For an object located at the $(x_o, y_o)$ plane, and a lens located at the $(x, y)$ plane, the image is produced at the $(x_i, y_i)$ plane. We can consider the image field $U_i$ to be the sum of the points of the object field $U_o$ with a weightage $h$. Written explicitly,

$$ U_i(x_i, y_i) = \int^\infty_{-\infty}\int^\infty_{-\infty} h(x_i, y_i; x_o, y_o) U_o(x_o, y_o) \,dx_o\,dy_o\, . $$

We obtain an expression of $h$ by first considering a point object $U_o(x_o, y_o) = \delta(x_o - x_p, y_o-y_p)$. Meanwhile, on the RHS of the equation, we can use our knowledge of the propagation of a point source light field through a lens to get an expression for $U_i$. For a thorough derivation refer to Introduction To Fourier Optics by Goodman Chapt 5. The expression for $h$ is thus given by

$$ h(x_i, y_i; x_o, y_o) \propto \int^\infty_{-\infty}\,dx\int^\infty_{-\infty}\,dy\, P(x, y) \, \\ \times \exp\left\{ -ik(n-1) \left[R_1 - R_2 + \sqrt{R_1 - x^2 + y^2} - \sqrt{R_2 - x^2 + y^2} \right] \right\}\\ \times \frac{exp\left\{ ik\sqrt{d_o^2 + (x-x_o)^2 + (y-y_o)^2} \right\}}{{d_o^2 + (x-x_o)^2 + (y-y_o)^2}}\\ \times \frac{exp\left\{ ik\sqrt{d_i^2 + (x_i-x)^2 + (y_i-y)^2} \right\}}{{d_i^2 + (x_i-x)^2 + (y_i-y)^2}}\,, $$

where $n$ is the refractive index of the lens, $R_1$ and $R_2$ are the radius of curvatures of each side of the lens. $d_o/d_i$ is the distance from the $(x_o, y_o)/(x_i, y_i)$ plane to the $(x,y)$ plane. $k$ is the wavenumber. All of which can just be treated as constants in the integral.

$P(x,y)$ is the pupil function, which basically adds limits to the infinite integrals. For a circular pupil of diameter $l$, we have $P(x,y) = \frac{circ (\sqrt{x^2 + y^2})}{(l/2)}$, the limits are changed from

$$ \int_{-\infty}^\infty \, dx \int_{-\infty}^\infty\, dy \, P(x,y) \rightarrow \int_{-l/2}^{l/2}\, dx \int_{-\sqrt{(l/2)^2 - x^2}}^{\sqrt{(l/2)^2 - x^2}} \, dy\, . $$

There is a much simpler expression of $h$, in which you have to take certain approximations about the relative magnitude of the distance between the lens plane and the object/image plane, and the size of the object being imaged, known as the Fresnel and Fraunhofer approximations. These approximations lead to $U_i$ being a simple convolution of $h$ and $U_o$.

I am interested in studying the case where these approximations do not hold, and how that affects the image. I have tried to compute for $U_i$ numerically, however I found that to be too computationally draining. Thus I seek for an analytical solution for $h$.

How do I solve for the integral of $h$ for a given lens circular pupil of diameter $l$?

  • $\begingroup$ try Huygens Principle--it does not use those approximations, although it has other problems. $\endgroup$ – user45664 Feb 24 at 17:13
  • $\begingroup$ @user45664 , I believe I am using the Huygens principle in my workings. The wavelets in each propagation are the sum of spherical waves of the form $\exp (ikr)/r^2$. Could you enlighten me as to what problems the Huygens Principle imposes? Especially for a high NA microscope objective $\endgroup$ – Tian Feb 25 at 6:58
  • $\begingroup$ Huygens Principle in its basic form has the problems of the wake and the backward wave. Also it used impulse type wavelets rather than sinusoidal ones. $\endgroup$ – user45664 Feb 25 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.