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Circle of least confusion

Short answer

The circle of least confusion occurs where marginal red and violet rays cross; its diameter d = (f_R − f_V)/(f_R+f_V) D ≈ ω D/2, where ω is the dispersive power and D the lens aperture. Focal length cancels, so d depends only on D and ω, not on focal length.

Show that the diameter of circle of least confusion depends on the diameter of lens aperture and dispersive power of the material of the lens but is independent of the focal length of the lens.

Circle of least confusion: because of chromatic aberration, a lens focuses violet light at F_V and red light at F_R. A screen placed anywhere between them shows a coloured disc rather than a point. The smallest such disc is the circle of least confusion.

 marginal rays cross at X, between F_V and F_R

  lens        F_V      X      F_R
   |           .      ||      .
 --|-----------\-----||-----/---> axis
   |            (circle of
   |             diameter d at X)

Derivation: let the lens aperture diameter be D = 2h, with focal lengths f_V (violet) and f_R (red), where f_R − f_V = ω f.

  1. The marginal violet ray starts at height h and converges to F_V. At distance x from the lens its height is h(1 − x/f_V). Beyond F_V this is negative (the ray has crossed the axis).
  2. The marginal red ray's height at distance x is h(1 − x/f_R), still positive for x < f_R.
  3. The narrowest section lies where these two marginal rays have equal and opposite heights: −h(1 − x/f_V) = h(1 − x/f_R) x (1/f_V + 1/f_R) = 2, so x = 2 f_V f_R / (f_V + f_R).
  4. Radius of the circle at this place: r = h(1 − x/f_R) = h (1 − 2f_V/(f_V + f_R)) = h (f_R − f_V)/(f_R + f_V).
  5. Diameter: d = 2r = D (f_R − f_V)/(f_R + f_V).
  6. Substituting f_R − f_V = ω f and f_R + f_V ≈ 2f:

d = D ω f / 2f = ω D / 2

The focal length cancels. The diameter of the circle of least confusion depends only on the aperture D of the lens and the dispersive power ω of its material, and not on the focal length. It is reduced by using a smaller aperture (stop) or a material of lower dispersive power.

✓ Answer checked by Claude Opus 5.5 (high effort)

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