Physics · Glossary

What is Compound microscope?

Definition 4.10 University Physics — Year 1 · Chapter 4 — Optical Instruments and the Eye

A microscope is an objective L1L_1 of short focal length f1f_1' forming a real, enlarged intermediate image of a nearby object in the object focal plane of an eyepiece L2L_2 (focal length f2f_2'), which acts as a magnifier. The distance Δ=F1F2\Delta = \overline{F_1'F_2} between the objective’s image focus and the eyepiece’s object focus is the optical interval, a standardized 160mm160\,\mathrm{mm} on classical instruments.

The microscope: the objective forms a real, enlarged, inverted intermediate image A_1B_1 in the object focal plane of the eyepiece; the eyepiece sends it to infinity, where the relaxed eye views it under a large angle.
The microscope: the objective forms a real, enlarged, inverted intermediate image A1B1A_1B_1 in the object focal plane of the eyepiece; the eyepiece sends it to infinity, where the relaxed eye views it under a large angle.

Examples

Example 4.12 (A student microscope)

Objective f1=4.0mmf_1' = 4.0\,\mathrm{mm}, eyepiece f2=25mmf_2' = 25\,\mathrm{mm}, Δ=160mm\Delta = 160\,\mathrm{mm}: γ1=40\gamma_1 = -40, G2=10G_2 = 10, G=400G = 400. The object sits at F1A=f12/Δ=16/160=0.10mm\overline{F_1A} = -f_1'^2/\Delta = -16/160 = -0.10\,\mathrm{mm} from F1F_1: 4.1mm4.1\,\mathrm{mm} from the objective. Focusing means moving the whole tube by fractions of a millimeter.

Example 4.9 (A watchmaker’s loupe)

f=25mmf' = 25\,\mathrm{mm}: G=250/25=10G = 250/25 = 10, sold as “10×10\times”. A 0.05mm0.05\,\mathrm{mm} detail then subtends 2×103rad2 \times 10^{-3}\,\mathrm{rad}, comfortably resolved. Pushing GG higher means ff' of a few millimeters, a tiny lens held against the eye — the limit of the single lens, and the reason the microscope exists.

Example 4.16 (An amateur’s 150-millimeter refractor)

D=150mmD = 150\,\mathrm{mm}, f1=1200mmf_1' = 1200\,\mathrm{mm}, eyepiece 25mm25\,\mathrm{mm}: G=48G = -48; exit pupil 3.1mm3.1\,\mathrm{mm}; light gain (150/7)2460(150/7)^2 \approx 460 against a dark-adapted pupil of 7mm7\,\mathrm{mm}; resolution 1.22×550nm/0.15m=4.5×106rad=0.91.22 \times 550\,\mathrm{nm}/0.15\,\mathrm{m} = 4.5 \times 10^{-6}\,\mathrm{rad} = 0.9'', so craters 1.7km1.7\,\mathrm{km} wide on the Moon; Gr=3×104/4.5×10667G_r = 3\times10^{-4}/4.5\times 10^{-6} \approx 67 — the 6mm6\,\mathrm{mm} eyepiece (G=200G = 200) shows no more detail, only a dimmer, bigger disk. The weekend problem designs this instrument.

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