DEVELOPING THE MAGNIFYING POWER OF A TELESCOPE

-Lew Thomas 5-30-2001

We all know that the magnifying power, M, of a telescope is determined by the ratio of the focal length of the objective, F, to that of the eyepiece, f, expressed as

M = F/f (1)

This is often stated as a fact without proof. Here is a simple method for developing this formula.

Consider the diagrams to the left. The top figure illustrates an objective imaging a subject of height, h. It is assumed that this object is extremely far away, as is the case with celestial objects. All the rays from all parts of the object are therefore very close to being parallel.

The distance to the object is noted by s and the image, h’, is therefore at the focal distance, F, from the objective.

When an eyepiece of focal length f is added as shown in the lower figure, it forms a virtual image, h", beyond the image formed by the objective h’. In other words the eyepiece acts as a magnifier to angularly enlarge the image formed by the objective. The magnifying power of the combination is defined as

angular size of the image

M = (2)

angular size of the object

h’/f

M = (3)

h/s

but by similar triangles

h/s = h’/F (4)

and substituting this into (3), we have

M = (h’/f)(F/h’)

or

M = F/f (5)

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Maintained by Ray Shapp Page last updated 06/13/2001 |