# Coinciding of 2F and C in ray diagram of lenses

1. Dec 13, 2013

### ovais

Hello All,

Guys in my recent lecture I discussed that unlike the focal length of spherical mirrors which is half the radius of curvature, the focal length of lenses depends not only on the geometry of lenses but also on the refractive indices of the media, and that here the relation f=R/2 is not correct. This means radius of curvature may not be equal to twice of focal length in case of lenses. And therefore the centre of curvature ( which is at distance of radius, from the optical centre) can be at a place other than 2F.

But all textbooks shows C and 2F coinciding, this makes matter not realistic and it provides unclear message as to how they apply rules for image formation on it and how they are declaring the results like " If the Object is at 2F1 image will form at 2F2."

What idea is there, behind making such ray diagrams? Any help will be highly appreciated!!!!

2. Dec 13, 2013

### Simon Bridge

For a general lens or mirror - the radius of curvature R is not constant across the surface of the element.
There is no need for the focal lengths to be the same front and back, or even for the optic axis to be a straight line. The exact situation, for mirrors as well as lenses, depends on the geometry of the optical element and the media the light moves through.

Many beginning text books work with exclusively spherical lenses and idealized thin lenses.
Not all of them restrict themselves to this.

Many lenses in the par-axial approximation are well modeled by a system of two spherical interfaces with a uniform section the fatness of the lens between them. Most school lenses can even be modeled by a thin lens.

The par-axial approximation is what is behind the ray diagrams students learn at secondary school level.

3. Dec 13, 2013

### ovais

Thanks SB, for the reply you really provide me a far closer look of the reality in very nice and precise way. But I want to know how they approximate 2F and C. Thanks a bunch.

4. Dec 13, 2013

### Simon Bridge

For a spherical bi-convex lens, radius of curvature R, refractive index n, max thickness t, in air - centered at the origin... is the intersection of two spheres ... $x^2+y^2+(z-(R-t/2))^2 = R^2 = x^2+y^2+(z+(R-t/2))^2$ where z is the optic axis.

You need to work out, using the law of refraction (twice), how a light ray parallel to the symmetry axis is bent.
This will give you a lot of formulas with sines and cosines in it ... in steps the par-axial approximation.

The intersection of the refracted ray with the axis is the focus.

5. Dec 14, 2013

### Staff: Mentor

Please give a specific example. I don't remember this being the case for any textbook that I've used.

6. Dec 14, 2013

### ovais

Ohh though I understand there something have been done for this approximation but at this stage it will be quite tough for me to formulate required formulas using the equation and laws of refraction, I will like to work out them in near future.

One last thing I want to ask is that, what theoretical explanation can I give to secondary school level students of the above case and do these ray diagrams( based on assuming 2F and C coincide) wrong( if yes to what degree, and if no, then are they acceptable?) and do I need to tell them that this approximation will give correct result only for refractive index µ=2, 1/f=(µ-1)[1/R1 - 1/R2]

Last edited: Dec 14, 2013
7. Dec 14, 2013

### ovais

Ncert Science 10th Book page 180 and 181 and i see it in many other books also

Here is the online book to be read
http://www.ncert.nic.in/ncerts/textbook/textbook.htm?jesc1=10-16

Also I will like to see your comment on this statements:
1. A line joining C and the point of incidence act as a normal at the point of incidence for the front refracting side of the (equiconvex or equiconcave) lens.

2. A line joining 2F and the point of incidence act as a normal at the point of incidence for the front refracting side of the (equiconvex or equiconcave) lens.

Last edited: Dec 14, 2013
8. Dec 14, 2013

### Staff: Mentor

Thanks for the link! The diagrams do indeed show C1 = 2F1 etc., and on page 176 is the statement:

However, it is simply not true that C = 2F for a lens. I'm amazed that this actually appears in textbooks. It certainly doesn't appear in any of the books that I've used, for introductory university level physics courses (Halliday/Resnick, Serway/Vuille, etc.) and for intermediate-level optics courses (Hecht, Pedrotti, Meyer-Arendt).

The focal length f of a thin lens depends on the radii of curvature of the two surfaces, R1 and R2, and on the index of refraction of the lens glass nlens (and on the index of refraction of the surrounding medium n0, which we usually take to be 1), via the lens-makers formula:

$$\frac{1}{f} = \frac{n_{lens} - n_0}{n_0} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

See here for example: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenmak.html which writes the equation using the lens power P = 1/f. Try plugging in e.g. R1 = 10 cm, R2 = -10 cm, nlens = 1.6 and n0 = 1. You will not get f = 5 cm.

You can get different f's for the same R's by using different values for nlens. Also, the two surfaces can have different R's, but the two focal points are still equal distances from the lens. That's why it makes no sense to me to say C = 2F (R = 2f) for a lens.

Added: Aha, now I see you know about the lens-maker's formula already. I didn't read your post before the one with the link, until after I wrote the stuff above.

Last edited: Dec 14, 2013
9. Dec 14, 2013

### Staff: Mentor

The distance 2f is nevertheless "special" because if the object is located at distance 2f on one side of the lens, then the image is located at 2f on the other side, and if the object is at > 2f, then the image is at < 2f, etc. So it's useful to mark the points 2F1 and 2F2 on a ray diagram. They're simply not the same as the centers of curvature.

10. Dec 14, 2013

### Simon Bridge

It's quite tricky - the result is given by someone else. You should have a go anyway - the process should be painful enough to help you.

Basically you can give them no justification for that statement because it is not really justified except on very special cases.

1. for a spherical lens only, the line along the normal passes through the center of curvature for the surface.

2. does not work in general ... i.e. the two statements 1 and 2 are not equivalent.

Note: I have checked your reference and I don't see a definition of the points C1 and C2 in terms of the radius of curvature. I notice point C is in the plane of the lens. It may just be a typo left from reusing the axes from the mirror section.

It does seem a common enough generalization.
Another teaching resource: http://www.trinityes.org/ftpimages/425/download/hhh.schaums_38.pdf
A math coach: http://www.icoachmath.com/physics/definition-of-radius-of-curvature.html

You will probably best serve your students by telling them that the text is wrong on this score.
You could set the task as an exercize: under what circumstances would the test be right?
How close is it for common substances - i.e. glass, water...

Last edited: Dec 14, 2013
11. Dec 14, 2013

### Simon Bridge

For a symmetric lens (R1=R2=R), refractive index n, in air ... $$f=\frac{R}{2(n-1)}$$

if we approximate it by $f'=R/2$

then the percentage deviation from true is $$d=100\frac{(f-f')}{f} = 100\frac{(2-n)}{(n-1)}$$

for (crown) glass n=1.5 d=100%
... i.e. a regular classroom lens, the approximation would be out by a whole length.

How close can we get this to zero ... n=2:
arsenic sulphide n=1.9 d=11%
cubic zirconia n=2.2 d= 17%

Normally we would not get the focus by measuring the curvature though.
Probably why this problem has not been picked up - may make an interesting practical lesson for the students... measure the curvature, measure the focal length, compare.

Last edited: Dec 14, 2013
12. Dec 16, 2013

### ovais

Woww very well explained!! Thanks again to both of you guys :-)

13. Dec 17, 2013

### Simon Bridge

Yeah - sorry about that: should have done it much earlier.
Still ... no worries?

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