Thin lenses

Each face of a lens refracts on its own: light crossing from index n₁ to n₂ at a surface of radius R obeys n₁/s + n₂/s′ = (n₂ − n₁)/R. Write that once for the front face and once for the back, let the image made by the first act as the object for the second, and shrink the thickness to zero so both surfaces sit in one plane. The intermediate distance cancels and what survives is 1/f = (n − 1)(1/R₁ − 1/R₂), with 1/sₒ + 1/sᵢ = 1/f. Concrete: crown glass of n = 1.52 ground to R₁ = +20.0 cm and R₂ = −20.0 cm, both faces bulging outwards, gives 1/f = 0.52 × (5.00 m⁻¹ − (−5.00 m⁻¹)) = 5.20 D, so f = 19.2 cm. Reverse the sign of both radii and the same glass becomes f = −19.2 cm, a diverging lens. Thin also means a ray's height barely changes inside the glass, which is the licence to draw the lens as a single vertical line.

An image point is where every ray from one object point reconvenes, so two rays fix it — and two of them have paths you can state in advance. A ray arriving parallel to the axis leaves through the far focal point F′. A ray aimed at the centre of the lens goes straight on, because there the two faces are parallel and a thin parallel slab shifts a ray sideways without turning it. A third, entering through the near focal point F and leaving parallel, is the check. For a diverging lens the same two rules hold with the focal points on the other sides: the parallel ray leaves as though it had come from F′, which now lies in front of the lens, and only the backward extensions of the outgoing rays ever meet. That difference is exactly the difference between a real image, where light truly arrives, and a virtual one, where it only seems to.

Stand an object of height hₒ at distance sₒ and let the two rays land at the image plane. The centre ray makes similar triangles on either side of the lens, so hᵢ/hₒ = −sᵢ/sₒ, the minus sign recording that this ray crosses the axis. The parallel ray makes a second pair sharing their vertex at F′: it is at height hₒ a distance f before F′ and at height hᵢ a distance sᵢ − f after it, so hᵢ/hₒ = −(sᵢ − f)/f. Set the two expressions equal, sᵢ/sₒ = (sᵢ − f)/f, multiply out to sᵢ f = sₒ sᵢ − sₒ f, and divide throughout by sₒ sᵢ f. That leaves 1/sₒ + 1/sᵢ = 1/f, and the ratio you started from is the definition of magnification, m = −sᵢ/sₒ. Nothing about the glass survives except f, which is why one equation serves every lens.

Declare the convention first: sₒ positive for an object in front of the lens, sᵢ positive for an image on the outgoing side, f positive for a converging lens. Now run a converging lens of f = +20.0 cm with a real object. At sₒ = 60.0 cm, beyond 2f, sᵢ = +30.0 cm and m = −0.500: real, inverted, reduced — a camera. At sₒ = 30.0 cm, between f and 2f, sᵢ = +60.0 cm and m = −2.00: real, inverted, enlarged — a projector. At sₒ = 10.0 cm, inside f, sᵢ = −20.0 cm and m = +2.00: virtual, upright, enlarged, 20.0 cm in front of the lens — a magnifying glass. Swap in a diverging lens, f = −20.0 cm at sₒ = 30.0 cm, and sᵢ = −12.0 cm with m = +0.400: virtual, upright, reduced, and it is never anything else at any object distance. One rule covers all four: for a real object the sign of sᵢ decides real from virtual, and real images are inverted while virtual ones are upright.

No optician quotes a focal length. They quote the power P = 1/f, in dioptres, where 1 D = 1 m⁻¹ and f must be in metres. A +2.50 D reading lens has f = 0.400 m; a −3.00 D lens for short sight has f = −0.333 m. The reciprocal is the better variable for two reasons. It is linear in what the grinder actually controls: the lensmaker's equation builds P out of the curvatures 1/R₁ and 1/R₂ with no powers or products, so doubling a curvature doubles its contribution. And thin lenses pressed into contact simply add their powers, P = P₁ + P₂, where 1/f = 1/f₁ + 1/f₂ looks like work: a +5.00 D and a −2.00 D in contact make +3.00 D, f = 0.333 m. Separate the two by any appreciable distance and the sum fails, which is the next topic's business, but in contact it is exact within the thin-lens model.

Clamp an object and a screen a distance D apart and ask where the lens must go. Add sₒ + sᵢ = D to 1/sₒ + 1/sᵢ = 1/f, whose left side is (sₒ + sᵢ)/(sₒ sᵢ), and you get sₒ sᵢ = fD. So sₒ and sᵢ are the two roots of x² − Dx + fD = 0, namely x = [D ± √(D² − 4fD)]/2. The square root is real only when D ≥ 4f, so a converging lens cannot throw a real image onto a screen closer to the object than 4f. With f = 20.0 cm nothing works below 80.0 cm, and at exactly 80.0 cm the single solution is sₒ = sᵢ = 40.0 cm with m = −1. Past that there are always two lens positions: at D = 100 cm the roots are 27.6 cm and 72.4 cm, giving m = −2.62 and m = −0.382, magnifications that are reciprocals. Slide the lens along the bench and the image comes sharp twice.