Lecture 3 Flashcards
numerical aperture (formula)
= nsinθ = (D/2)/(object distance)
grating transmission
T(x) = 1/2(1+cos(kx))
period d = 2π/k
Diffracted order at:
x = 0 ± zλ/d
frequency content
object can be decomposed into its component Fourier components
these are not transmitted perfectly to the image
high spatial frequencies are filtered out at the system’s aperture
leads to loss of resolution and contrast in the image
spatial frequency in image (formula)
fXi = 1/di = s/ziλ
diffracted orders at
x = 0, ± zoλ/do
spatial period in image
di = ziλ/(zoλ/do) = zi/(zo/do) = Mdo
spatial frequency only resolved
if diffraction orders corresponding to that frequency pass through the aperture of the system
otherwise information about that frequency is lost
Image amplitude
Ui(X,Y) = h(X,Y)*Ug(X,Y)
Image intensity
Ii(X,Y) = |h(X,Y)|^2 * Ig(X,Y)
Convolution theorem
Fourier transform of convolution is the product of the Fourier transforms
to obtain the frequency content of an image take
the fourier transform
Fourier transform of intensity (formula)
Ii = H . Ig
where H is the optical transfer function (OTF)
what is attributed to the Optical Transfer Function
filtering and degradation of the system attributed to the OTF
it transmits low spatial frequencies and attenuates high spatial frequencies
Optical Transfer Function describes
how optical system filters or modifies spatial frequencies that would otherwise contribute to the image
the magnitude of OTF, |H|, corresponding
to contrast