Antireflection Coating Problem

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In summary, the minimum thickness of the coating that will cause reflection rays R2 and R4 to interfere destructively is 1.38 microns. The minimum thickness that will cause reflection rays R1 and R3 to interfere constructively is 1.52 microns.
  • #1
superspartan9
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Homework Statement


"A glass lens (nglass = 1.52) has an antireflection coating of MgF2 (n = 1.38). (a) For 517-nm light, what minimum thickness of the coating will cause reflection rays R2 and R4 to interfere destructively assuming normal incidence? (b) Interference will also occur between forward moving rays R1 and R3. What minimum thickness will cause these rays to interfere constructively?

We are given a figure (attached as temp.jpg).


Homework Equations



Δphase = π*(2k + 1) for destructive, Δphase = 2kπ for constructive, Δphase = (2*thickness*n / λ) * 2π + π for part (a) because the rays go from the air to the back wall of the coating an reflect back, part (b) has got me stumped

The Attempt at a Solution



Ok, so attempting to solve part (a), I've solved the equations for Δphase (setting them equal to each other), I got 2k = ((4 * 1.38) / (517 e -9)) * thickness. Now I don't know where to go from here. As far as part (b), I don't understand what the rays are actually doing according to the figure.
 
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  • #2
The figure is not shown.ehild
 
  • #3
Weird, I thought I attached it to the original post. Regardless here it is.
 

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  • #4
Ok, I solved for part (a), and it makes sense. Now going to part (b), I face a problem concerning the thickness of the glass and how it would play into the phase change.

So far, I've concluded that, using constructive interference equation, Δphase = 2kπ = [(thickness * n) / λ] * 2π + 2π + [(thickness or distance traveled in glass * nglass) / λ] * 2π

I get this equation through analyzing the three phase changing elements in the problem. Firstly, the phase change through the film, which only goes through once hence it's just 1*thickness of the film. Secondly, since the light goes from n = 1 to n = 1.38 to n = 1.52, there are going to be two π shifts as the light goes from low n to high n. Lastly and the strange part, the phase change in the glass. I originally thought there would be no change in phase here, but that yielded the same equation as part (a), so unless the point of this part is to show that the same thickness of film will reflect destructive waves from the glass and let constructive waves through, I don't know where to go.
 
  • #5
No need to calculate with the phase change in the glass. In the reflected light, the directly reflected ray interferes with the one reflected from the back boundary of the film and traversing the film twice. Both reflected rays change phase upon reflection as they are reflected from a higher refractive index material. The phase changes (both pi) cancel, and the phase difference is simply (4pi/lambda)*n* (thickness).

In the transmitted light, one ray travels through directly, the other reflects back from the back interface of the MgF2 layer and changes phase by pi; then goes back to the air+layer interface, reflects back (with no phase change this time) and goes through the layer again. So you have a ray which went through the layer once without reflection and the other which went through the layer three times and reflected twice, once from a higher index material, once with a lower index one. What is the phase difference between these two rays then?

ehild
 
  • #6
[(thickness * n * 2∏) / λ] - ([(3 * thickness * n * 2∏) / λ] + ∏) = 2k∏ ?
 
  • #7
Actually, I'd probably want to switch the two around to get a positive phase change, right?
 
  • #8
Correct, and make it positive:smile:

ehild
 

1. What is an antireflection coating and why is it important?

An antireflection coating is a thin layer applied to the surface of a material to reduce the amount of light reflected off of it. This is important because it helps to improve the clarity and contrast of images, as well as reduce glare and increase the efficiency of optical devices.

2. How does an antireflection coating work?

An antireflection coating works by utilizing the principles of interference and phase cancellation. When light passes through a material, some of it is reflected off the surface due to the difference in refractive index between the material and air. The antireflection coating is designed to have a refractive index that is between these two values, causing the reflected light waves to cancel each other out and reducing the amount of light reflected.

3. What are the common problems associated with antireflection coatings?

One common problem with antireflection coatings is the degradation of the material over time due to environmental factors such as exposure to heat, moisture, and chemicals. Another issue is the potential for the coating to crack or peel, which can reduce its effectiveness. Additionally, improper application or handling of the coating can lead to defects that can impact its performance.

4. How do you troubleshoot antireflection coating problems?

The first step in troubleshooting antireflection coating problems is to identify the specific issue. This can be done through visual inspection, measuring the reflectivity of the coated surface, or performing durability tests. Once the problem is identified, it can be addressed by adjusting the coating thickness or composition, optimizing the fabrication process, or improving the environmental conditions during application and use.

5. What are some current research and developments in the field of antireflection coatings?

There is ongoing research in the development of new materials and fabrication techniques for antireflection coatings, with a focus on improving their durability and performance. Some advancements include the use of nanostructures and metamaterials to achieve even lower levels of reflectivity, as well as the development of self-cleaning coatings that can repel dirt and water. Other research is focused on incorporating antireflection coatings into new applications, such as solar cells and virtual reality devices.

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