It does filter frequencies. Not time frequencies, though.
Science and Technology Quiz 3
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It does filter frequencies. Not time frequencies, though.
Mise
isle of lucy
Spatial frequencies? i.e. wavenumbers / lengths? Surely that's the same as filtering time frequencies or colour?
I can't think of any other types of frequency
I can't think of any other types of frequency
Spatial frequencies? i.e. wavenumbers / lengths? Surely that's the same as filtering time frequencies or colour?
I can't think of any other types of frequency
Yes, spatial frequencies is right and no, that's not the same as filtering time frequencies.
Now you only have to consider which spatial frequencies are filtered, and what this means for the image.
This isn't going anywhere, so I'll solve:
Assuming that the optical axis is the z-direction, the disks filter the spatial frequencies of the object in x- and y-direction. One can visualise one cycle of these frequencies to be a black-white line pair.
Fourier transforming the object will give the object in k-space (k_x and k_y). As the total value of k is given by the wavelength, k_z must also be adjusted for different spatial frequencies. This means that those rays emitted along the z-axis, correspond to very low spatial frequencies and those emitted at a high angle to the z-axis correspond to high spatial frequencies.
In case a) the low spatial frequencies are filtered and the high spatial frequencies get through. This means, that only the contours of the features of the object can be seen on the image. When this is done in a microscope, this is called "darkfield microscopy".
In case b) it's the other way around: Now the contours and fine features are blurred, leaving only the main features on the image. This can clear up artifacts on the image (at the cost of losing resolution).
Open floor
Assuming that the optical axis is the z-direction, the disks filter the spatial frequencies of the object in x- and y-direction. One can visualise one cycle of these frequencies to be a black-white line pair.
Fourier transforming the object will give the object in k-space (k_x and k_y). As the total value of k is given by the wavelength, k_z must also be adjusted for different spatial frequencies. This means that those rays emitted along the z-axis, correspond to very low spatial frequencies and those emitted at a high angle to the z-axis correspond to high spatial frequencies.
In case a) the low spatial frequencies are filtered and the high spatial frequencies get through. This means, that only the contours of the features of the object can be seen on the image. When this is done in a microscope, this is called "darkfield microscopy".
In case b) it's the other way around: Now the contours and fine features are blurred, leaving only the main features on the image. This can clear up artifacts on the image (at the cost of losing resolution).
Open floor
thetrooper
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Open floor?
Okay, what's this sentence supposed to give you info about:
How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics.
Okay, what's this sentence supposed to give you info about:
How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics.
ParadigmShifter
Random Nonsense Generator
It's the (first few digits of the) decimal expansion of pi (number of letters in each word).
thetrooper
Misanthrope
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Indeed, too easy eh?
Enough decimals even for NASA I suppose...
Your turn
Enough decimals even for NASA I suppose...
Your turn
Que j'aime a faire apprendre un nombre utile aux sages.
Glorieux Archimede, artiste ingenieux!
Toi de qui Syracuse aime encore la gloire.
Soit ton nom conserve par de savants grimoires.
Jadis, mysterieux, un probleme existait.
Tout l'admirable procede (l'oeuvre etonnante!)
Que Pythagore decouvrit aux anciens Grecs:
O Quadrature! Vieux tourment du philosophe! Sibylline rondeur!
Trop longtemps vous avez defie Pythagore et ses imitateurs!
Comment integrer l'espace plan circulaire?
Thales tu tomberas! Platon tu desesperes!
Apparait Archimede:
Archimede inscrira dedans un hexagone:
Appreciera son aire fonction du rayon;
Pas trop ne s'y tiendra!
Dedoublera chaque element anterieur,
Toujours de l'orbe calculee approchera
Laquelle limite donne l'arc,
La longueur de cet inquietant cercle,
Ennemi trop rebelle!
Professeur, enseignez son probleme avec Zele ...
the 10-letter words represent a 0.
Glorieux Archimede, artiste ingenieux!
Toi de qui Syracuse aime encore la gloire.
Soit ton nom conserve par de savants grimoires.
Jadis, mysterieux, un probleme existait.
Tout l'admirable procede (l'oeuvre etonnante!)
Que Pythagore decouvrit aux anciens Grecs:
O Quadrature! Vieux tourment du philosophe! Sibylline rondeur!
Trop longtemps vous avez defie Pythagore et ses imitateurs!
Comment integrer l'espace plan circulaire?
Thales tu tomberas! Platon tu desesperes!
Apparait Archimede:
Archimede inscrira dedans un hexagone:
Appreciera son aire fonction du rayon;
Pas trop ne s'y tiendra!
Dedoublera chaque element anterieur,
Toujours de l'orbe calculee approchera
Laquelle limite donne l'arc,
La longueur de cet inquietant cercle,
Ennemi trop rebelle!
Professeur, enseignez son probleme avec Zele ...
the 10-letter words represent a 0.
ParadigmShifter
Random Nonsense Generator
How I wish I could calculate pi
Is good enough for me (although it ends with a bad rounding
)
Describe the form of all integers that are sandwiched inbetween a perfect square and a perfect cube.
(i.e. all x such that a*a*a+1 = x = b*b-1 or a*a+1 = x = b*b*b-1). [with a, b integers too of course]
Who first proved this?
Is good enough for me (although it ends with a bad rounding
)Describe the form of all integers that are sandwiched inbetween a perfect square and a perfect cube.
(i.e. all x such that a*a*a+1 = x = b*b-1 or a*a+1 = x = b*b*b-1). [with a, b integers too of course]
Who first proved this?
Mise
isle of lucy
pie*
was it one of fermat's theorems?
was it one of fermat's theorems?
ParadigmShifter
Random Nonsense Generator
The only number I can find that suffices this condition is 26 (5^2+1=3^3-1)=26
ParadigmShifter
Random Nonsense Generator
Keep looking
ParadigmShifter
Random Nonsense Generator
That is correct! -1^3 + 1 = 0 = 1^2 - 1, and 5^2 + 1 = 26 = 3^3 - 1.
Those are the only numbers satisfying those conditions.
I'm looking for my number theory book which I think has a proof.
Those are the only numbers satisfying those conditions.
I'm looking for my number theory book which I think has a proof.
That was the clue for me. I was thinking why you would specifically say integers instead of natural numbers, when a negative number couldn't be sandwiched by a square (except -1, but that's easy enough to rule out). Then I realised you don't consider 0 a natural number.
ParadigmShifter
Random Nonsense Generator
You're up again then, with your fancy-pants Excel spreadsheets 
Can't find my book I might have a look on tinternet.
Can't find my book I might have a look on tinternet.
thetrooper
Misanthrope
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ParadigmShifter said:
I have a book by Simon Singh on Fermat's Last Theorem. Have you read it?
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