How do dark spot and tiny hole filters affect Fourier optics images?

  • Thread starter Thread starter George444fg
  • Start date Start date
  • Tags Tags
    Fourier Optics
AI Thread Summary
Dark spot and tiny hole filters significantly impact Fourier optics images by altering the frequency components of the light used in imaging. The tiny hole filter eliminates high-frequency light, resulting in a limited but clear image by focusing on low-frequency components. Conversely, the dark spot filter removes the concentrated beam, allowing only high-frequency diffracted and refracted beams, which leads to a less clear image. The discussion emphasizes the importance of understanding these filtering processes theoretically, alongside practical experimentation. Overall, the filtering techniques directly influence image clarity and detail in Fourier optics.
George444fg
Messages
25
Reaction score
4
Homework Statement
George44fg
Relevant Equations
Fourier Transform Equations
I was assigned an experiment of Fouriers optics where I have to use different Filters. One of them was the dark spot and the tiny hole. As of my understanding, for tiny hole, we cut off all high-frequency light related to diffraction and refraction, thus using only the low freuency part of the initial image. The result would be a limited yet very clear image on the screen. While for the black spot we cut off the concentrated beam and project only the high frequency diffracted and refracted beams. Thus we get a rather unclear image? Thank you in advance
 
Physics news on Phys.org
How about you run the experiment and then any problems or results you have can be discussed if needed?
 
Tom.G said:
How about you run the experiment and then any problems or results you have can be discussed if needed?
I've already done the experiment, and everything is fine. However, I would like to be sure about the theoretical understanding of the filtering process.
 
George444fg said:
Homework Statement:: George44fg
Relevant Equations:: Fourier Transform Equations

...for tiny hole, we cut off all high-frequency light related to diffraction and refraction, thus using only the low freuency part of the initial image.

...the black spot we cut off the concentrated beam and project only the high frequency diffracted and refracted beams.

I agree!
 
Tom.G said:
I agree!
I thank you a lot.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top