Telescopes

Point a small pair of lenses at the night sky and Jupiter's moons snap into view — not because the telescope makes anything bigger in the way a magnifying glass enlarges a coin, but because it enlarges the angle the object subtends at your eye. A distant star is always a point of light; what a telescope changes is how much of your retina a whole scene (a planet's disc, a pair of close stars) can spread across. An astronomical refracting telescope does this with two lenses — a large-aperture objective and a small eyepiece; a reflecting telescope replaces the objective with a curved mirror. This section builds the magnifying-power formulas for the refracting case, in both the relaxed-eye (image at infinity) and strained-eye (image at near point) settings, and explains — with concrete reasons, not just a slogan — why every large research telescope on Earth today is a reflector.

The core insight

A telescope magnifies the angle subtended by a distant object using an objective of large focal length paired with an eyepiece of small focal length, giving $M=f_o/f_e$ in normal adjustment; reflecting telescopes use a mirror instead of an objective lens because a single curved surface is easier to build large, free of chromatic aberration, and self-supporting.

The core idea

Two-lens layout. The astronomical refracting telescope uses two converging lenses on a common axis: the objective, of large focal length $f_o$ and large aperture, facing the distant object; and the eyepiece, of small focal length $f_e$ , close to the eye. Because the object is effectively at infinity, parallel rays from any one point on it converge to a real, inverted, diminished image exactly at the objective's focal plane. The eyepiece then simply re-images this intermediate image as a magnifier — its job is not to make the picture bigger, but to let the eye view it from very close up, so it fills a larger angle.

Normal adjustment — final image at infinity. If the eyepiece is positioned so the intermediate image sits exactly at the eyepiece's own focal point, the emergent rays are parallel again and the eye views with no strain (relaxed accommodation). The two focal points then coincide, and the tube length is simply the sum of the focal lengths:

L=f_o+f_e

where: $L$ is the separation between objective and eyepiece, $f_o$ the objective's focal length, $f_e$ the eyepiece's focal length. Magnifying power $M$ is defined as the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ the object itself would subtend (essentially the same viewing position as the objective, since the object is so distant):

M=\frac{\beta}{\alpha}=\frac{f_o}{f_e}

where: $f_o$ is the objective focal length and $f_e$ is the eyepiece focal length; both measured in the same unit, so $M$ is dimensionless. The larger $f_o$ is relative to $f_e$ , the bigger the magnifying power — the opposite emphasis to a compound microscope, where it is the objective's focal length that must be small.

Final image at the near point $D$ . Moving the eyepiece slightly closer forms a virtual final image at the near point $D$ (about 25 cm) instead of infinity, which trades a small strain on the eye for extra magnification:

M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)

where: $D$ is the near point distance (about $25\ \text{cm}$ for a standard eye), and the other symbols are as above. Since $f_e/D$ is positive, this near-point value is always slightly larger than the normal-adjustment value $f_o/f_e$ .

Figure

Ray diagram of an astronomical refracting telescope in normal adjustment showing parallel rays from a distant object forming a real inverted image at the shared focal point of objective and eyepiece then leaving the eyepiece as parallel rays to a relaxed eye

Reflecting telescope. A large concave mirror (paraboloidal, to avoid spherical aberration) replaces the objective lens, collecting and focusing light to a real image; a secondary mirror or an eyepiece then redirects this focused light out to the observer, since placing the eyepiece straight at the primary's focus would sit in the incoming light's path. The common Cassegrain design uses a small convex secondary mirror to send the converging beam back through a hole in the primary to an eyepiece behind it.

Why reflectors dominate large astronomy. A mirror has only one surface to figure and polish, versus two curved surfaces plus internal glass for a lens, so very large apertures are far easier to manufacture accurately. A mirror can be supported fully from behind, whereas a large lens can only be held at its rim and sags under its own weight. Reflection does not depend on wavelength, so a mirror shows no chromatic aberration, unlike a lens where different colours refract by slightly different amounts and blur the image. These three reasons together, not any single one, are why every major research telescope today is a reflector.

Derivation AMagnifying power of a telescope in normal adjustment

The goal is to show, from nothing more than the definition of magnifying power as a ratio of angles and simple small-angle geometry at the two lenses, that the magnifying power of an astronomical refracting telescope with its final image at infinity works out to the clean ratio $M=f_o/f_e$ , with no dependence on the object's actual distance or size.

Step 1. Set up the angle at the objective. The object is effectively at infinity, so parallel rays from one edge of it arrive at the objective inclined at the small angle $\alpha$ to the axis — this is also, to good approximation, the angle the object subtends at an unaided eye positioned near the objective. These parallel rays converge to a point in the objective's focal plane at height $h=f_o\tan\alpha\approx f_o\alpha$ from the axis, using the small-angle approximation $\tan\alpha\approx\alpha$ (in radians).

Step 2. Place this image at the eyepiece's focus. In normal adjustment the intermediate image, of height $h$ , sits exactly at the eyepiece's own focal point, since that is what makes the outgoing rays parallel again (relaxed eye). The eyepiece therefore views this height $h$ from a distance equal to its own focal length $f_e$ .

Step 3. Write the angle at the eyepiece. The final image, still at height $h$ , subtends an angle $\beta$ at the eyepiece given by $\tan\beta\approx\beta=h/f_e$ , again using the small-angle approximation, since $\beta$ is small for any sensibly designed eyepiece.

Step 4. Take the ratio and substitute. By definition, $M=\beta/\alpha$ . Using $h=f_o\alpha$ from Step 1 and $\beta=h/f_e$ from Step 3, $M=\dfrac{h/f_e}{\alpha}=\dfrac{f_o\alpha/f_e}{\alpha}=\dfrac{f_o}{f_e}$ . The object's angle $\alpha$ and the intermediate image height $h$ both cancel out, leaving a result that depends only on the two focal lengths — which is why a telescope's magnifying power is quoted as a fixed instrument specification, independent of what it happens to be pointed at.

Worked example

A refracting astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm, viewing the Moon, which subtends an angle of about $0.5°$ at the unaided eye. Find (a) the magnifying power in normal adjustment, (b) the angle the Moon's image subtends through the telescope in that setting, and (c) the magnifying power if the eyepiece is instead adjusted for a final image at the near point ( $D=25\ \text{cm}$ ).

Step 1. Magnifying power in normal adjustment, from $M=f_o/f_e$ : $M=100/5=20$ . The telescope makes the Moon's image subtend an angle 20 times larger than to the naked eye.

Step 2. Angle subtended by the image: since $M=\beta/\alpha$ , the image angle is $\beta=M\alpha=20\times0.5°=10°$ . A lunar disc that ordinarily spans half a degree now spans ten degrees of your field of view — vastly more detail is resolvable, even though the Moon itself has not moved or grown.

Step 3. Magnifying power with the final image at the near point, from $M=(f_o/f_e)(1+f_e/D)$ : $M=\dfrac{100}{5}\left(1+\dfrac{5}{25}\right)=20\times1.2=24$ .

Answer: In normal adjustment $M=20$ and the Moon's image spans about $10°$ ; with the final image brought to the near point, $M$ rises to $24$ , at the cost of a slightly strained (non-relaxed) eye.

Common traps

Confusing telescope and microscope magnification formulas. Telescope (infinity case): $M=f_o/f_e$ , a plain ratio of two focal lengths. Microscope (infinity case) involves the tube length and $D$ as well — the formulas are not interchangeable between instruments.
Thinking a large objective focal length alone gives a compact telescope. A larger $f_o$ raises magnification but also lengthens the tube ( $L=f_o+f_e$ in normal adjustment) — magnification and instrument size trade off together, they don't improve independently.
Assuming reflecting telescopes are free of all aberrations. They specifically eliminate chromatic aberration; they can still suffer from other aberrations such as coma, managed (not eliminated) by a paraboloidal shape and careful secondary-mirror design.
Forgetting the final image in an astronomical telescope is inverted. The objective forms a real, inverted image, and the eyepiece (acting as a simple magnifier) does not flip it back — acceptable for astronomy, which is why "terrestrial" telescopes need an extra erecting system.

Ask the tutor: want the ray-diagram derivation walked through step by step on a whiteboard, or a side-by-side comparison of the telescope and microscope magnification formulas?

Why does a bigger objective focal length give more magnification instead of the eyepiece doing that job

Magnifying power is the ratio $f_o/f_e$ , so either a larger $f_o$ or a smaller $f_e$ raises it, but the two lenses play different roles. The objective's large focal length (and large aperture) mainly determines how big and bright the intermediate image is and how much fine detail the telescope can resolve; the eyepiece then decides how closely that image is viewed. Increasing $f_o$ without limit also stretches the tube length $L=f_o+f_e$ , so real designs balance magnification against a workable instrument size rather than chasing focal length alone.

Does the near-point setting always beat normal adjustment since it gives a higher M

Numerically yes, $M$ is always a little larger with the final image at $D$ than at infinity, since the factor $(1+f_e/D)$ exceeds 1. But normal adjustment lets the eye relax completely, as if looking at a very distant object, which is comfortable for long observing sessions; the near-point setting forces the eye to focus at 25 cm, which strains it over time. Astronomers mostly use normal adjustment for exactly this comfort reason, accepting the slightly lower magnification.

If reflecting telescopes have no chromatic aberration, why not use them for everything, including small telescopes

Reflectors are strongly preferred once apertures get large, where a lens's weight and two-surface manufacturing become impractical, but small refracting telescopes remain common too — they are simpler in having no secondary mirror or support structure obstructing the incoming light, and chromatic aberration is a smaller practical problem at modest apertures and can be reduced with an achromatic doublet. The choice depends on the aperture size and budget, not just on which design is theoretically flawless.

Telescopes

The core idea

Worked example

Common traps

Before this

Where this leads