Saturday, December 24, 2016

Relation between image and resolution and wavelength required for taking non-fuzzy images

Q: 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. Reference https://www.physicsforums.com/threads/wavelength-and-image-resolution.204452/ A: 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; c = f λ c=fλ which can also be written c = ω k c=ωk Where omega is the angular frequency and k is the wavevector. If we rearrange the previous equation thus; k 2 = c 2 ω 2 k2=c2ω2 Remember that k is a vector and can be broken down into its constituent components; c 2 ω 2 = k 2 = k 2 x + k 2 y + k 2 z c2ω2=k2=kx2+ky2+kz2 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; − c ω < k x < c ω −cω Original thread (Answer by Cluade): https://www.physicsforums.com/threads/wavelength-and-image-resolution.204452/ WIKI: https://en.wikipedia.org/wiki/Diffraction-limited_system

No comments:

Post a Comment

Ubuntu 12.04, 14.04, 16.04 - auto start an app or script before login

To run a command or application at startup, even before the user has logged in, you can use this file: /etc/rc.local The commands entered...