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For compound lenses, the image formed by first lens acts as the imaginaryobject for the second lens.

In telescopes, the objective lens projects an image on its focal point which works as the object for the eyepiece. Per the property of convex lenses, the eyepiece magnifies the image. If the focal length of the eyepiece is smaller we'll get a higher magnification.

Now if the focal length of the objective lens is increased, it'll again project a small image on its focal point. So for two objective lenses with different focal length, it seems the image size should about the same.

So why does the magnification change?

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It's like leverage. The longer the distance from the objective lens to the virtual image, the larger the virtual image.

Imagine there's a piece of frosted glass at the focal point. It will show the virtual image.

Now the eyepiece looks at that virtual image with a magnifying glass. That also makes it look bigger.

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any reference for this? –  articlestack May 3 '13 at 2:56
    
@articlestack: Wikipedia –  Mike Dunlavey May 3 '13 at 11:41
    
So in simple words, it is because of angular magnification. When focal length of objective lens increases, angle subtended by the objective lens gets decreased. Hens magnification gets increased. Am I correct? –  articlestack May 4 '13 at 13:27
    
@articlestack: Suppose you have two stars, where the angular distance between them is .01 radian. Then if you have an objective lens that focuses them to an image plane, how big is the image of the two stars? It will be .01 times the focal length, right? 10cm -> .1cm, 100cm->1cm, 1000cm -> 10cm. So then if you look at that image plane with a keplerian eyepiece (magnifying glass), you're examining a small piece of a bigger image. –  Mike Dunlavey May 4 '13 at 18:34
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