Plano-convex lens with phase changes

  • Thread starter Thread starter jisbon
  • Start date Start date
  • Tags Tags
    Lens Phase
Click For Summary
SUMMARY

The discussion centers on calculating the radius of curvature for a plano-convex lens using the formula ##2nt=m \lambda## for dark fringes, where m=75, n=1, and the wavelength is 633nm. Participants emphasize the importance of understanding the geometry of intersecting chords in circles to derive the radius from the thickness of the gap between the lens and a glass plate. The calculations led to a radius of curvature exceeding 400 cm, which is acceptable for experiments requiring a long focal length. The conversation highlights the significance of accurate diagram representation in optical physics.

PREREQUISITES
  • Understanding of plano-convex lens properties
  • Familiarity with the formula for dark fringes in thin films
  • Knowledge of geometry related to intersecting chords of a circle
  • Basic principles of optics, including reflection and refraction
NEXT STEPS
  • Study the derivation of the dark fringe formula in thin film interference
  • Learn about the geometry of intersecting chords and its applications in optics
  • Explore the relationship between radius of curvature and focal length in lenses
  • Investigate the impact of lens thickness on optical performance
USEFUL FOR

Optics students, physicists, and engineers involved in lens design and optical experiments will benefit from this discussion, particularly those focusing on interference patterns and lens curvature calculations.

jisbon
Messages
475
Reaction score
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
 
Physics news on Phys.org
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?
 
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?
 
Is this correct? Pretty new to this:

1579848862399.png
 
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.
 
1579850872262.png

Are the two intersecting chords in red?
 
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.
 
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
 

Similar threads

Question about whether the intensity of visible wavelengths is intensified
  • · Replies 3 ·
Replies
3
Views
1K
Image with an Plano-Convex lens
  • · Replies 7 ·
Replies
7
Views
3K
Single lens creating a diffraction pattern?
  • · Replies 6 ·
Replies
6
Views
2K
AP physics 2: Phase change in thin film interference
  • · Replies 1 ·
Replies
1
Views
898
Double Slit Question with Thin Lens concept
  • · Replies 6 ·
Replies
6
Views
4K
Inserting thick lenses into a thin lens system and deducing values
  • · Replies 1 ·
Replies
1
Views
2K
Does a Silvered Plano-Convex Lens Act Like a Concave Mirror?
  • · Replies 5 ·
Replies
5
Views
5K
Optics (dealing with 2 sources of wavelength)
  • · Replies 1 ·
Replies
1
Views
1K
Deriving back focal length using Fermat's principle
  • · Replies 1 ·
Replies
1
Views
2K
A problem involving thin film interference
  • · Replies 2 ·
Replies
2
Views
2K