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Qs: How do I show that for a point object, for a ray at a large angle from the optical axis, spherical aberration (SA) produces a distorted focal length $f_{SA}$ that is shorter than the normal focal length $f$ (i.e. spherical aberration $\to$ more strongly bent ray)?

My attempt. Consider the diagram below.

From geometry, applying the cosine law to the triangle that links $l_1$ with $\phi$ $$l_1=\sqrt{(s_0+R)^2+R^2-2(s_0+R)Rcos(\phi))}$$ Similarly for $l_2$ $$l_2=\sqrt{R^2+\left(\frac{R-t(0)}{cos(\alpha)}\right)^2-2R\left(\frac{R-t(0)}{cos(\alpha)}cos(\alpha- \phi)\right)}$$ For $l_3$, used pythagoras $$l_3=\sqrt{(s_i-t(0))^2+(R-t(0))^2tan^2(\alpha)}$$ I want to find an expression that relates $l_1,l_2,l_3,s_i,s_0,R, n_1,n_2$ using $$OPL=n_1l_1+n_2l_2+n_1l_3$$, and (Fermat) $$\frac{dOPL}{d\phi}=0$$ so that I can show that $s_i$ is smaller than the normal focal length. However, I'm struggling to find the relation bewteen $\phi$ & $\alpha$.

Is there a simpler way to do this? enter image description here (Spherical lens with radius R, centre C and thickness $t(0)$ . The ray hits the optical axis at the point P. $s_o$ and $s_i$ are measured from the vertex)

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    $\begingroup$ (a) If you're interested in focal length, why aren't you considering a ray parallel to the axis? (b) Probably not what you had in mind, but why not do numerical calculations for a given value of $R$ and different values of $h$? Much easier, I'd have thought, than trying to do the general case, but will prove the point just as well $\endgroup$
    – Philip Wood
    Commented Mar 28, 2021 at 21:27
  • $\begingroup$ @Philip Wood (a) because I also want to show that the spherical aberration ( from a point source at finite distance) for a curveside lens (incident on the curveside, exiting on the flat side) is stronger than that of a flatside lens (incident on the flatside, exiting on the curve side), but this does not seem to be true any more if the incident rays are parallel (i.e. this time the flatside prodcues more aberration due to larger angle of deviation). (b) That's actually worth trying, thanks. $\endgroup$
    – Chern-Simons
    Commented Mar 28, 2021 at 21:49
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    $\begingroup$ Re my (a)... Your use of focal length confused me. 'Image distance' would be better. But for my part I should have read your question more carefully, as you did talk about a point object. $\endgroup$
    – Philip Wood
    Commented Mar 28, 2021 at 21:55

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