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Suppose we have a light of power $P$ distributed on a plane $(x,y)$. The distribution of the power is of the form: $$P=f(x,y)$$ If we have a lens conjugating every point of the plane $(x,y)$ in another plane $(X,Y)$ (focal plane) in a way that for every point in the plane $(x,y)$ we have a circle of radius $r$ in $(X,Y)$, what is the distribution of the power in $(X,Y)$? Thanks.

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The wave amplitude in the focal plane is the Fourier transform of the wave amplitude in the pupil plane of the lens $L$. Assuming we are dealing with a diffraction-limited lens, the relationship between object amplitude $A_o$ and image amplitude $A_i$ is thus a convolution:

$$A_i(x,y) = \iint_{\mathbb{R}^2} dx' dy' \, L(x',y') A_o(x-x',y-y')$$

Note that $A_0(x,y) = \sqrt{f(x,y)} e^{i \phi(x,y)}$, where $\phi$ is some undetermined phase function - it could be a constant or zero, but it might not be. (This may be specified or determined through experiment.) The power distribution in the conjugate plane would then be $|A_i(x,y)|^2$.

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  • $\begingroup$ I understand you supposed the lens 'focuses' the light of a point in the object plane in a point on the image plane. In this case, the focal plane isn't located in the focus of the lens; it's defocused. What changes in your analysis? $\endgroup$ – Riccardo.Alestra Dec 2 '14 at 15:29
  • $\begingroup$ @Riccardo.Alestra: Not much. Defocus introduces a quadratric (or more complicated for faster lenses) phase variation in the lens function $L$. The expression in general remains the same. This is true even in the presence of reasonably small lens aberrations. $\endgroup$ – Ron Gordon Dec 2 '14 at 15:31

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