Physics · Astrophysics · A2 Level

Refracting Telescopes

The refracting telescope is the classic image most people picture — a long tube with a lens at each end — and it was the design Galileo used to first turn a telescope on the night sky. Understanding how its two lenses work together explains not just how it magnifies, but why its magnification has hard physical limits.

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Two lenses, one focused image

An astronomical refracting telescope is built from two converging (convex) lenses mounted at opposite ends of a tube, sharing the same optical axis. The lens closer to the distant object is the objective lens, with focal length $f_o$; the lens closer to the observer's eye is the eyepiece lens, with focal length $f_e$. Both lenses do their job through refraction — bending incoming light rays to bring them to a focus — which is where this design gets its name.

Light from a very distant astronomical object arrives at the objective lens as an essentially parallel bundle of rays, since the source is so far away that any spreading of the rays over the width of the objective is negligible. These parallel rays are not all travelling exactly along the optical axis, though — rays from different points across the object arrive at slightly different angles, and it's this small spread of angles that the eyepiece must eventually convert into a magnified image.

Forming and viewing the image

The objective lens brings the incoming parallel rays to a focus at its focal point, a distance $f_o$ from the lens, forming a real, inverted intermediate image of the distant object. This intermediate image is small, but crucially it is a physical image sitting inside the tube — the eyepiece lens now treats this intermediate image as its own object.

In the telescope's normal adjustment, the eyepiece is positioned so that the intermediate image falls exactly at the eyepiece's own focal point, a distance $f_e$ from the eyepiece. With the object placed at the focal point of a converging lens, the rays leaving that lens emerge parallel again — so in normal adjustment, the final rays reaching the eye are parallel, meaning the final image is formed at infinity. This is comfortable for prolonged viewing, since a relaxed, unaccommodated eye focuses naturally on objects at infinity.

A ray diagram of an astronomical refracting telescope in normal adjustment, drawn as a horizontal tube. On the left, three parallel rays arrive from a distant object at a small angle to the optical axis, pass through a large converging objective lens, and converge to a point on the optical axis at distance f_o from the objective — this point is labelled as the intermediate image. A small eyepiece lens is positioned further to the right, at a distance f_e beyond the intermediate image point, so the intermediate image sits exactly at the eyepiece's focal point. From that point, the rays diverge, pass through the eyepiece lens, and emerge on the right travelling parallel to one another again but now at a much steeper angle to the optical axis than the original incoming rays. The angle of the incoming rays to the axis is labelled alpha (a small angle), and the angle of the outgoing rays on the eyepiece side is labelled beta (a noticeably larger angle), with the tube length labelled as f_o plus f_e. — In normal adjustment, parallel rays entering the objective are brought to a focus at the intermediate image, which sits at the eyepiece's focal point — so parallel rays leave the eyepiece too, but at a much larger angle β than the incoming angle α.

Angular magnification

Because the final image of a distant object is still effectively at infinity, magnification for a telescope isn't measured as a ratio of image size to object size the way it is for a simple magnifying lens. Instead, it's measured as angular magnification: how much larger the image appears, in terms of the angle it subtends at the eye, compared with the angle the object itself subtends when viewed directly, without the telescope.

$$M = \dfrac{\text{angle subtended by the image at the eye}}{\text{angle subtended by the object at the unaided eye}}$$

DerivationAngular magnification in normal adjustment

Let the intermediate image formed by the objective have height $h$.

This image sits a fixed distance $f_o$ from the objective lens and, in normal adjustment, also a fixed distance $f_e$ from the eyepiece lens — it is the shared object/image linking the two lenses.

The angle $\alpha$ subtended by the distant object at the unaided eye is the same as the angle subtended by the intermediate image at the objective lens.

Since the object is effectively at infinity, the ray direction reaching the eye without the telescope matches the ray direction converging to form the intermediate image via the objective. For small angles, $\alpha \approx \dfrac{h}{f_o}$.

The angle $\beta$ subtended by the final image at the eye, looking through the eyepiece, is the angle at which rays leave the eyepiece from the same intermediate image.

Since the intermediate image sits at the eyepiece's focal point, for small angles $\beta \approx \dfrac{h}{f_e}$.

Angular magnification is the ratio of these two angles:

$$M = \dfrac{\beta}{\alpha} = \dfrac{h/f_e}{h/f_o} = \dfrac{f_o}{f_e}$$ The height $h$ of the intermediate image cancels, since it is common to both angles.

M = f_o / f_e

This result, $M = f_o/f_e$, is remarkably simple: the magnification of an astronomical refracting telescope in normal adjustment depends only on the ratio of the two focal lengths, not on the size of either lens or the brightness of the object.

A large objective focal length $f_o$ increases magnification, and also spreads the intermediate image over a larger physical size for a given angular size of object, which can help with resolving fine detail in that intermediate image.
A small eyepiece focal length $f_e$ also increases magnification, since the same intermediate image is viewed at a steeper angle when the eyepiece sits closer to it — though very short focal lengths bring practical difficulties, such as a very small, uncomfortable eye relief.

In normal adjustment, the physical length of the telescope tube — the distance from the objective to the eyepiece — is the sum of the two focal lengths, $f_o + f_e$, since the intermediate image sits at both $f_o$ from the objective and $f_e$ from the eyepiece, on the same point of the optical axis.

Why this matters

Because $M = f_o/f_e$, high magnification pushes a refracting telescope toward a very long objective focal length, a very short eyepiece focal length, or both — and a long $f_o$ means a long, unwieldy physical tube. This is one of several practical reasons observatories eventually moved away from very long refractors and toward reflecting telescopes, covered in the next lesson, which can achieve a long effective focal length without needing a correspondingly long tube.

Key takeaways

An astronomical refracting telescope uses two converging lenses: an objective (focal length $f_o$) and an eyepiece (focal length $f_e$).
The objective brings distant, effectively parallel rays to a focus, forming a real intermediate image.
In normal adjustment, the intermediate image sits at the eyepiece's focal point, so rays leave the eyepiece parallel again and the final image forms at infinity.
Angular magnification is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye.
Using the small-angle approximation on both sides of the shared intermediate image gives $M = f_o/f_e$ in normal adjustment.
A larger $f_o$ or a smaller $f_e$ both increase magnification.
The tube length in normal adjustment equals $f_o + f_e$, since the intermediate image is the shared point of focus for both lenses.

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Introduction to Telescopes

Reflecting Telescopes