Optical instruments and angular magnification

A relaxed eye is focused on infinity; contracting the ciliary muscle thickens the lens and pulls the focus in, and the closest point it can hold sharply is the near point. For a young eye that is about 25 cm, which costs 1/0.25 m = 4 D of accommodation on top of the roughly 60 D the eye already has. The retina responds to angle: a 4.0 mm object at 25 cm subtends 0.016 rad and, with the eye's nodal point about 17 mm from the retina, makes a retinal image 0.016 × 17 mm = 0.27 mm tall. Now the crucial point. You cannot get a bigger retinal image simply by stepping closer, because inside the near point the eye can no longer focus and the extra angle buys only blur. That is what an instrument is for: it supplies the extra angle while handing the eye a bundle it can still focus. Everything below is bookkeeping on that one idea, and 25 cm is a convention, not physics — a presbyopic near point of 75 cm makes every quoted magnification larger for that observer.

Take a converging lens of focal length f and put the object exactly at its front focal point. Every ray leaving one point of the object emerges parallel, so the virtual image sits at infinity and the eye stays relaxed. The tilt of that emergent bundle is θ′ = h/f, while the best the unaided eye could do was θ = h/Nₚ. The h cancels: M = θ′/θ = Nₚ/f, independent of how big the object is. A 2.5 cm loupe therefore gives 250 mm/25 mm = 10×. Slide the object slightly inside the focal point instead and the virtual image comes back to the near point, where the angle is θ′ = h/f + h/Nₚ and M = 1 + Nₚ/f — 11× for the same loupe, a 10% gain paid for with full accommodation and a tired eye after ten minutes. Note what the formula demands: short focal length, which means a strongly curved, small lens. A single uncorrected lens is swamped by aberrations beyond about 3× or 4×, so even that 10× loupe is a cemented doublet — and past that the microscope has to split the job between two lenses.

The objective is a short-focal-length lens with the specimen just outside its front focal point, so it forms a real, inverted, strongly enlarged image inside the tube. With the object barely beyond the focal point, that image lands a distance fₒ + L behind the objective, and the linear magnification works out as mₒ = −L/fₒ, where L is the optical tube length measured between the two inner focal points. The eyepiece is then a plain magnifier used on that intermediate image: place the image at the eyepiece's front focal plane and the final bundle leaves parallel, giving Mₑ = Nₚ/fₑ. Because the first stage delivers a real object to the second, the factors simply multiply: M = mₒ × Mₑ. A standard fₒ = 4.0 mm objective with L = 160 mm gives mₒ = −40, and a 25 mm eyepiece adds Mₑ = 10, for 400× inverted. The minus sign is not decoration — it is why a microscope slide must be nudged left when you want the view to move right.

Here the object is already at infinity, so there is no near point to beat — the reference angle is simply the angle θ the object subtends in the sky. Parallel rays arriving tilted by θ are brought to a focus in the objective's rear focal plane, at a height h′ = fₒ tanθ ≈ fₒ θ off the axis. Set the eyepiece so that its front focal plane coincides with that same plane, exactly one eyepiece focal length further along, and the light leaves as a parallel bundle tilted by θ′ ≈ h′/fₑ. Divide: M = θ′/θ = fₒ/fₑ, inverted. The tube is therefore fₒ + fₑ long — 925 mm for a 900 mm objective and a 25 mm eyepiece. Two consequences follow that students rarely expect. Aperture does not appear in M at all, so a wide objective buys light and detail but not power. And that 900 mm objective makes an image of the Moon only about 8.2 mm across, which is tiny — the instrument is not enlarging the Moon, it is enlarging the angle.

A camera has a fixed lens and a fixed sensor plane, so it focuses by translation: 1/s + 1/s′ = 1/f. A 50 mm lens focused at infinity sits 50.00 mm from the sensor; focus it on a subject 1.50 m away and s′ = 1/(1/50 − 1/1500) = 51.72 mm, a barrel travel of 1.7 mm for the whole range from infinity to a metre and a half. The aperture is described by the f-number N = f/D, because image illuminance depends on the cone angle rather than on D alone; halving N from f/5.6 to f/2.8 quadruples the light and lets the shutter run four times faster. The eye solves the same problem the opposite way, holding the image distance fixed at roughly 22 mm and changing its own lens power by about 4 D. When that range shifts, a thin correcting lens restores it: an eye whose far point has come in to 0.50 m needs P = −1/0.50 = −2.0 D to push infinity back onto the far point, and a presbyopic near point of 0.75 m needs P = 1/0.25 − 1/0.75 = +2.7 D to read at 25 cm.

Three separate limits close in as M rises, and none of them appears in M = fₒ/fₑ. First, the exit pupil: the eyepiece forms an image of the objective's aperture, of diameter D/M, and that is where you must put your eye. A 200 mm telescope at 400× delivers a 0.5 mm exit pupil, narrow enough that floaters in your own eye start to show; at the other end, an exit pupil wider than your dark-adapted 7 mm pupil means the iris simply blocks the outer ring of light you paid for. Second, surface brightness: the flux collected is fixed, so spreading it over M² times the apparent area dims every extended object as 1/M². Third, detail. Diffraction at the aperture sets a floor on resolvable angle, so beyond roughly 2× per millimetre of aperture the extra magnification enlarges a blur disc and nothing else — empty magnification, taken up properly in the next topic. Field of view shrinks too: a 50° eyepiece shows 50/36 = 1.4° at 36× but only 0.56° at 90×, barely more than the Moon's 0.52°.