Waves: Superposition - Thickness of a Reflective coating

AI Thread Summary
The discussion focuses on determining the optimal thickness of an MgF2 film on glass that results in strong reflection for light at a wavelength of 531 nm. The key concept is that for constructive interference, the extra distance traveled by the reflected wave must equal an integer multiple of the wavelength. Participants emphasize the importance of understanding the phase relationship between the two reflected waves, which requires calculating the wavelength of light in the medium using its refractive index. A common mistake noted is overlooking the requirement for the two waves to be in phase for constructive interference to occur. Overall, the thread highlights the necessity of grasping wave interference principles to solve the problem accurately.
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?
 
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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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