Plano-convex lens with phase changes

[jisbon]

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

$2nt=m \lambda$

By drawing arrows, I can find out the following:

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

 

[haruspex]

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?

 

[jisbon]

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:

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?

 

[haruspex]

So I assume I will have the diagram below as shown:

View attachment 255982

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?

Draw the whole vertical circle.
https://en.m.wikipedia.org/wiki/Intersecting_chords_theorem

 

[jisbon]

Is this correct? Pretty new to this:

 

[haruspex]

Is this correct? Pretty new to this:

View attachment 256025

I can't read the bit after "t x".
Identify the two intersecting chords.

 

[jisbon]

Are the two intersecting chords in red?

 

[haruspex]

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.

 

[jisbon]

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?

 

[haruspex]

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.

 

[jisbon]

If the vertical chord is in the centre (as shown in orange lines), won't it be weird considering the equations I wrote above?

 

[haruspex]

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?

 

[jisbon]

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

 

[haruspex]

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?

 

[jisbon]

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?

 

[haruspex]

r+(r-t) =2r-t

So (2r-t)*t = 1.5*1.5?

Yes.

 

[jisbon]

So I carried on, and found r to be abnormally big in this case..

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

 

[haruspex]

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.

 

[jisbon]

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

 

[haruspex]

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.

 

[jisbon]

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?

 

[haruspex]

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.

 

[jisbon]

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