2
$\begingroup$

I am trying to follow the derivation of lens maker's formula from the textbook "University Physics", p.1133 (https://books.google.com.hk/books?id=nQZyAgAAQBAJ&pg=PA1133#v=onepage&q&f=false)

I can understand the first equation because it is just the object–image relationship for spherical refracting surface. But for the second equation, why the left hand side is nb/s2+nc/s'2 instead of nc/s2+nb/s'2? s2 is the first image's distance and it is on the nc side. In addition, on the right hand side why it is nc-nb on the numerator instead of nb-nc? If we follow strictly the formula for spherical refracting surface, the nb should be the lens side and nc is the air side.

A more fundamental question is, why this kind of superposition principle can be applied? I mean why the lens can be expressed as two lens added together? In many books they directly apply the object–image relationship for spherical refracting surface twice and added together. But this formula is only for single spherical surface (e.g. one side is air only and the other side is water only). If it is a lens it is air on both sides but lens in the middle. Why the solution for single spherical surface can be superposed like this?

$\endgroup$
5
  • 3
    $\begingroup$ The Google Books don't seem to be freely available. $\endgroup$ Commented Aug 8, 2015 at 13:54
  • $\begingroup$ Generally, a lens is composed of two boundaries between different environments. So the direction of a light ray is changing twice. You calculate how it changes on the first boundary - that's some insight that says that one creates some image of the original object and the redirected photons seem to come from the image. And these redirected photons change their direction once again, on the other boundary, when they leave the lens again. So you apply the same transformation once again. One must be a bit careful to distinguish the object and image, and signs, and it gives what it gives. $\endgroup$ Commented Aug 8, 2015 at 13:57
  • $\begingroup$ @Luboš Motl , I have got two pictures of it: dropbox.com/s/mk4vxpdc7nk7t00/1.jpg?dl=0 dropbox.com/s/5jvq3siyntpwmvq/2.jpg?dl=0 Thank you $\endgroup$
    – Kelvin S
    Commented Aug 9, 2015 at 2:53
  • 1
    $\begingroup$ Good, Kelvin. Is there still a problem? The role of the two indexes may look reversed because the first interface is air-to-glass and the second one is glass-to-air, so you have to first flip the whole picture from the left to the right, to get the mirror image, and then derive all the rules of refraction. There may be hypothetically mistakes in partial explanations but when you do the whole calculation, which you're encouraged, you will surely get the same results for the whole lens as the result widely quoted. $\endgroup$ Commented Aug 9, 2015 at 9:40
  • $\begingroup$ @Luboš Motl, thanks. I have just figured out what the authors were trying to do. The hardest part is to get the sign in every terms correct. $\endgroup$
    – Kelvin S
    Commented Aug 9, 2015 at 16:39

1 Answer 1

0
$\begingroup$

The superposition is only approximately correct and the easiest way to understand both (1) why it works and (2) when it can be applied is to think in terms of waves and wavefronts, not rays. Since your link isn't working, let's write the equation down:

$$P_{lens}\approx \frac{n_{lens}-n_0}{n_o}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

With the fields thought of as waves, the lens surfaces become phase masks and the superposition of the two lens curvatures holds because the phase delays imparted by the phase masks one after the other are additive as long as the wavefront curvature (i.e. lateral phase distribution of the field) does not change much between the phase masks.

The power of the lens is the reciprocal of its focal length $f$, i.e. it is the reciprocal of the radius $f$ of curvature of the wavefronts that are output from the lens when a plane wave is input. A spherical wave of this radius converges to its diffraction limited focus after having propagated through this distance. So think of a plane wave input: the first surface represents a phase mask with phase delay as function of distance $r$ from the optical axis given by:

$$\frac{2\,\pi}{\lambda} (n_{lens}-n_0) \frac{r^2}{2 R_1}$$

the phase mask function for the second is:

$$-\frac{2\,\pi}{\lambda} (n_{lens}-n_0) \frac{r^2}{2 R_2}$$

and so the total phase delay is simply:

$$\frac{2\,\pi}{\lambda} (n_{lens}-n_0) \frac{r^2}{2} \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

Now ask yourself: what is the radius of curvature of the spherical wave that shows this (to terms of second order and lower) phase distribution. You'll find it is the reciprocal of the power given by the lensmaker's formula.

I give the full details of these kinds of calculations in this answer here. I show the calculation above as well as the transformer matrix method where one thinks of lens surfaces and other optical processing elements as operators in the group $SL(2,\,\mathbb{R})$.

$\endgroup$
2
  • $\begingroup$ as mentioned above I have got two pictures of it: dropbox.com/s/mk4vxpdc7nk7t00/1.jpg?dl=0 dropbox.com/s/5jvq3siyntpwmvq/2.jpg?dl=0 Thank you $\endgroup$
    – Kelvin S
    Commented Aug 9, 2015 at 2:54
  • 1
    $\begingroup$ @KelvinS The answer still stands. The equations you show amount to the same phenomenon: the additive approximation to the composition of two transformations: the approximation holding because the wavefront curvature is assumed not to change too much between the transformations - thus we need the lens to be thin and negligible wavefront curvature change between surfaces is the definition of "thin" for this problem. $\endgroup$ Commented Aug 9, 2015 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.