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Big aperture of telescope |
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| Apr17-12, 04:19 AM | #1 |
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Big aperture of telescope
1. The problem statement, all variables and given/known data
My prof mentioned that if a telescope has a bigger aperture, you can generally resolve the image better. He did mentioned the reason but I kinda forget the reason so can someone tell me why bigger aperture allows you to better resolve images?? |
| Apr17-12, 05:47 AM | #2 |
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| Apr17-12, 07:55 PM | #3 |
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Since telescope images faraway objects, at the aperture the incident field is the Fourier transform of the (2D) object to be imaged, an aperture serves to FILTER this spectrum of the object, the bigger the aperture, the more spatial spectrum is allowed, and the lens serve to RECONSTRUCT the object from its filtered spectrum. Since a point source, after filtering, spreads out a little, the heavier the filtering, the more spread-out the image is.
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| Apr18-12, 02:21 AM | #4 |
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Big aperture of telescope
Well yea I know the diffraction limit. But my prof mentioned something about filtering high frequency will filter out the details of your image resulting in a blurred image. I just don't understand why a bigger lens will allow high frequency to pass through.
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| Apr18-12, 02:53 PM | #5 |
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The wavefront U(p,q) at the plane of the aperture is the Fourier transform of the object plane U0(x,y),
U(p,q)~∫ ∫ dx dy U0(x,y)exp(i(p*x+q*y)) The aperture A(p,q) of radius a such that A(p,q)=1, if √(p^2+q^2)<a; A(p,q)=0 otherwise will multiply the wavefront U, i.e., U'(p,q)=A(p,q)*U(p,q), effectively filter out higher frequencies than a. The lens does the job of inverse FT from U'(p,q) back to U0'(x,y), which is low-pass filtered with respect to U0. |
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