Anti-reflective lenses and interference

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SUMMARY

The discussion centers on the principles of anti-reflective (AR) coatings and their effectiveness at specific wavelengths, particularly λ=500nm. The key mechanism involves creating a new reflection that interferes destructively with the original light wave, achieved through precise control of the coating's thickness. The equation (k+1/2)λ=2nw+λ/2 illustrates how the ideal thickness of the coating is dependent on the wavelength of light. The participants clarify that the zero reflectance at λ=500nm is a result of optimizing the coating's thickness for that specific wavelength.

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  • Knowledge of the relationship between wavelength and material thickness
  • Basic grasp of light reflection and refraction concepts
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joshwarner
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Homework Statement
Anti-reflective coatings and interference
Relevant Equations
While learning about wave interference in anti-reflective lenses, I came across something mentioning these lenses being only efficient for a certain wavelength. I know these lenses appear blue or purple because, when adding the layer, the reflectance for a wavelength over 600nm isn't worth 0 but is for a wavelength around 500nm. Why is that?
reflectance ar.png
 
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How do you think anti reflectance works?
 
haruspex said:
How do you think anti reflectance works?
To remove the reflection on the glasses, the additional layer creates a new reflection of the light in order for the two waves reflected to interfere destructively when having a phase difference of an odd multiple of π. So using the distance traveled by the two waves, we get, at normal incidence, with w being the width of the coating layer : (k+1/2)λ=2nw+λ/2. However I still don't see how this explains the reflectance being worth 0 for λ=500nm.
 
joshwarner said:
To remove the reflection on the glasses, the additional layer creates a new reflection of the light in order for the two waves reflected to interfere destructively when having a phase difference of an odd multiple of π. So using the distance traveled by the two waves, we get, at normal incidence, with w being the width of the coating layer : (k+1/2)λ=2nw+λ/2. However I still don't see how this explains the reflectance being worth 0 for λ=500nm.
Now I am not sure what your question is.
In post #1, I thought you were asking why the reflectance is reduced more at one wavelength than another. That should be evident from your equations in post #3; the ideal thickness of the coating depends on the wavelength.
But now you seem to be asking how it manages to achieve zero reflectance, even at the wavelength the thickness is optimised for.
Which is it?
 
haruspex said:
Now I am not sure what your question is.
In post #1, I thought you were asking why the reflectance is reduced more at one wavelength than another. That should be evident from your equations in post #3; the ideal thickness of the coating depends on the wavelength.
But now you seem to be asking how it manages to achieve zero reflectance, even at the wavelength the thickness is optimised for.
Which is it?
Sorry for the confusion, I was asking why it was specifically for λ=500nm, I understand how it achieves a zero reflectance and why that only applies for a certain wavelength, but I didn't know it was for λ=500nm because of the thickness of the layer. Many thanks.
 

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