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My understanding is that the top rays spend less time inside the lens where the speed of light is slow and the middle rays spend more time inside the lens so the time is kind of compensated . But I'm looking for, like a mathematical proof. I'm asking in the context of Huygens' principle

Try to not involve quantum mechanics.

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As is pointed out in another answer you can numerically perform the computation and you will find that rays close to the principle axis all arrive approximately in phase at a point but the difference becomes larger for rays further away from the principal axis.
This defect of a lens is called spherical aberration.

Making suitable approximations it is relatively easy to show that two parallel rays incident on a plano-convex lens do take the same time to reach he focal point.

The curved surface of the lens has a radius of curvature is $R$ and the material of which it is made has a refractive index $n$.

enter image description here

We need to show that the time from $A$ to $F$ is the same as the time from $P$ to $B$ plus the time from $B$ to $F$.

If the speed of light is $c$ then $ \dfrac{\sqrt{(d^2+f^2)}}{c}=\dfrac{nx}{c} + \dfrac {f-x}{c}$

Using the intersecting chord theorem $d^2 = x(2R-x) \approx x2R $ if $R\gg x$ and the binomial expansion $\sqrt{(d^2+f^2)} \approx f\left (1+ \dfrac{d^2}{2f^2} +. . . . . . \right )$ if $f\gg d$ results in $\dfrac 1 f \approx (n-1) \dfrac 1 R$ which is the lens-makers formula for a plano-convex lens.

Update as a result of a comment from the OP

Noting the assumptions made before I think that the analysis above can be extended for all incoming parallel rays and also for a particular example for a biconvex lens.

enter image description here

In the left hand diagram the two blue parallel rays and the ray along the principal axis all traverse a thickness $y'$ of the lens before reaching the part of the lens shaded red and so take the same time to reach the 2red" lens.
That part of the lens shaded red is a smaller version of the lens which was considered originally and provided that $y'$ is small such that $PF\approx P'F$ those rays will satisfy the equal time condition.
This all parallel rays will reach point $F$ at the same time.

Developing this analysis with more and more approximations illustrate how spherical aberration plays a part in the functioning of a lens.

For the biconvex lens and a special case where $O$ and $I$ are the focal points of the individual plano-convex lenses which make up the biconvex lens it can be shown that $\dfrac 1 u + \dfrac 1v = (n-1) \left ( \dfrac {1}{R_1}+\dfrac {1}{R_2}\right ) $ which in the lens maker's is equal to the reciprocal of the focal length of a biconvex lens.

Probably this is the end of the line for "hand waving" and to show in general the equality of time resulting in the lens maker's formula and the lens equation requires a numerical approach?

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  • $\begingroup$ is it okay to assume a biconvex lens as 2 plano-convex lenses placed together with similar faces joining each other? $\endgroup$ – Tilak Maddy Jan 23 at 5:22
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Of course, the slowing of light in the lens does introduce a phase difference, but light going through different thicknesses of glass also goes along paths of different total length.

And light that travels longer paths (And still arrives at the focus) goes through less glass.

It isn't obvious that the two contributions come out equal, but they do.


Taken in light of Fermat's Principle, the fact that a continuous bundle of rays come to a focus guarantees that this is the case because if it were not only light following the least time path would arrive at the putative "focus".

And of course, the spherical aberration that Farcher mentions manifests as a failure to come to a precise focus.

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  • $\begingroup$ "It isn't obvious that the two contributions come out equal, but they do." exactly my textbooks states facts as though its very obvious and i don't get it $\endgroup$ – Tilak Maddy Jan 23 at 4:46
  • $\begingroup$ I'm afraid that the only thing to do at this point is to start calculating unless you are willing to swallow Fermat's principle whole and work the problem from that side. But how to think about and apply Fermat's principle didn't become clear to me until long after I had mastered ray optics on the basis of refraction and reflection. $\endgroup$ – dmckee Jan 23 at 4:49
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You can simply use geometry to determine how much distance the ray is travelling in the glass and how much in air for the various rays. Then find the times with the different velocities in air and glass. Finally sum it all up. Will be tedious. Rest assured it will all compensate.

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