Multi-element imaging and sign conventions

Work in the order the light travels, and finish one element before looking at the next. Take f₁ = 12.0 cm with the object 20.0 cm in front of it: 1/dᵢ₁ = 1/12.0 − 1/20.0 = 1/30.0, so dᵢ₁ = 30.0 cm and m₁ = −30.0/20.0 = −1.50. Now forget the original object entirely. A second lens, f₂ = 8.0 cm, stands 45.0 cm beyond the first, so its object is that 30.0 cm image, 15.0 cm in front of it: 1/dᵢ₂ = 1/8.0 − 1/15.0 gives dᵢ₂ = 17.1 cm and m₂ = −17.1/15.0 = −1.14. The final image is 17.1 cm past lens 2 with m = (−1.50)(−1.14) = +1.71. Nothing in stage 2 refers to stage 1 except the single number dᵢ₁.

Write the contract down before you write any numbers, because every sign in the answer descends from it. In the real-is-positive convention this course uses: dₒ is positive for an object on the side the light comes from; dᵢ is positive for an image on the side the light goes to, and such an image is real; f is positive for a converging element and negative for a diverging one; and m = −dᵢ/dₒ, positive meaning upright relative to that element's own object. The convention is a contract for the whole calculation, not for one line of it. Half the wrong answers in multi-element problems come from a student who used one convention for lens 1 and remembered a different one by lens 2. A mirror in the chain needs one extra care: it reverses the direction of travel, so the side the light goes to is now back the way it came.

Move the second lens closer and something new happens. With f₁ = 10.0 cm and the object at 15.0 cm, dᵢ₁ = 30.0 cm; put lens 2, of focal length f₂ = 15.0 cm, only 20.0 cm along, and dₒ₂ = 20.0 − 30.0 = −10.0 cm. Nothing has broken. Lens 2 has intercepted a cone that was still converging, and the point it was converging towards lies 10.0 cm beyond the lens, on the outgoing side — a virtual object, which the convention records as a negative dₒ. Feed it in: 1/dᵢ₂ = 1/f₂ − 1/dₒ₂ = 1/15.0 + 1/10.0, so dᵢ₂ = 6.00 cm. A converging lens handed a converging cone forms a real image closer than its own focal length, 6.00 cm against f₂ = 15.0 cm, which no real object could ever make it do.

Each stage scales the height it is given, so the heights compose by multiplication: h₂ = m₂h₁ = m₂m₁h₀, and the system magnification is m = m₁m₂. Carry the signs and the orientation comes free. In the case above m₁ = −2.00 and m₂ = −6.00/(−10.0) = +0.600, so m = −1.20: a 4.00 mm object becomes a 4.80 mm image, inverted with respect to the original object. Notice what the positive m₂ is telling you, and then get the rule exactly right. Rewrite the stage magnification as m₂ = f₂/(f₂ − dₒ₂): it is positive precisely when f₂ and f₂ − dₒ₂ carry the same sign. A converging element handed a virtual object always qualifies, since dₒ₂ < 0 < f₂, so that stage does not invert again and the single inversion from lens 1 survives to the end. A diverging element is the exception worth carrying: f₂ = −8.00 cm with dₒ₂ = −4.00 cm gives m₂ = +2.00, but the same lens with dₒ₂ = −12.0 cm gives dᵢ₂ = −24.0 cm and m₂ = −2.00, which does invert. And a stage contributes an inversion only when it takes a real object to a real image, so two such stages in a row invert twice and the final image stands upright.

Four cheap tests catch nearly every slip. Back-substitute: 1/dₒ₂ + 1/dᵢ₂ must return 1/f₂ exactly, which catches a sign flipped in the algebra. Push the separation to zero: two thin lenses in contact must give 1/f = 1/f₁ + 1/f₂, the dioptre rule, and any general result that fails this limit is wrong. Test the system against its own focal length: for f₁ = 10.0 cm, f₂ = 15.0 cm and L = 5.0 cm, 1/f = 1/10.0 + 1/15.0 − 5.0/150 = 1/7.50, so f = 7.50 cm, while an object at infinity gives dᵢ₁ = 10.0 cm, dₒ₂ = −5.0 cm and dᵢ₂ = 3.75 cm — and 3.75 cm is exactly f(1 − L/f₁), the back focal distance, not f itself. Finally, watch the pole: as dₒ approaches f the image distance runs away like f²/(dₒ − f), so near there a 1% error in dₒ is not a 1% error in dᵢ.

There is a second way through, and it never needs an image at all. Describe a paraxial ray by its height y above the axis and its small angle θ, then note that each piece of the system does something linear to that pair. A gap of length L leaves the angle alone and shifts the height: y → y + Lθ. A thin lens leaves the height alone and bends the angle: θ → θ − y/f. Send a ray in parallel at y = 1.00 cm through f₁ = 10.0 cm, L = 5.0 cm, f₂ = 15.0 cm. After lens 1, θ = −0.100 rad. At lens 2, y = 1.00 − 5.0 × 0.100 = 0.500 cm, and θ becomes −0.100 − 0.500/15.0 = −0.1333 rad. It reaches the axis 0.500/0.1333 = 3.75 cm later — the back focal distance again — and yᵢₙ/|θₒᵤₜ| = 1.00/0.1333 = 7.50 cm is the system focal length.