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Reverse of diffraction

Does diffraction contribute to the Black Drop effect?Why does a CD work as a diffraction grating even with light from a light bulb?Diffraction Grating in spectrophotometerDiffraction problem - How do I interpretBasic rule for diffractionx-ray diffraction of crystalsTotal number of primary maxima in diffraction gratingDiffraction of monochromatic lightExplanation of single and double slit diffraction, and Frensel diffraction in light of Feynman's method of probability amplitudesDiffraction pattern vs Interference pattern

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4

$begingroup$

Can we arrange a practical in such a way that the dark and bright bands in diffraction grating be allowed to pass through the same slit to get the original light (i.e the incident light before diffraction), just like we reverse the arrows in reflection and refraction

optics visible-light reflection refraction diffraction

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edited 7 hours ago

Qmechanic♦

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  $begingroup$
  You might want to look at a zone plate.
  $endgroup$
  – M. Enns
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Can we arrange a practical in such a way that the dark and bright bands in diffraction grating be allowed to pass through the same slit to get the original light (i.e the incident light before diffraction), just like we reverse the arrows in reflection and refraction

optics visible-light reflection refraction diffraction

share|cite|improve this question

edited 7 hours ago

Qmechanic♦

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asked 8 hours ago

Syed Hasnain AhmedSyed Hasnain Ahmed

2111 bronze badge

New contributor

Syed Hasnain Ahmed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You might want to look at a zone plate.
  $endgroup$
  – M. Enns
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Can we arrange a practical in such a way that the dark and bright bands in diffraction grating be allowed to pass through the same slit to get the original light (i.e the incident light before diffraction), just like we reverse the arrows in reflection and refraction

optics visible-light reflection refraction diffraction

share|cite|improve this question

edited 7 hours ago

Qmechanic♦

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asked 8 hours ago

Syed Hasnain AhmedSyed Hasnain Ahmed

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Syed Hasnain Ahmed is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 7 hours ago

Qmechanic♦

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share|cite|improve this question

edited 7 hours ago

Qmechanic♦

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edited 7 hours ago

Qmechanic♦

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edited 7 hours ago

Qmechanic♦

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asked 8 hours ago

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3 Answers
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$begingroup$

Optics is time-reversal invariant, so the short answer is "yes".

As a practical matter it is essentially impossible to set up the exact reversed situation, but we can set up good approximations in some cases (and in particular in electromagnetic bands not including visible light). Once the approximation is good enough it becomes useful.

A "phased array" transmitter is exactly a mechanism for focusing electromagnetic waves by emitting them from many places with carefully arranged phase off-sets. They are difficult to build, tune, and operate effectively; but they are a thing.

share|cite|improve this answer

answered 8 hours ago

dmckee♦dmckee

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add a comment | 

2

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Reversing diffraction is precisely what is done in Fourier Optics!

There, instead of placing a screen after the slit to see the diffraction pattern, a lens is put there instead. If you arrange the diffraction image to lie at the back focal plane of the lens, you will see the image of the slit itself at the front focal plane of the lens.

This procedure of reversing diffraction relies on a fundamental Fourier transform property of a thin lens. The idea is that placing an object exactly one focal length $+f$ away from the lens will produce an image of the object one focal length $-f$ on the opposite side that is the Fourier transform of the object.

This is shown in the image below

share|cite|improve this answer

answered 4 hours ago

KF GaussKF Gauss

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add a comment | 

1

$begingroup$

for light beams, you might also consider something called a corner reflector or retro-reflector. this is a glass prism cut in such a way that any beam of light entering it is reflected back at almost exactly the same angle that it made going into the prism. if you constructed a flat plate containing a large number of small corner reflectors and placed it at the location of the screen in the experiment you describe, then all the light striking the plate across its width would be reflected back to the slit.

I do not know how that collection of beams coming back into the slit would behave when they all meet!

share|cite|improve this answer

answered 5 hours ago

niels nielsenniels nielsen

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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glass corner cubes aren't so simple. Rays do not counterpropagate on themselves. And each of the six possible paths of a ray through a corner cube affects the phase of the light differently. Metallic reflecting corner cubes help, but since metals are not perfect conductors, there is still a small phase/polarization effect, but there is energy lost.
  $endgroup$
  – garyp
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Optics is time-reversal invariant, so the short answer is "yes".

As a practical matter it is essentially impossible to set up the exact reversed situation, but we can set up good approximations in some cases (and in particular in electromagnetic bands not including visible light). Once the approximation is good enough it becomes useful.

A "phased array" transmitter is exactly a mechanism for focusing electromagnetic waves by emitting them from many places with carefully arranged phase off-sets. They are difficult to build, tune, and operate effectively; but they are a thing.

share|cite|improve this answer

answered 8 hours ago

dmckee♦dmckee

76k66 gold badges139139 silver badges277277 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Optics is time-reversal invariant, so the short answer is "yes".

As a practical matter it is essentially impossible to set up the exact reversed situation, but we can set up good approximations in some cases (and in particular in electromagnetic bands not including visible light). Once the approximation is good enough it becomes useful.

A "phased array" transmitter is exactly a mechanism for focusing electromagnetic waves by emitting them from many places with carefully arranged phase off-sets. They are difficult to build, tune, and operate effectively; but they are a thing.

share|cite|improve this answer

answered 8 hours ago

dmckee♦dmckee

76k66 gold badges139139 silver badges277277 bronze badges

$endgroup$

add a comment | 

$begingroup$

Optics is time-reversal invariant, so the short answer is "yes".

As a practical matter it is essentially impossible to set up the exact reversed situation, but we can set up good approximations in some cases (and in particular in electromagnetic bands not including visible light). Once the approximation is good enough it becomes useful.

A "phased array" transmitter is exactly a mechanism for focusing electromagnetic waves by emitting them from many places with carefully arranged phase off-sets. They are difficult to build, tune, and operate effectively; but they are a thing.

share|cite|improve this answer

answered 8 hours ago

dmckee♦dmckee

76k66 gold badges139139 silver badges277277 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

dmckee♦dmckee

76k66 gold badges139139 silver badges277277 bronze badges

answered 8 hours ago

dmckee♦dmckee

76k66 gold badges139139 silver badges277277 bronze badges

answered 8 hours ago

dmckee♦dmckee

76k66 gold badges139139 silver badges277277 bronze badges

2

$begingroup$

Reversing diffraction is precisely what is done in Fourier Optics!

There, instead of placing a screen after the slit to see the diffraction pattern, a lens is put there instead. If you arrange the diffraction image to lie at the back focal plane of the lens, you will see the image of the slit itself at the front focal plane of the lens.

This procedure of reversing diffraction relies on a fundamental Fourier transform property of a thin lens. The idea is that placing an object exactly one focal length $+f$ away from the lens will produce an image of the object one focal length $-f$ on the opposite side that is the Fourier transform of the object.

This is shown in the image below

share|cite|improve this answer

answered 4 hours ago

KF GaussKF Gauss

1,61811 gold badge1010 silver badges2626 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Reversing diffraction is precisely what is done in Fourier Optics!

There, instead of placing a screen after the slit to see the diffraction pattern, a lens is put there instead. If you arrange the diffraction image to lie at the back focal plane of the lens, you will see the image of the slit itself at the front focal plane of the lens.

This procedure of reversing diffraction relies on a fundamental Fourier transform property of a thin lens. The idea is that placing an object exactly one focal length $+f$ away from the lens will produce an image of the object one focal length $-f$ on the opposite side that is the Fourier transform of the object.

This is shown in the image below

share|cite|improve this answer

answered 4 hours ago

KF GaussKF Gauss

1,61811 gold badge1010 silver badges2626 bronze badges

$endgroup$

add a comment | 

$begingroup$

Reversing diffraction is precisely what is done in Fourier Optics!

There, instead of placing a screen after the slit to see the diffraction pattern, a lens is put there instead. If you arrange the diffraction image to lie at the back focal plane of the lens, you will see the image of the slit itself at the front focal plane of the lens.

This procedure of reversing diffraction relies on a fundamental Fourier transform property of a thin lens. The idea is that placing an object exactly one focal length $+f$ away from the lens will produce an image of the object one focal length $-f$ on the opposite side that is the Fourier transform of the object.

This is shown in the image below

share|cite|improve this answer

answered 4 hours ago

KF GaussKF Gauss

1,61811 gold badge1010 silver badges2626 bronze badges

$endgroup$

share|cite|improve this answer

answered 4 hours ago

KF GaussKF Gauss

1,61811 gold badge1010 silver badges2626 bronze badges

answered 4 hours ago

KF GaussKF Gauss

1,61811 gold badge1010 silver badges2626 bronze badges

answered 4 hours ago

KF GaussKF Gauss

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1

$begingroup$

for light beams, you might also consider something called a corner reflector or retro-reflector. this is a glass prism cut in such a way that any beam of light entering it is reflected back at almost exactly the same angle that it made going into the prism. if you constructed a flat plate containing a large number of small corner reflectors and placed it at the location of the screen in the experiment you describe, then all the light striking the plate across its width would be reflected back to the slit.

I do not know how that collection of beams coming back into the slit would behave when they all meet!

share|cite|improve this answer

answered 5 hours ago

niels nielsenniels nielsen

23.3k55 gold badges3232 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glass corner cubes aren't so simple. Rays do not counterpropagate on themselves. And each of the six possible paths of a ray through a corner cube affects the phase of the light differently. Metallic reflecting corner cubes help, but since metals are not perfect conductors, there is still a small phase/polarization effect, but there is energy lost.
  $endgroup$
  – garyp
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

for light beams, you might also consider something called a corner reflector or retro-reflector. this is a glass prism cut in such a way that any beam of light entering it is reflected back at almost exactly the same angle that it made going into the prism. if you constructed a flat plate containing a large number of small corner reflectors and placed it at the location of the screen in the experiment you describe, then all the light striking the plate across its width would be reflected back to the slit.

I do not know how that collection of beams coming back into the slit would behave when they all meet!

share|cite|improve this answer

answered 5 hours ago

niels nielsenniels nielsen

23.3k55 gold badges3232 silver badges6666 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glass corner cubes aren't so simple. Rays do not counterpropagate on themselves. And each of the six possible paths of a ray through a corner cube affects the phase of the light differently. Metallic reflecting corner cubes help, but since metals are not perfect conductors, there is still a small phase/polarization effect, but there is energy lost.
  $endgroup$
  – garyp
  4 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

for light beams, you might also consider something called a corner reflector or retro-reflector. this is a glass prism cut in such a way that any beam of light entering it is reflected back at almost exactly the same angle that it made going into the prism. if you constructed a flat plate containing a large number of small corner reflectors and placed it at the location of the screen in the experiment you describe, then all the light striking the plate across its width would be reflected back to the slit.

I do not know how that collection of beams coming back into the slit would behave when they all meet!

share|cite|improve this answer

answered 5 hours ago

niels nielsenniels nielsen

23.3k55 gold badges3232 silver badges6666 bronze badges

$endgroup$

share|cite|improve this answer

answered 5 hours ago

niels nielsenniels nielsen

23.3k55 gold badges3232 silver badges6666 bronze badges

answered 5 hours ago

niels nielsenniels nielsen

23.3k55 gold badges3232 silver badges6666 bronze badges

answered 5 hours ago

niels nielsenniels nielsen

23.3k55 gold badges3232 silver badges6666 bronze badges