Waves: Superposition - Thickness of a Reflective coating

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Homework Help Overview

The problem involves determining the thickness of a magnesium fluoride (MgF2) film on glass that results in strong reflection for light of a specific wavelength. The context is within the study of wave interference, particularly in thin films.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for constructive interference of reflected waves and question the relationship between film thickness and wavelengths. There is an exploration of the equations related to standing waves and the conditions for interference.

Discussion Status

Some participants have offered guidance on focusing on the conditions for constructive interference, while others express uncertainty about relating wave amplitudes and wavelengths. Multiple interpretations of the problem are being explored, particularly regarding the necessary conditions for reflection and interference.

Contextual Notes

There is mention of a lack of clarity in the original poster's understanding of wave interference concepts, as well as references to external resources that may provide additional context. The discussion reflects an ongoing exploration of the problem without a definitive resolution.

Prophet029
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Homework Statement



What is the thinnest film of MgF2 (n=1.21) on glass that produces a strong reflection for the light with a wavelength of 531 nm?


Homework Equations



Open-Closed Standing wave

Fn= (nv/4L)
v = Wave Length * F


The Attempt at a Solution



Basically, I tried to plug in the data. I solved for Frequency by using velocity for light (300,000,000 m/s) = (5.31*10^-7m) * F
Then used the equation Fn= (nv/4L)
using n = 1.21
Alternatively the eq can be simplified into

L= ((n*wave length)/4)
I get 1.606*10^-7m, but I get the problem wrong. I think I overlooked something, but I'm not sure, can anyone help out. Most appreciated.
 
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Prophet029 said:

Homework Equations



Open-Closed Standing wave

Fn= (nv/4L)
This is not what you need. Instead, think about the two reflections that must constructively interfere. What must twice the thickness of the film--the extra distance traveled by one of the reflections--be in terms of wavelengths?
 
Last edited:
Doc Al said:
This is not what you need. Instead, think about the two reflections that must constructively interfere. What must twice the thickness of the film--the extra distance traveled by one of the reflections--be in terms of wavelengths?

Ok, I see. If I remember correctly, then 2 waves that constructively interfere have Amplitudes that add up to the Resultant wave amplitudes. But I'm having trouble relating this to the wavelengths. Pardon my lack of knowledge with interfering waves, my lecturer skipped over it and only talked about it briefly. So thank you very much for helping out.
 
Reading this discussion might help: http://www.physicsclassroom.com/Class/light/u12l1c.cfm"

The key is that the wave that reflects from the bottom surface must end up exactly in phase with the wave that reflects from the top surface in order for them to constructively interfere. So that extra distance--which is twice the film thickness--must equal how many wavelengths?

And how do you calculate the wavelength of light in a medium with refractive index n?

Another discussion that you might find useful is this: http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/thinfilm.html#c1"
 
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