Anti reflective coating, finding the 'correct thickness'

In summary, the conversation discusses the correct thickness of a coating on glass lenses to reduce reflections. The thickness is determined by the wavelength of the light passing through the coating and the refractive index of the coating and the glass. This thickness should cause a slight shift in the phase of the light, leading to constructive interference and reducing reflections. The formula used to calculate the thickness is t = λ0/4n, where λ0 is the vacuum wavelength and n is the refractive index of the coating.
  • #1
Glorzifen
25
0

Homework Statement


To reduce reflections from glass lenses (n ≈ 1.5), the glass surfaces
are coated with a thin layer of magnesium fluoride (n ≈ 1.39).
What is the correct thickness of the coating for green light (510 nm
vacuum wavelength)?

Homework Equations


[tex]\lambda[/tex] = 2nd/(m+1/2)

The Attempt at a Solution


I don't really know where to start. It says "correct thickness"...what defines the correct thickness? Is it when a certain value is 0? Or when a certain amount of light gets through? I'm confused as to where to start because I'm not sure what this means...if anyone could give me a hint, that would be awesome. Thanks.
 
Physics news on Phys.org
  • #2
I've been studying this a bit - is the correct thickness they're looking for one that gives off destructive interference? Normally, to get destructive interference...I would do this:

r1 has a phase shift of pi when its reflected off of the coating (in my notes that is also written as lambda/2...not sure why that is). r2 continues on through the coating and reflects off the other edge, travels back through the coating and out into the air where it interferes with r1.

In a previous example I did, there was air on the other side of the thin film and so there was no phase shift when r2 reflects off of the other edge. In that case, I would want the phase of r2 to remain unchanged or changed only by k[tex]\lambda[/tex]/2n. Right?

But in this case, there is an additional phase shift off the outer edge because the light is going from the coating to the glass (n=1.39 to n=1.5). I'm confused as how to fit that into my solution.

Any help would be appreciated.
 
  • #3
Okay. Did a bit more thinking on this:

r1 is shifted by [tex]\lambda[/tex]/2 (or pi, still not entirely sure how those two equate. The only formula I have concerning them is k=2pi/[tex]\lambda[/tex]...where does the k go?)

Anyway, since there is this issue going to the coating --> glass (low to high again) there is ANOTHER phase shift in r2 by [tex]\lambda[/tex]/2. Normally there wouldn't be the same shift so we could just have a thickness that leaves the phase of r2 untouched...but in this case they have the same shift, so if we have a thickness that leaves the phase of r2 untouched...they will be the same and therefore constructive interference will occur. Right?

So then we want to have a thickness that causes a slight shift in the phase of r2. The smallest shift we can do is by 1/2 a wavelength...so:

2L = [tex]\lambda[/tex](1/2) <-- the [tex]\lambda[/tex] we want is the wavelength of the light once its going through the coating though, so:
2L = [tex]\lambda[/tex]0/n (1/2)
L = t = [tex]\lambda[/tex]0/2n (1/2)
t = [tex]\lambda[/tex]0/4n
= 501nm/4(1.39)
= 92µm (the correct answer)

That is the right answer, but there are a few questions I put in there along the way. I want to make sure I'm understanding this pretty comprehensively so if there is anything anyone wants to clarify for me I would be grateful.
 
  • #4
Glorzifen said:
r1 has a phase shift of pi when its reflected off of the coating (in my notes that is also written as lambda/2...not sure why that is).

It's because if you phase shift any sine or cosine function by pi radians you move some integer multiple of lambda/2.

Glorzifen said:
k[tex]\lambda[/tex]/2n. Right?

If k is 2pi/lambda, then that's not the correct equation. If k is some integer then that's correct.

Glorzifen said:
The only formula I have concerning them is k=2pi/[tex]\lambda[/tex]...where does the k go?)

Again you shouldn't be using k like this, probably some notation confusion going on here.

Glorzifen said:
Anyway, since there is this issue going to the coating --> glass (low to high again) there is ANOTHER phase shift in r2 by [tex]\lambda[/tex]/2. Normally there wouldn't be the same shift so we could just have a thickness that leaves the phase of r2 untouched...but in this case they have the same shift, so if we have a thickness that leaves the phase of r2 untouched...they will be the same and therefore constructive interference will occur. Right?

Yes. If you shift anything by a integer number of wavelengths the waves will still line up like they were the same wave (different from shifting by integer number of half wavelengths).

To remember this:

Integer number of wavelengths --> Doesn't change the initial phase difference
Integer number of half-wavelengths --> Switches the initial phase difference (in phase to out of phase, vice versa)
 

1. How does anti-reflective coating work?

Anti-reflective coating works by reducing the amount of light that is reflected off the surface of a material, such as glass or plastic. It does this by using layers of materials with different refractive indexes to cancel out the reflections. This allows more light to pass through the material, resulting in clearer and more transparent viewing.

2. Is there a specific thickness that is considered the "correct thickness" for anti-reflective coating?

The "correct thickness" for anti-reflective coating depends on several factors, such as the type of material being coated, the refractive index of the coating materials, and the desired level of reflection reduction. Generally, a thickness between 100-150 nanometers is considered optimal for most applications.

3. What are the benefits of using anti-reflective coating?

Anti-reflective coating has several benefits, including reducing glare and reflections, improving contrast and clarity, and increasing light transmission. It can also make materials more scratch and smudge resistant, and can enhance the overall aesthetic of the material.

4. Can anti-reflective coating be applied to any type of material?

Anti-reflective coating can be applied to a wide range of materials, including glass, plastic, and even certain metals. However, the success of the coating may vary depending on the specific material and its properties. It is best to consult with a professional to determine the suitability of anti-reflective coating for a particular material.

5. How long does anti-reflective coating last?

The longevity of anti-reflective coating can vary depending on the quality of the materials used and the conditions it is exposed to. Generally, it can last for several years with proper care and maintenance. However, it may need to be reapplied or touched up over time to maintain its effectiveness.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top