Thank you Dr. Jordan for the videos. IMHO, ‘wt’ can be dropped off because all the sin waves form standing waves, since for each component kx, there is a counterpart -kx component.
SO every one talks about Fourier optics but what about Laplace to encode the Transmission loss as is done in EE. Also, are you going to talk about optical phase space? And I would love to see that discussed and how the Lagrange optical invariant is conserved in and when it's not conserved say when the object is larger then what the overall system smallest transmissible object by Cheif and Marginal Ray analysis. And your videos are a great supplement to Born & Wolf that spearheads the next gen of optical enginers
Thanks! The Laplace transform is actually implicitly used in Fourier optics, because the k-vector can have an imaginary part. These are called “evanescent waves” and are generally treated separately. But if you are propagating through lossy media your Fourier Transform must turn into a Laplace transform. I will definitely be covering the Lagrange invariant.
@@JordanEdmundsEECS Sweet; what about optical phase space (www-optica.inaoep.mx/escuela2011/docs/Markus_Testorf_INAOE/Markus_Testorf_INAOE_1.pdf) ? Cause it's easier to wrap one's head about this than how it's typically done accelerator physics or Classical Hamilton Mech .
I was just wondering how the x - component of the k vector became ksin(theeta) and the z-component kcos(theeta). Should'nt the x-component be kcos(theeta) and z-component be ksin(theeta)? This is what I get when I construct a triangle with the angle theeta.
Depends on how you define theta xD The z-component is defined to point away from the screen (from left to right), and should be maximum when theta = 0.
@@JordanEdmundsEECS You defined theta to be the angle between the z=0 axis and how much k points off axis from z-direction. the cosine and sine should be switched for sure.
SingularAttitude it looks like in the diagram defined theta to be the angle between z=0 (the x axis) and the wavefront. The k vector is normal to the wavefront, so theta in terms of k is the angle between the z axis and the k vector. So like he said in his comment, the z component is maximum when theta=0, so it is the cos term.
I'm a nanoparticle chemist trying to wrap my head around how TEM creates contrast. This was helpful, thank you.
This was exactly what I needed today. Thank you!
Thank you Dr. Jordan for the videos. IMHO, ‘wt’ can be dropped off because all the sin waves form standing waves, since for each component kx, there is a counterpart -kx component.
im no in this field but i came here to understand cmb angular power spectrum and I think I can understand it now. thanks
Great video, thank you.
SO every one talks about Fourier optics but what about Laplace to encode the Transmission loss as is done in EE. Also, are you going to talk about optical phase space? And I would love to see that discussed and how the Lagrange optical invariant is conserved in and when it's not conserved say when the object is larger then what the overall system smallest transmissible object by Cheif and Marginal Ray analysis. And your videos are a great supplement to Born & Wolf that spearheads the next gen of optical enginers
Thanks! The Laplace transform is actually implicitly used in Fourier optics, because the k-vector can have an imaginary part. These are called “evanescent waves” and are generally treated separately. But if you are propagating through lossy media your Fourier Transform must turn into a Laplace transform. I will definitely be covering the Lagrange invariant.
@@JordanEdmundsEECS Sweet; what about optical phase space (www-optica.inaoep.mx/escuela2011/docs/Markus_Testorf_INAOE/Markus_Testorf_INAOE_1.pdf) ? Cause it's easier to wrap one's head about this than how it's typically done accelerator physics or Classical Hamilton Mech .
Definitely. Phase space is magical
Never made the connection of the numerical aperture type limit in this derivation of angular spectrum decomposition :)
I was just wondering how the x - component of the k vector became ksin(theeta) and the z-component kcos(theeta). Should'nt the x-component be kcos(theeta) and z-component be ksin(theeta)? This is what I get when I construct a triangle with the angle theeta.
Depends on how you define theta xD The z-component is defined to point away from the screen (from left to right), and should be maximum when theta = 0.
@@JordanEdmundsEECS You defined theta to be the angle between the z=0 axis and how much k points off axis from z-direction. the cosine and sine should be switched for sure.
SingularAttitude it looks like in the diagram defined theta to be the angle between z=0 (the x axis) and the wavefront. The k vector is normal to the wavefront, so theta in terms of k is the angle between the z axis and the k vector. So like he said in his comment, the z component is maximum when theta=0, so it is the cos term.
Who is theta _x