Image formation in an astronomical telescope

An astronomical telescope uses two converging lenses. The objective first forms an image of a distant object. The eyepiece then directs light from that image into the eye.

Light from each point of a very distant object reaches the objective as an approximately parallel bundle of rays. The objective brings each bundle to a focus, forming a real intermediate image at its image-side focal plane. The image is real because the rays actually meet there. It is inverted relative to the object.

Normal adjustment

In normal adjustment, the telescope is arranged so that the intermediate image also lies at the eyepiece’s object-side focal plane. For each point on that image, the eyepiece makes the rays emerge parallel. The final image is therefore at infinity and can be viewed by a relaxed eye.

The final image remains inverted relative to the original object. Although rays from any one image point emerge parallel to one another, bundles from different image points emerge in different directions.

The intermediate image is one objective focal length, $$f_o$$, from the objective and one eyepiece focal length, $$f_e$$, from the eyepiece. For thin lenses in normal adjustment, the lens separation is therefore:

Rays through the telescope

In a ray diagram, draw a parallel bundle arriving from a point on the distant object. The objective brings the rays together at a point on the intermediate image. From there, the rays spread towards the eyepiece, which makes them emerge parallel again.

A bundle from another object point arrives at a different angle and is focused at a different point on the intermediate image. The emerging bundles show the directions in which the eye sees the final image at infinity.

Angular magnification of an astronomical telescope

A telescope makes a distant object appear larger by increasing its angular size: the angle between the directions from the eye to opposite edges of the object. Angular magnification compares the angular size of the final image seen through the telescope with the angular size of the object seen by the unaided eye.

Here, $$M$$ is the magnitude of the angular magnification. For example, $$M = 10$$ means that the image subtends ten times the angle that the object subtends when viewed without the telescope.

Magnification in normal adjustment

In normal adjustment, the objective forms an intermediate image in the focal plane of the eyepiece. Light leaves the eyepiece so that the final image appears at infinity.

For a distant object, let the intermediate image have height $$h$$. Using the small-angle approximation, its angular size without the telescope is approximately $$h/f_o$$, while the angular size of the final image through the eyepiece is approximately $$h/f_e$$. Dividing these angles gives:

In this expression, $$f_o$$ is the focal length of the objective and $$f_e$$ is the focal length of the eyepiece. A larger objective focal length or a smaller eyepiece focal length gives a greater angular magnification.