Waves: Superposition - Thickness of a Reflective coating

In summary, the thinnest film of MgF2 (n=1.21) on glass that produces a strong reflection for the light with a wavelength of 531 nm is 1.606*10^-7m.
  • #1
Prophet029
11
0

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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  • #2
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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  • #3
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.
 
  • #4
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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1. What is wave superposition?

Wave superposition is a phenomenon where two or more waves overlap in a medium, resulting in the combined amplitude of the waves at each point in the medium.

2. How does wave superposition work?

When two waves overlap, their amplitudes are added together at each point in the medium. This can result in constructive interference, where the amplitudes add up to a larger value, or destructive interference, where the amplitudes cancel each other out.

3. What is the thickness of a reflective coating?

The thickness of a reflective coating refers to the distance between the top and bottom layers of the coating, which determines how much light is reflected off the surface.

4. How does the thickness of a reflective coating affect wave superposition?

The thickness of a reflective coating can affect wave superposition by altering the path length that the waves must travel, which can result in changes in interference patterns and ultimately affect the amount of light reflected off the surface.

5. Why is the thickness of a reflective coating important?

The thickness of a reflective coating is important because it can impact the quality and efficiency of the reflective surface. A coating that is too thin or too thick may not produce the desired reflection or may cause unwanted interference effects.

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