On 5/14/2009 10:40:07 AM, Jbryant61 wrote:
>This is why I kept with the
>Virendra equations for so long
>- I couldn't get my head
>around this MTF theory.
The concept of MTF is easy. Take an object with a perfect sinusoidal intensity going from light (intensity=1) to dark (intensity=0). If that is imaged through a perfect optical system we will get a sinusoidal image going from intensity=1 to intensity=0. No optical system is perfect though, if only because of diffraction. So the sinusoid is blurred, and energy is moved from the peaks to the valleys and the valleys to the peaks. The sinusoid therefore no longer goes from 0 to 1, but from something greater than zero to something less than 1. In a diffraction limited case the blurring depends on the spatial frequency of the sinusoid and the wavelength of the light. To represent the intensity variation as a number we can calculate (max-min)/(max+min), which is the modulation at the frequency of the sinusoid, for a specific wavelength. If we divide the image modulation by the object modulation we have the MTF for a specific spatial frequency and a specific wavelength. When the optical system has aberrations they will also blur the image and the MTF will be lower than the diffraction limited case. A graph of MTF vs spatial frequency is very useful, because it shows you the resolving power in a far more detailed way than just looking at a simple formula like the Rayleigh criterion. Any wavefront distortion will lower the MTF though, so if the MTF is low you do not know why, only that it is. You can get the MTF either from measurements or by calculation in a software package like Zemax (look up MTF in the manual: it can use more than one method to calculate it).
> I have
>been reading up and thats
>where the confusion is. I read
>that you can divide the
>spectra to get the MTF's as
>well as divide the modulation
>depths (or contrast as you
>term it). It dosen't make
>sense to me the modulation
>depth of the object.
The spectra are just the amplitude vs frequency, so in either case, for a sinusoid, you are just calculating the ratio of the amplitudes.
The modulation depth of the object is just the contrast of the object: for sinusoidal objects.
> So in your
>worksheet why is the frequency
>f/Npts?.
I fixed that so f is now just f, in pixels.
> Why doesn't NA come
>into this, after all this
>defines resolution (0.61*lamda
>/ NA)?
It does, in the theory. This is a good introductory write up of MTF:
http://www.mellesgriot.com/products/optics/os_2_2.htm>I read that there is a cut off
>frequency associated with the
>diffraction limit. But again
>there seems to be a strange
>way to represent this on a
>graph where both x and y-axes
>are 0 to unity.
No, there is no real cutoff. The MTF is obviously wavelength dependent, and sometimes it would be nice to have a theoretical expression that removes that dependence, for example when plotting the diffraction limited MTF. It's the same concept as plotting the Airy disk with x-axis units of wavelengths. The MTF drops with increasing spatial frequency, so we can pick an upper limit to that frequency, and say that we don't care about anything higher. Then we normalize the spatial frequencies to that upper frequency. If the diffraction limited MTF graph is plotted with normalized frequencies the x-axis goes from 0 to 1 and the graph is the same for all wavelengths. We could plot the graph for values greater than 1, but since we decided that higher frequencies are not important we don't. there are several ways to calculate the upper frequency: see the link above.
Richard