Fourier Optics

Figure 2: Schematic representation of the optical train highlighting the relationship between the beam geometry in the input and focal planes. Monochromatic light, with wavevector $\vec k$, is incident on the input plane. A lens of focal length $f$ projects the Fourier transform of the incident light's wavefront onto the focal plane.

A planar array of optical tweezers can be described by the intensity distribution, $I^f(\vec \rho)$, of laser light in the focal plane of a microscope's objective lens. This pattern is determined by the electric field of light incident at its input plane, as depicted in Fig. 2. Suppose that the input plane is illuminated by monochromatic light of wavelength $\lambda$. Its wavefront at the input plane, $E^{in}(\vec r)$, contains both phase and amplitude information,

$$E^{in}(\vec r) = A^{in}(\vec r) \exp[i \Phi^{in}(\vec r)],$$ (1)

where the amplitude, $A^{in}(\vec r)$, and phase, $\Phi^{in}(\vec r)$, are real-valued functions. The electric field in the focal plane has a similar form,

$$E^f(\vec \rho) = A^f(\vec \rho) \exp[i \Phi^f(\vec \rho)],$$ (2)

so that $I^f(\vec \rho) = \vert E^f(\vec \rho)\vert^2 = \vert A^f(\vec \rho)\vert^2$. These fields are related by the Fourier transform pair

$$E^f(\vec \rho) = \frac{k}{2\pi f} \, e^{i \theta(\vec \rho)} \, \int d^2 r \, E^{in}(\vec r) \, e^{-ik \vec r \cdot \vec \rho/f}$$ (3)

$$\equiv {\cal F} \{ E^{in}(\vec r) \}, \qquad \mathrm{and}$$ (4)

$$E^{in}(\vec r) = \frac{k}{2 \pi f} \, \int d^2 \rho \, e^{-i\theta(\vec \rho)} E^f(\vec \rho) \, e^{ik \vec r \cdot \vec \rho/f}$$ (5)

$$\equiv {\cal F}^{-1}\{E^f(\vec \rho) \},$$ (6)

where $f$ is the focal length of the lens and $k = 2\pi/\lambda$ is the wavenumber of the incident light. The additional phase profile, $\theta(\vec \rho)$, due to the lens' geometry does not contribute to $I^f(\vec \rho)$ and may be ignored without loss of generality [12].