## Abstract

IN light optics and electron optics one normally measures the image intensity and attempts to infer from these measurements the object structure. In light optics these intensity measurements are insufficient to evaluate the detailed structure of the object and information on the phase of the transmitted light is of prime importance^{1}; the interaction of the light or electrons in transmission through the object can be only satisfactorily explained by wave optics. The phase problem in optics is analogous to the X-ray diffraction phase problem, that is, whereas an object can be reconstructed from the phase information (even with incorrect amplitudes), it is not possible to define an object from amplitude measurements only^{1}. We suggest here how a normal optical system can be used to determine both the amplitude and phase from image intensity measurements in bright-field optics. In bright-field optics, where the main (on-axis) beam is allowed to interfere with the scattered wave, the transmitted object wavefunction Ψ_{0}(*r*_{0}) can be written as where Ψ_{s}(*r*_{0}) represents the effect of the object on the incident light or electron wave at the point *r*_{0} = (*x*_{0}, *y*_{0}) in the object plane; Φ_{s} carries information not only on the amplitude attenuation of the wave but also phase shifts introduced by refractive index differences (potential differences in electron optics) across the object. The 1 represents the amplitude of the transmitted wave that is not scattered in the object, and this unscattered contribution will give a background in the image. The wavefunction Φ_{0}(*r*_{0}) transmitted by the object may be subsequently affected by lens defects and restricting apertures in the optical system, particularly in electron optics, but for simplicity we shall assume that the wave at the image plane Φ_{i}(*r*_{i}) is equivalent to Φ_{0}(*r*_{0}). We record not Φ_{i}(*r*_{i}) but the image intensity ∣Φ_{i}(*r*_{i})∣^{2} ≡∣Φ_{0}(*r*_{0})∣^{2}; the image intensity is from equation (1) where *Re* and *Im* denote the real and imaginary parts of Φ_{s}. At best, if the interaction between the light or electron wave and the object is small and the squared terms in equation (2) can be neglected, we can only determine the real part of Φ_{s} from image intensity measurements, and there is no direct way of evaluating the imaginary part of Φ_{s}, hence determining both the amplitude and phase. But instead of using a normal circular aperture in the back focal plane (Fourier plane) of the objective lens, we can use a semicircular (half-plane) aperture to cut off one-half of the Fourier plane of the object wave-function. The Fourier plane, which corresponds to the diffraction or scattering plane, displays the spatial frequencies of the object structure, and the idea is to eliminate initially one-half of the object frequencies from the image. Now there is an explicit relationship between the real and imaginary parts of Φ_{s}. We maintain the bright-field situation by having a small centre section cut out from the aperture to allow the main beam through to interfere with the scattered wave—this is strictly necessary in electron optics to avoid electrical charging problems. Suppose we omit the negative half of the Fourier plane (spatial frequencies ν_{x}<0), but allow all positive frequencies to contribute to the image; then the image intensity corresponding to this situation (Fig. 1*a*) is, omitting subscripts i, The idea of using a half-plane aperture is not new^{2}, but the important relationship between $\mathit{\text{Im}}{\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ and $\mathit{\text{Re}}{\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ has not been given. The relationship between the real and imaginary parts of ${\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ can be derived from standard Fourier theory^{3} and the analyticity of Φ_{s}^{4}, and in order to avoid detailed mathematics, we state the result that the imaginary part of ${\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ is a one-dimensional Hilbert transform of the real part; that is, taking the Cauchy principal value of the integral. From an image recorded using a semicircular or half-plane aperture we have an explicit relation between $\mathit{\text{Re}}{\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ and $\mathit{\text{Im}}{\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$. But the squared terms in equation (3) prevent a direct calculation of $\mathit{\text{Re}}{\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ from the image intensity ${j}_{{\nu}_{x}\mathit{\text{>0}}}$. We have solved this problem by initially neglecting these squared terms and we calculate $\mathrm{Re}{\mathit{\text{(}}{\Phi}_{s}\mathit{\text{)}}}_{{\nu}_{x}}\mathit{\text{>0}}$ from ½(${j}_{{\nu}_{x}\text{>0}}$>0−1); using this approximation to $\mathit{\text{Re}}{\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ we evaluate $\mathit{\text{Im}}{\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ using a form of equation (4) which avoids the problem of the singularity at *z*=*x* and also takes account of the finite extent of the actual image; again this is a mathematical detail. We now have values for both the real and imaginary parts of ${\text{(}{\text{\Phi}}_{\mathrm{s}}\text{)}}_{{{\nu}_{\mathrm{x}}}_{\text{>0}}}$ and a correction is applied to equation (3) for the squared terms, *Re*^{2}+*Im*^{2}; it may be necessary to apply this correction several times, depending on the relative contributions of the squared and linear terms to the image intensity ${j}_{{\nu}_{x}\text{>0}}$. In numerical tests, contributions from these squared terms pf 25% of the value of ${j}_{{\nu}_{x}\text{>0}}$ can be systematically corrected in 1–5 iterations. This magnitude for the correction is well within the normal conditions of electron microscopy, where image contrast is only 20% (of which squared terms contribute 10%); *Re*^{2}+*Im*^{2} = 0.25 (relative to the unity of the unscattered beam) would correspond to a very thick object with a strong interaction between the incident wave and the object, corresponding to about 25% scattering from the incident beam.

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## Author information

### Author notes

- D. L. MISELL

Present address: Biophysics Division, National Institute for Medical Research, The Ridgeway, Mill Hill, London, NW7 1AA

### Affiliations

#### Physics Department, Queen Elizabeth College, Campden Hill Road, London, W8 7AH

- D. L. MISELL
- , R. E. BURGE
- & A. H. GREENAWAY

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