Wavelength and Image Resolution


by chingkui
Tags: image, resolution, wavelength
chingkui
chingkui is offline
#1
Dec13-07, 04:31 PM
P: 195
I have been trying to find the explaination why the wavelength of a wave is the limit of the resolution of an image (i.e. one cannot see the detail smaller than the wavelength), which is sited as the reason why we cannot see an atom using visible light. I don't think I really understand why this is so. can someone please explain to me the physics behind resolving detail with wave and the reason behing this limit? Thank you.
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Claude Bile
Claude Bile is offline
#2
Dec13-07, 05:42 PM
Sci Advisor
P: 1,465
Firstly, the diffraction limit is wavelength/2 (approximately) - actually a more precise formula is 0.67*wavelength/NA where NA is the numerical aperture of the imaging apparatus and has a practical maximum of about 1.4.

When we image something, the light we collect is spatially modulated. That means if I measured the field as a function of distance, in the x direction for example - call it E(x), E(x) would vary with x, as opposed to being a constant. The best resolution we can get actually depends on the Fourier transform of E(x) - what is sometimes called the spatial frequency spectrum in the direction of x, and the label we give to spatial frequency is k (so the Fourier transform E(x), would be E'(k_x) since we are concerned with the x direction only).

To further illustrate the above point - take transmission through an aperture at z = 0. E(x) will be a rectangular function at z = 0, and it's spatial frequency spectrum, E'(k_x) will be a sinc function. If we decrease the width of the aperture, the width of the spatial frequency spectrum increases. This is because higher spatial frequency components are needed to "resolve" smaller objects.

So where does the diffraction limit come from? It comes from the fact that only a certain range of spatial frequencies can propagate in a vacuum. Take the dispersion relation of free space;

[tex]c = f\lambda[/tex]

which can also be written

[tex]c = \omega k[/tex]

Where omega is the angular frequency and k is the wavevector. If we rearrange the previous equation thus;

[tex]k^2 = \frac{c^2}{\omega^2}[/tex]

Remember that k is a vector and can be broken down into its constituent components;

[tex]\frac{c^2}{\omega^2} = k^2 = k_x^2 + k_y^2 + k_z^2[/tex]

Only components of k that are real will reach the far-field. Imaginary components result in evanescent waves that die off exponentially with distance and thus do not reach the far-field. From the above equation, it is easy to see that;

[tex]-\frac{c}{\omega} < k_x < \frac{c}{\omega} [/tex]

The resultant image of a sub-wavelength object (obtained by taking the inverse Fourier transform of E'(k_x)) is therefore the Fourier transform of the rectangular function of half-width [itex]c/\omega[/itex], which is a sinc function of width [itex]\pi.c/\omega[/itex]. It is then noted that;

[tex]\frac{\pi c}{\omega} = \frac{\pi f \lambda}{2\pi f} = \frac{\lambda}{2}[/tex]

Which is the resolution limit for diffraction. To get the more precise expression, one needs to simply repeat the above process with the 2D case.

Claude.


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