Newton's Rings and Thin Film Interference

I'm sorry, I am not sure how to answer that question. In my understanding, the airspace has no thickness, as it is just the gap between the lens and the surface it is resting on. Can you please provide a hint or further explanation?So how much thicker is the lens in the centre than at the edge?In summary, the question is asking for the difference in thickness between the center and the edge of the lens, not the thickness of the airspace. Since the 31st dark ring is at the edge of the lens, the thickness of the lens at the edge is 8.5 micrometers, while the thickness at the center is 0 micrometers. Therefore
  • #1
Abu

Homework Statement


A total of 31 bright and 31 dark Newton's rings (not counting the dark spot at the center) are observed when 550-nm light falls normally on a planoconvex lens resting on a flat glass surface. How much thicker is the center than the edges?

Homework Equations


2t = mλ
2t = (m+1/2)λ

The Attempt at a Solution


20021028002405__newtonrings1.png


The above image is just one I found online. As for my attempt:
Since the incident ray is being reflected off of a surface where it would have gone faster had it not been reflected, there is no phase shift. However, when it reflects off of the flat glass, it is being reflected off of a surface where it would have gone slower had it not been reflected, thus giving a phase shift.

That means that the formula I will be using is:
2t = mλ for destructive interference
2t = (m+1/2) for constructive interference.

As it is asking how much thicker the center is than the end, my first instinct is to find the thickness of the center, than find the thickness at the very edge, and find the difference.

Since the center is destructive however, the formula used is 2t = mλ
The center means that m = 0
That means that the thickness of the air film between the curved surface is zero (which I think makes sense because there is no room for air at the center)

The furthermost point of the Newton rings is destructive interference (NOT SURE ABOUT THIS) because the last ring would have color, and beyond that there is no more color.

So I am tempted to use this formula:
2t = mλ for destructive interference
So...
2t = 31(550)
t = 8525 nm

Can someone tell me if my reasoning is right? Specifically how I assume the furthermost point is destructive interference and also how I assume that the thickness in the center is zero. Also, why does the question say how much thicker is the center than the edges when in reality the center has zero thickness (value for t was zero)

Thank you for your time.
 

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  • #2
Abu said:
how much thicker is the center than the edges when in reality the center has zero thickness
The question is asking about the thickness of the lens. The zero thickness you found is for the airspace under the lens.
 
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  • #3
haruspex said:
The question is asking about the thickness of the lens. The zero thickness you found is for the airspace under the lens.
The answer for the question is 8.5 micrometers in the textbook. Since I got 8525 nm from my attempt, this makes 8.525 micrometers, which is the thickness of the airspace under the lens at the edges. So that is why I found it confusing why they said how much thicker is the center than the edges. But I do realize that the width of the lens in the center is larger than at the edges, so it makes me wonder if there is a typo in the textbook question or if I am misunderstanding something
 
  • #4
Abu said:
The answer for the question is 8.5 micrometers in the textbook. Since I got 8525 nm from my attempt, this makes 8.525 micrometers, which is the thickness of the airspace under the lens at the edges. So that is why I found it confusing why they said how much thicker is the center than the edges. But I do realize that the width of the lens in the center is larger than at the edges, so it makes me wonder if there is a typo in the textbook question or if I am misunderstanding something
I am not sure whether you understood what I wrote.
There is no typo. When the question asks how much thicker at the centre than at the edge it is referring to the lens thickness, not the airspace thickness. It is not really even ambiguous. If it meant the thickness of the airspace it would have said so, or, more likely, it would have asked for the depth of the airspace. "Thickness" suggests something solid.
 
  • #5
haruspex said:
I am not sure whether you understood what I wrote.
There is no typo. When the question asks how much thicker at the centre than at the edge it is referring to the lens thickness, not the airspace thickness. It is not really even ambiguous. If it meant the thickness of the airspace it would have said so, or, more likely, it would have asked for the depth of the airspace. "Thickness" suggests something solid.
Okay, sorry sir. I am only expected to be using 2t = mλ and 2t = (m+1/2) . I realize that the question is asking for the thickness of the lens. But first, am I wrong in saying that t in the formula 2t = mλ means the thickness of the airspace? If t in the formula 2t = mλ is in fact the thickness of the airspace, then I have no idea how to find the thickness of the lens. All I know is that the final answer for the question in the textbook answer sheet is 8.5 micrometers. The final answer I got was 8.525 micrometers, but it was for the thickness of the airspace underneath the edge.

Thank you for your patience.
 
  • #6
Abu said:
If t in the formula 2t = mλ is in fact the thickness of the airspace, then I have no idea how to find the thickness of the lens.
In relation to the diameter of the whole lens, where do you think the 31st dark ring is?
 
  • #7
haruspex said:
In relation to the diameter of the whole lens, where do you think the 31st dark ring is?
The 31st dark ring is at the edge of the lens. It is the furthermost point from the center.
 
  • #8
Abu said:
The 31st dark ring is at the edge of the lens. It is the furthermost point from the center.
Right, so what is the thickness of the lens there?
How much deeper is the airspace at the edge than in the centre?
So how much thicker is the lens in the centre than at the edge?
 
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  • #9
haruspex said:
Right, so what is the thickness of the lens there?
How much deeper is the airspace at the edge than in the centre?
So how much thicker is the lens in the centre than at the edge?
I just realized. The airspace at the edge is the thickness of the lens at the center, because the airspace at the center is zero. If that is correct, than holy cow I am either extremely dumb or just really tired haha.

Besides that, were my other assumptions in my original attempt at the question correct?

Thank you sir.
 
  • #10
Abu said:
If that is correct, than holy cow I am ... really tired haha.
Yes.
 
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1. What are Newton's Rings?

Newton's Rings are a type of interference pattern observed when monochromatic light is reflected between a convex lens and a flat glass surface. They appear as a series of concentric, alternating light and dark rings.

2. How are Newton's Rings formed?

Newton's Rings are formed when light waves reflected from the top and bottom surfaces of a thin, air-filled gap between a convex lens and a flat glass surface interfere with each other. This interference creates the alternating bright and dark rings.

3. What is the difference between Newton's Rings and Thin Film Interference?

Newton's Rings are a specific type of thin film interference, where the film is formed between a convex lens and a flat glass surface. Thin film interference, on the other hand, can occur in any situation where light is reflected from a thin film, such as a soap bubble or a layer of oil on water.

4. How do we use Newton's Rings and Thin Film Interference in scientific research?

Newton's Rings and Thin Film Interference can be used to study the properties of thin films, such as their thickness and refractive index. They are also used in the manufacturing of optical devices, such as lenses and mirrors, to ensure the quality and accuracy of the thin films used.

5. Are Newton's Rings and Thin Film Interference only observed with visible light?

No, Newton's Rings and Thin Film Interference can also occur with other types of electromagnetic radiation, such as infrared and ultraviolet light. In fact, they can even be observed with waves of particles, such as electrons.

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