravani
2008-11-02 18:44:34

If anybody wants to get a clear explanation on thin film interference, they can go through

http://physics.bu.edu/~duffy/PY106/Diffraction.html

It really gives a very good explanation with a step by step procedure and figures.

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wangjj0120
2008-08-27 08:29:21

No~
the change in wavelength is not $\lambda$ /2.
When the incident wave (I) touches the coating, the first reflecting wave (R1) occurs and its phase was changed by pi, while the transmitting waves enter coating (T1) without changing phase. The second reflecting wave (R2) occurs when T1 touches glass, and phase of R2 was changed by pi (assum n(glass) > n(coating) > n(air) ). R2 touches interface between coating and air and then transmits into air, that is T2. T2 is in phase with R2. (phase of transmission at interface does't change). So the phase of R2 changes by pi due to interface happens only ONCE.

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madfish
2007-11-02 16:41:41

this question is ambiguous. It really needs to state that the coating is on BOTH sides on the glass, otherwise the math does not work out (since there wouldn't be a 180 degree phase shift on the back wall). However the problem states "it is necessary to coat a glass with A non-reflecting layer (A = one)" those ETS assholes.

jesford
2008-04-01 21:39:16

The coating is only on one side of the glass. There is a 180 degree phase shift upon reflection at BOTH the air-coating interface and the coating-glass interface, because in both cases the index of refraction is greater in the material that the light is reflecting off of, than it is in the material in which the light propagates. In other words, n(coating)>n(air), and n(glass)>n(coating).

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Mindaugas
2007-09-28 08:16:07

Some explanation http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

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irishroogie
2007-09-04 23:07:11

I still don't get the line, "The change in wavelength occurs over twice the thickness t of the coating, thus " Why 2x the thickness?

jive
2007-10-02 22:39:02

I believe it's because the light passes through the thickness twice -- once going in, and once coming out after being reflected.

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nitin
2006-11-16 12:35:58

"Destructive interference is thus given by a half-integer wavelength change $m\lambda /2$ , where the smallest change is $\lambda/2$ ."

Hmm.. well, the uninformed or careless would be inclined to think that your $m$ would assume any positive integral value, which will not be true for destructive interference. I guess you should have pointed out that $m$ should be an odd number, or simply use the form $(m+\frac{1}{2})\lambda$ , $m=0,1,2,...$ instead.

This question seems to presume a minimum thickness for the coating (I guess that is what is meant by "best choice"), which would then justify taking $m=0$ , or in your case $m=1$ .

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Some explanation http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

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