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In a thin plano-convex lens, if the wave front is perpendicular to the lens axis, all of the rays are in phase and are focused at the focal point. Can we use the lens geometry to mathematically prove that all rays take the same time to reach the focal point?

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  • $\begingroup$ Duplicate $\endgroup$
    – Farcher
    Feb 22 '19 at 10:17
  • $\begingroup$ do you mean 'wave front is perpendicular to the lens axis' $\endgroup$
    – user45664
    Feb 22 '19 at 17:00
  • $\begingroup$ Yes, it was a mistake. I edited the question. $\endgroup$
    – Karim Samy
    Feb 22 '19 at 17:10
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If you know how to use Snell's law and do a bit of trigonometry, you can calculate the distance a ray travels in glass and in air , starting at each point on the plane surface of the lens and ending at the focus. The time spent in the glass is the distance traveled in the glass, times the refractive index, divided by the speed of light. Same in the air, using distance and index in the air. Write a formula for the total time as a function of radial position on the lens, and you will find that the radial position cancels out, making time independent of radial position.

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  • $\begingroup$ Thanks for your answer. I did reach a formula depending on the vertical distance of the ray at the point the ray impinges the lens. However, it is not easy to have this dependence canceled. I will try to express it as a function of radial position. $\endgroup$
    – Karim Samy
    Feb 22 '19 at 15:33
  • $\begingroup$ Because there is spherical aberration, the light does not go to a perfect focus. As a result, the path calculated as I described is not exactly right. If your formula describes the difference between the time for light to travel along the axis and the time to take a path from near the edge of the lens to the focus, it will be "almost* correct. If you use a Taylor series expansion, you'll find that when the field of view of the system is small enough, you will find that the term containing radial position is very small. $\endgroup$
    – S. McGrew
    Feb 22 '19 at 16:27
  • $\begingroup$ My formula derives the exact distance or in another view the exact time taken by any ray impinging the lens at any point on the lens given that these rays are all parallel to the horizontal axis. As a fact of the thin lens, this time should be independent of any variable. But, I think this is only possible according to the par-axial approximation. If the spherical aperture increases significantly, the time formula becomes more dependent on the radial distance. Do you think my understanding is true ? $\endgroup$
    – Karim Samy
    Feb 22 '19 at 17:30
  • $\begingroup$ Yes, your understanding is correct. If the lens is an asphere with exactly the right shape, it's possible to obtain a perfect focus for a plane wavefront that is exactly parallel to the planar face of the lens; but off-axis the focus will not be perfect. $\endgroup$
    – S. McGrew
    Feb 22 '19 at 18:59
  • $\begingroup$ Does this mean that at any point where the wave front impinges the lens the rays add constructively at the focal point? or as far as the position on the lens follows the par-axial approximation where the angles are assumed to be very small? $\endgroup$
    – Karim Samy
    Feb 22 '19 at 20:55
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$\def\rP{{\rm P}}$ IMO there's need to clarify the matter, as I'm seeing different aspects superimposed.

1) You'd not confuse behaviour of a beam of rays parallel to optical axis and that of a parallel beam at angle with optical axis. In former case it happens what you found: in paraxial approximation all rays converge to the focus, but if you consider rays too far away from axis this is no longer true. Generally the distant rays intersect optical axis before focus - the whole beam after traversing the lens traces a figure known as "caustic". This phenomenon is known as "spherical aberration".

2) Spherical aberration may be corrected if rear spherical surface is replaced by an "aspherical" one. For your particular case - plano-convex lens, source at infinity - this can be obtained with a hyperbolic surface.

3) Once one succeeds - e.g. by aspherical surface - to get exact convergence of all rays in one point, the propagation time is the same along every ray. This can be seen observing equal-time (equal-phase) surfaces.

4) It may look improper to speak of phase in geometrical optics - that's why I wrote equal-time. Equal-time surfaces may be rigorously defined within geometrical optics as follows.

Consider a ray starting from source S and choose a time $t$. On that ray mark the point P such that propagation time from S to P equals $t$. Repeat the construction for all rays, keeping $t$ fixed. The set of all points so obtained is a surface, which we'll call "equal-time surface" for time $t$. The same construction may be repeated for different $t$'s and we get a family of equal-time surfaces.

You may also view $t$ as a function of P, for fixed S. This function (usually multiplied by $c$) is called eikonal: $$c\,t = W(\rP).$$ Thus equal-time surfaces are the level surfaces of $W$. The value of $W(\rP)$ is also called the optical path from S to P. It can be shown that the equal-time surface through P is orthogonal to the ray through P.

5) Now we may give the proof of property 3). If all rays do pass through one point Q, then an equal-time surface, being orthogonal to all rays, is a sphere centred in Q. For any point P on a given sphere the distance PQ is the same, so it's obvious that time from S to Q is the same along any ray, qed.

6) If spherical aberration is present there is no point Q, equal-time surfaces aren't spherical. Of course you may always choose a point P at will, and there is always a ray passing through P, with a well defined optical path. But generally such ray is unique or there are only a few rays through P - it depends on P's position wrt caustic. In special cases (e.g. if P is on optical axis of your lens) it's clear by symmetry that there are infinite rays through P. Infinite, but by no means all.

7) About off-axis rays, the discourse is different. Even if spherical aberration was corrected, there still remain "extra-axial" aberrations, much harder both to treat theoretically and to correct in practice. Outgoing rays will behave in an asymmetrical way, the caustic will take a complicated shape. Reducing the width of incoming beam isn't so effective for extra-axial aberrations as it's for spherical aberration. That's why designing wide-angle optical systems is especially difficult.

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  • $\begingroup$ Thanks very much.It is summarizing the main points. $\endgroup$
    – Karim Samy
    Feb 23 '19 at 17:58

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