Multi-element systems and optical instruments

Solve each element with the same thin-lens equation, in the order the light meets them, and let the geometry set the object distance for each stage. Take a converging lens of f₁ = 10 cm with an object 15 cm in front of it, and a second converging lens of f₂ = 5.0 cm placed d = 20 cm behind the first. Stage one: 1/s₁′ = 1/10 − 1/15, so s₁′ = 30 cm and m₁ = −30/15 = −2. That image sits 10 cm beyond the second lens, so the object distance for stage two is s₂ = d − s₁′ = 20 − 30 = −10 cm: a virtual object. Nothing is wrong. The rays leaving lens 1 are converging towards a point 10 cm past lens 2, and lens 2 intercepts them before they get there. Stage two: 1/s₂′ = 1/5.0 − 1/(−10) = 0.30 cm⁻¹, so s₂′ = 3.3 cm, real and just behind lens 2, with m₂ = −s₂′/s₂ = −3.33/(−10) = +0.33. Overall, m = m₁m₂ = −0.67: inverted once, and smaller than the object although stage one doubled it. Draw the intermediate image whether or not it physically forms; it is the token that passes from one stage to the next.

Hold a coin closer and it looks bigger, until at the near point, about 0.25 m for a young adult, the eye can no longer focus. The largest unaided view of an object of height h therefore subtends θ = h/(0.25 m). A converging lens with the object at its focal point sends the eye parallel light, an image at infinity that a relaxed eye focuses easily, subtending θ′ = h/f. The ratio M = θ′/θ = (0.25 m)/f is the angular magnification: 5 for f = 5.0 cm, 10 for f = 2.5 cm. Pull the object slightly inside f so the virtual image lands at the near point and the eye, now fully accommodated, sees M = 1 + (0.25 m)/f, which is 6 for f = 5.0 cm, at the cost of strain. The lateral magnification m = −s′/s is unbounded here and says nothing useful. A simple magnifier is limited by its own focal length: 10× needs f = 2.5 cm, a small, strongly curved lens whose aberrations grow quickly, which is why higher powers are built from two stages instead.

A corrective lens does not magnify; it relocates. A myopic eye whose far point is 50 cm cannot focus anything beyond it, so the lens must take parallel light from a distant object and deliver a virtual image at 50 cm: s = ∞, s′ = −0.50 m, hence f = −0.50 m and a power P = 1/f = −2.0 D (1 D = 1 m⁻¹). The prescription is a diverging lens whose focal point coincides with the far point. A presbyopic eye whose near point has receded to 100 cm needs an object at the comfortable 25 cm to appear at 100 cm: 1/f = 1/0.25 − 1/1.00 = +3.0 D, so f = 33 cm. Distances are measured from the lens, not the eye, and that matters: spectacles sitting 1.5 cm in front of an eye with a 50 cm far point must image at 48.5 cm, P = −2.06 D, and for a far point of 12.5 cm the −8.0 D contact lens becomes −9.5 D as spectacles 2.0 cm away. Keep the sign convention you used for lenses; the eye is just the last element in the train.

A compound microscope is two stages of the sequential method. The objective, a short-focus lens, has the specimen just outside its focal point and forms a real, enlarged intermediate image a tube length L beyond its back focal point; L is 160 mm in a traditional stand. Its lateral magnification is exact: with s₁′ = fₒ + L, the lens equation gives s₁ = fₒ s₁′/(s₁′ − fₒ), so mₒ = −s₁′/s₁ = −(s₁′ − fₒ)/fₒ = −L/fₒ. For fₒ = 4.0 mm, s₁′ = 164 mm, s₁ = 4.1 mm and mₒ = −40. The eyepiece is a magnifier applied to that intermediate image; placed so the image sits at its focal point, it delivers parallel light with angular magnification Mₑ = (0.25 m)/fₑ = 10 for fₑ = 25 mm. The overall angular magnification is the product, M = mₒ Mₑ = −400: the specimen is seen 400 times larger in angle than it would be at the near point, and inverted. The two lenses are L + fₒ + fₑ = 189 mm apart, and each obeys nothing beyond the single-lens equation.

For a distant object the input is a bundle of parallel rays tilted at θ to the axis, and the ray through the objective's centre fixes the intermediate image height h = fₒ tan θ in the focal plane. Set the eyepiece so that plane is its own focal plane, a separation fₒ + fₑ, and every ray from the image point leaves parallel, tilted at θ′ with tan θ′ = h/fₑ. Hence tan θ′/tan θ = fₒ/fₑ, and in the paraxial limit M = θ′/θ = −fₒ/fₑ, the sign recording that the bundle leaves tilted the opposite way: the image is inverted. A 900 mm objective with a 25 mm eyepiece gives M = −36 in a 925 mm barrel, turning a star 0.10° from the axis into one seen 3.6° from it. The exit bundle is narrower than the objective by the same ratio: a 90 mm objective feeds the eye through an exit pupil of 90 × 25/900 = 2.5 mm. Doubling M by halving fₑ halves that pupil and spreads the same light over four times the retinal area, so higher power is always dimmer for an extended object.

Nothing in the ray equations contains a wavelength, so they let magnification grow without limit and predict a point image for a point object. Light is a wave, and passing through a circular aperture of diameter D it spreads into an Airy pattern whose first dark ring lies at 1.22 λ/D; two points closer than θₘᵢₙ = 1.22 λ/D blur into one. For the 90 mm telescope at λ = 550 nm, θₘᵢₙ = 7.5 × 10⁻⁶ rad, about 1.5 arcseconds. The eye resolves roughly one arcminute, 2.9 × 10⁻⁴ rad, so a magnification of about 39 already lifts the finest detail the objective can form to the eye's threshold; a few times that makes viewing comfortable, and beyond it every extra power is empty magnification, enlarging the blur. The microscope obeys the same rule: an objective of numerical aperture NA resolves dₘᵢₙ = 0.61 λ/NA, which is 0.52 μm for NA = 0.65, and lifting 0.52 μm to the 73 μm the eye separates at 25 cm needs M ≈ 140, so useful powers run to a few hundred times NA. The correspondence check is the limit λ/D → 0, where the wave result collapses back onto the ray model's point image.