How do you determine the magnification of a telescope?
mxtheif asked:
I have a 10 inch F/6.13 Schmidt Cassegrain with a 30 mm eye piece. Also a 16 inch Schmidt Cassegrain 30 mm eye piece. (Was not given the focal length).
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I have a 10 inch F/6.13 Schmidt Cassegrain with a 30 mm eye piece. Also a 16 inch Schmidt Cassegrain 30 mm eye piece. (Was not given the focal length).
I read the Focal Length is in mm, so is 6.13 the mm? How would I figure the focal length of the 16 inch telescope?
If you could please show step by step how you figured the magnification, it would be greatly appreciated! Thanks in advance.
February 5th, 2009 at 7:05 am
objective focal length / eyepiece focal length
February 7th, 2009 at 12:04 pm
First of all, I’ve never heard of anyone making a 10″ f/6.13 Schmidt-Cassegrain. Probably you mean a 10″ f/6.3, which Meade made for a while. To calculate the focal length of this telescope, multiply the aperture (in millimetres) by the focal ratio, i.e. 254 mm x 6.3 = 1600 mm. The magnification is the focal length of the telescope divided by the focal length of the eyepiece, i.e. 1600 mm / 30 mm = 53x.
The only 16″ Schmidt-Cassegrain on the market is an f/10 made by Meade. Its focal length is 406 mm x 10 = 4060 mm. The 30 mm eyepiece will yield 4060 mm / 30 mm = 135x.