Thin-lens imaging

The previous topic refracted light at one spherical surface and, applied twice with the thickness dropped, produced the lensmaker's equation 1/f = (n − 1)(1/R₁ − 1/R₂). From here on the surfaces disappear and only f survives: a converging lens of n = 1.52 with R₁ = +120 mm and R₂ = −120 mm has 1/f = 0.52 × 2/(120 mm), so f = 115 mm, and nothing downstream needs to know the radii again. Before using f, declare the convention and keep it for the whole problem. Light travels left to right; s is positive for a real object on the incoming side; s′ is positive for a real image on the outgoing side and negative for a virtual image on the incoming side; f is positive for a converging lens and negative for a diverging one; heights are positive above the axis. Under this convention the first and second focal points sit at equal distances |f| on either side of the lens, whichever way round the lens is placed — a thin lens has one f, not two.

Three rays from the tip of an object of height y can be drawn with a ruler alone. The ray parallel to the axis refracts through the second focal point F′. The ray through the optical centre is undeviated, because at the centre the two faces are parallel and a thin plate shifts a ray sideways by an amount proportional to its thickness, which is zero. The ray through the first focal point F emerges parallel to the axis. Two rays fix the image; the third is a check on the drawing. For a diverging lens the same three rules hold with the focal points swapped over: the parallel ray leaves as if it had come from F′ on the incoming side, and the ray aimed at F on the far side leaves parallel. The construction is the equation solved with a pencil. The central ray gives similar triangles with |y′|/y = s′/s; the parallel ray, crossing the axis at F′, gives |y′|/y = (s′ − f)/f. Equating the two, s′f = ss′ − sf, and dividing through by ss′f leaves 1/s + 1/s′ = 1/f, with m = −s′/s once the inversion is given its sign.

Rewrite the law as 1/s′ = 1/f − 1/s and it becomes a map you can walk along. For a converging lens with f = +10.0 cm: an object at s = 30.0 cm images at s′ = 15.0 cm with m = −0.500; at s = 20.0 cm the image is also at 20.0 cm with m = −1.00; at s = 15.0 cm it is at 30.0 cm with m = −2.00; at s = 12.0 cm it is at 60.0 cm with m = −5.00. As s approaches f the image recedes to infinity, and for s < f the reciprocal turns negative: at s = 6.00 cm, 1/s′ = 1/10 − 1/6 = −1/15, so s′ = −15.0 cm and m = +2.50 — virtual, erect, enlarged, on the object's side, which is the magnifier. A diverging lens never leaves that regime: with f = −10.0 cm and s = 15.0 cm, 1/s′ = −1/10 − 1/15 = −1/6, so s′ = −6.00 cm and m = +0.400, and every real object gives a virtual, erect, reduced image inside F′. Plot 1/s′ against 1/s and all of this is one straight line of slope −1 with both intercepts at 1/f, which is how f is best extracted from bench data. In the Newtonian variables x = s − f and x′ = s′ − f the same line reads x x′ = f².

Hold f fixed and differentiate 1/s + 1/s′ = 1/f with respect to s: −1/s² − (1/s′²) ds′/ds = 0, so ds′/ds = −(s′/s)² = −m². The derivative is never positive: pushing the object away always lowers s′, which pulls a real image in toward the lens and pushes a virtual image further out on the object's side. Its magnitude is the transverse magnification squared, so an object imaged at m = −2 that shifts 1.0 mm along the axis sends its image 4.0 mm the other way, while a camera lens of f = 50 mm focused on a subject at 5.0 m, where m = −0.0101, moves its image only about 0.1 mm when the subject walks a full metre closer. That asymmetry is depth of field: distant scenes share one focus because m² is tiny, and a photomicroscope stage must be positioned to micrometres because m² is enormous. It also says that a cube is never imaged as a cube. Its faces are scaled by m, its depth by m², so only at |m| = 1 do the two agree. Near s = f the derivative diverges, which is the quantitative reason an image near infinity cannot be focused by moving the object.

Because the law is linear in reciprocals, the natural measure of a lens is P = 1/f in dioptres, 1 D = 1 m⁻¹: a +4.00 D lens has f = 0.250 m and a −2.00 D lens has f = −0.500 m. Put two thin lenses in contact and the image of the first is the object of the second at the same plane. For the first lens 1/s + 1/s₁′ = P₁; for the second the object distance is s₂ = −s₁′, because an image that the first lens would form on the far side arrives at the second as a converging bundle, a virtual object with negative s. So −1/s₁′ + 1/s′ = P₂, and adding the two lines cancels s₁′ to leave 1/s + 1/s′ = P₁ + P₂. Powers add; focal lengths do not. A +5.00 D lens touching a −2.00 D lens acts as a single +3.00 D lens of f = 0.333 m, and two f = 20.0 cm lenses in contact make f = 10.0 cm, not 40.0 cm. This is why a reading addition of +2.00 D on a +1.50 D distance prescription is simply written +3.50 D. Separate the lenses by a gap d and a cross term −d P₁P₂ appears; that belongs to the next topic.

Each assumption fails by a calculable margin. Thickness first: keeping t in the two-surface derivation gives 1/f = (n − 1)[1/R₁ − 1/R₂ + (n − 1)t/(nR₁R₂)], so for an equiconvex lens the thin result is wrong by the fraction (n − 1)t/(2nR). With n = 1.52, R = 120 mm and t = 6.0 mm that is 0.86%: f = 116.4 mm, not 115.4 mm, and s and s′ must be measured from two principal planes that sit about t/(2n) = 2.0 mm inside each face. Paraxial next: the replacement sin θ → θ is wrong by θ²/6, which is 0.17% for a ray meeting a 100 mm lens 10 mm off-axis and grows as h², so the rim of a wide lens focuses short of the paraxial point — spherical aberration. Monochromatic last: from the lensmaker's equation, Δf/f = −Δn/(n − 1), and crown glass changes its index by about 0.008 between the blue F and red C lines, so f differs by about 1.5% across the visible — chromatic aberration, which a single thin lens cannot remove. Of the three, thickness is the smallest for a well-made lens; aperture and colour set the real limits.