The Compound Microscope
Two lenses, two stages of magnification, multiplied.
Lesson at a glanceShow the key points
- Objective (short focal length, small aperture) faces the object; eyepiece (ocular) near the eye.
- Object just beyond f₀ → real, inverted, magnified image at v₀; the eyepiece views it as a simple microscope.
- Normal adjustment: first image at the eyepiece's focus, final image at infinity; final image inverted relative to the object.
- m = (v₀/u₀)(D/fₑ) (19.3) in normal adjustment; m = (v₀/u₀)(1 + D/fₑ) (19.4) for the image at the near point.
- With f₀ small and v₀ ≈ L (tube length): m ≈ −(L/f₀)(D/fₑ). Real microscopes use several lenses in each part.
One lens can only magnify so much before the image falls apart. Use two, and let the second magnify the image the first one made.
Construction
A compound microscope has two converging lenses on a common axis. The objective, facing the object, has a small aperture and a very short focal length f₀. The eyepiece (ocular), near the eye, has a larger aperture and focal length fₑ. Screws adjust the distance between them.
How it forms the image
- The object sits just beyond f₀. The objective forms a real, inverted, magnified image at distance v₀ behind it.
- The eyepiece views that image as a simple microscope. In normal adjustment the first image sits at the eyepiece’s focus and the final image is at infinity. Or the eyepiece can be set so the final image is at the near point, for more magnification.
The final image is inverted relative to the object.
Magnifying power
The naked eye sees the object at best at angle θ₀ = h/D. Through the microscope the final image subtends θ = h′/uₑ, where h′ is the first image’s height and uₑ its distance from the eyepiece. So
D/uₑ is just the eyepiece’s magnifying power as a simple microscope:
Usually f₀ is tiny, so v₀/f₀ ≫ 1 and v₀/u₀ = v₀/f₀ − 1 ≈ v₀/f₀; and the first image is near the eyepiece, so v₀ ≈ L, the tube length:
A microscope has an objective of focal length 1.5 cm and an eyepiece of 3.0 cm. The object is 1.8 cm from the objective, in normal adjustment. Find the magnifying power and tube length.
- 1Objective: short f, forms a real magnified image; eyepiece: magnifier for that image.
- 2m = (v₀/u₀)(D/fₑ) in normal adjustment; (v₀/u₀)(1 + D/fₑ) with the image at the near point.
- 3Approximately m ≈ −(L/f₀)(D/fₑ); the final image is inverted.
- 4Normal adjustment: first image at the eyepiece focus; tube length v₀ + fₑ.
Check yourself
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