Plano-convex lens with phase changes

In summary, the person is asking for help with finding the radius of curvature of a lens and has provided a diagram of the intersecting chords of a circle. They are not sure what to do next and ask for help.
  • #1
jisbon
476
30
Homework Statement
A plano-convex lens is placed on a glass plate with the convex surface in
contact. Light with a wavelength of 633 nm is illuminated from above, which creates Newton’s rings. The rings are composed of 75 concentric dark rings with the largest one at the
outer edge of the lens. Calculate the radius of curvature R of the convex surface
of the lens. The refractive index of the lens is 1.52, and its radius is 1.50 cm.
The glass plate has an index of refraction of 1.55.
Relevant Equations
##2nt=m \lambda##
1579768883746.png


By drawing arrows, I can find out the following:

1579768984933.png

Since they are in different phases, I get:

##2nt=m \lambda## for dark fringes.
In this case, they want me to find the radius of curvature. I'm not sure what to really proceed on here. What I figured out so far is that I will be able to find the thickness of the gap between the lens and the glass plate using the formula when m=75 since the largest one is at the outer edge of the lens. Any tips to proceed on with this question?

Thanks
 
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  • #2
jisbon said:
I will be able to find the thickness of the gap
If you know the thickness of the gap at the edge and the radius of the top surface of the lens then finding the radius of the lens is just a bit of standard geometry.
Do you recall anything about the segment lengths of intersecting chords of a circle?
 
  • #3
haruspex said:
If you know the thickness of the gap at the edge and the radius of the top surface of the lens then finding the radius of the lens is just a bit of standard geometry.
Do you recall anything about the segment lengths of intersecting chords of a circle?
So I assume I will have the diagram below as shown:

1579785553651.png


I'm not sure what I can do here other than finding the length of the dotted line using pythagoa's theoroem. Am I missing out any rules on circles here?
 
  • #5
Is this correct? Pretty new to this:

1579848862399.png
 
  • #6
jisbon said:
Is this correct? Pretty new to this:

View attachment 256025
I can't read the bit after "t x".
Identify the two intersecting chords.
 
  • #7
1579850872262.png

Are the two intersecting chords in red?
 
  • #8
jisbon said:
View attachment 256026
Are the two intersecting chords in red?
Yes, but it bothers me that you have written 'r' for the upper portion of the vertical one. It is not the radius.
 
  • #9
haruspex said:
Yes, but it bothers me that you have written 'r' for the upper portion of the vertical one. It is not the radius.
Yes, it will be very obvious that r is not the radius curvature. In this case after finding r as shown in the picture, how can I find the radius of the curvature?
 
  • #10
jisbon said:
Yes, it will be very obvious that r is not the radius curvature. In this case after finding r as shown in the picture, how can I find the radius of the curvature?
The vertical chord is central.
 
  • #11
1579913281802.png

If the vertical chord is in the centre (as shown in orange lines), won't it be weird considering the equations I wrote above?
 
  • #12
jisbon said:
View attachment 256061
If the vertical chord is in the centre (as shown in orange lines), won't it be weird considering the equations I wrote above?
The vertical is in the centre but the horizontal is not. (Delete the horizontal orange line, but keep the point you have marked as the centre and the distance you are now calling r.)
How far is the centre of the circle from the point of intersection of the red lines, in terms of the variables you now have?
 
  • #13
haruspex said:
The vertical is in the centre but the horizontal is not. (Delete the horizontal orange line, but keep the point you have marked as the centre and the distance you are now calling r.)
How far is the centre of the circle from the point of intersection of the red lines, in terms of the variables you now have?
r-t

1580091330350.png
 
  • #14
jisbon said:
r-t
Right.
The vertical chord is in two parts: the part above the point of intersection and the part below. The part below has length t. What is the length of the part above?
 
  • #15
haruspex said:
Right.
The vertical chord is in two parts: the part above the point of intersection and the part below. The part below has length t. What is the length of the part above?
r+(r-t) =2r-t

So (2r-t)*t = 1.5*1.5?
 
  • #16
jisbon said:
r+(r-t) =2r-t

So (2r-t)*t = 1.5*1.5?
Yes.
 
  • #17
So I carried on, and found r to be abnormally big in this case..
1580364588626.png

I calculated t to be 0.0024cm.. Is there any mistakes? I took m=75, n=1, and wavelength to be 633nm

EDIT: 2.25 came from 1.5*1.5
 
  • #18
jisbon said:
So I carried on, and found r to be abnormally big in this case..
View attachment 256316
I calculated t to be 0.0024cm.. Is there any mistakes? I took m=75, n=1, and wavelength to be 633nm

EDIT: 2.25 came from 1.5*1.5
What units for r?

By the way, just noticed your diagram of the interfering rays is wrong. One should be internally reflected in the lens, while the other is off the top surface of the glass plate.
 
  • #19
haruspex said:
What units for r?

By the way, just noticed your diagram of the interfering rays is wrong. One should be internally reflected in the lens, while the other is off the top surface of the glass plate.
In this case, I calculated r in cm since t is 0.0024 is in cm and 2.25 is in cm^2
 
  • #20
jisbon said:
In this case, I calculated r in cm since t is 0.0024 is in cm and 2.25 is in cm^2
Then I agree with your answer.
 
  • #21
haruspex said:
Then I agree with your answer.
It is possible that the radius of the curvature to be so big 400+cm as compared to the length of the lens?
 
  • #22
jisbon said:
It is possible that the radius of the curvature to be so big 400+cm as compared to the length of the lens?
Length? You mean the diameter?
The radius of curvature can be as large as you like. Depends what it is for. If it is for this experiment, it needs to be a long focal length, as mentioned at https://vlab.amrita.edu/?sub=1&brch=189&sim=335&cnt=1.
 
  • #23
haruspex said:
Length? You mean the diameter?
The radius of curvature can be as large as you like. Depends what it is for. If it is for this experiment, it needs to be a long focal length, as mentioned at https://vlab.amrita.edu/?sub=1&brch=189&sim=335&cnt=1.
Yep I meant diameter.
Didn't know it could be that big compared to the diameter. Thanks for the insight too
 

FAQ: Plano-convex lens with phase changes

What is a plano-convex lens with phase changes?

A plano-convex lens with phase changes is a type of lens that has a curved surface on one side and a flat surface on the other. This lens also has the ability to change the phase of light passing through it, which can be used for various applications such as beam steering and optical communication.

How does a plano-convex lens with phase changes work?

This type of lens works by utilizing a material with a varying refractive index across its surface. As light passes through the lens, it undergoes a change in phase due to the different refractive index, which can be controlled by applying an external electric field or temperature change.

What are the applications of a plano-convex lens with phase changes?

Plano-convex lenses with phase changes have a wide range of applications in the field of optics. They can be used for beam steering in laser systems, adaptive optics in telescopes, and in the development of optical switches and modulators for communication and data processing.

How is a plano-convex lens with phase changes different from a regular plano-convex lens?

A regular plano-convex lens has a constant refractive index throughout its surface, while a plano-convex lens with phase changes has a varying refractive index. This allows for the manipulation of light passing through the lens, making it more versatile and useful for different applications.

What are the advantages of using a plano-convex lens with phase changes?

One of the main advantages of this type of lens is its ability to control the phase of light passing through it. This can lead to improved performance in various optical systems, as well as the development of new technologies. Additionally, these lenses are compact, lightweight, and can be easily integrated into existing optical setups.

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