In Young’s double slit experiment, can there be an even number of maxima in the central envelope maximum?

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In Young’s double slit experiment, can we get an even number of maxima in the central envelope maximum? If so how and why the regular modelling of the double split interfaces is with an odd number like in the picture

Source of the pic: Hyperphysics

In Young’s double slit experiment, can we get an even number of maxima in the central envelope maximum? If so how and why the regular modelling of the double split interfaces is with an odd number like in the picture

Source of the pic: Hyperphysics

1

In Young’s double slit experiment, can we get an even number of maxima in the central envelope maximum? If so how and why the regular modelling of the double split interfaces is with an odd number like in the picture

Source of the pic: Hyperphysics

In Young’s double slit experiment, can we get an even number of maxima in the central envelope maximum? If so how and why the regular modelling of the double split interfaces is with an odd number like in the picture

Source of the pic: Hyperphysics

optics waves double-slit-experiment

edited Nov 28 at 22:32

knzhou

39.1k9109189

39.1k9109189

asked Nov 28 at 18:20

Learnero

232

232

3 Answers3

active

oldest

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In order for two maximum interference fringes to appear, it is necessary to break the reflection symmetry between the two slits. If this symmetry remains unbroken, the phases associated with paths through the top and bottom slits always match, and the amplitudes interfere constructively as mentioned by Tausif Hossain. There are many ways to break this symmetry, most of which involve doing so at a classical level (this is the case, for example, in Pieter’s answer pertaining to double slit interference of photons.)

In the case of double slit interference of electrons, it is possible to break the symmetry in a way that is intrinsically quantum mechanical (preserving the reflection symmetry at a classical level) using the Aharonov Bohm effect. If you insert a thin magnetic flux through and perpendicular to the electron beam and tune it to just the right value, you should be able to obtain an even number of maxima. (Whether you can get more than two maxima is another story, and probably depends more sensitively on how the magnetic flux is distributed.)

One can get an even number of fringes by covering the pair of slits with a wedge of glass or plastic that gives half a wavelength more delay at one slit than at the other.

Notice that if it is strictly a double slit experiment with a coherent light source then there will always be a central bright fringe where the path difference between the waves is zero(which is that, there is a point in the screen which is equidistant from the slits which lies on the perpendicular bisector of the slits). Hence the constructive interference at the center and the bright fringe.

Thus, as long as that center fringe exists and also as there is symmetry ofcourse between the path differences on both sides of the central bright fringe. So the other bright fringes occur in pairs (like staring from 1 wavelength path difference from each side and so on).So, the number of fringes is always $$1+2n$$ (Where 1 is for the central bright fringe). Hence an odd number of bright fringes.

• I had doubts about it cause I read this question “In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maximum of the single slit pattern? (d=1mm)” The solution was like this Distance between the slits = d = 1 mm Path difference = a sinθ ≅ aθ = λ ⇒ θ = λ /a Width of central maximum of the single slit = 2 λ/a Width of the 20 maxima = 20 x fringe spacing = 20 x λ/d Width of central maximum of the single slit = Width of the 20 maxima of double slit 2 λ/a= 20 x λ/d a = d/10 = 0.1 mm So how is that possible?
– Learnero
Nov 28 at 18:57

• What I think the question meant is there are 20 Maxima apart from the central maxima. Hence you can say there are 10 Maxima on each side of the central fringe.
– Tausif Hossain
Nov 28 at 19:00

• Thank you Tausif
– Learnero
Nov 28 at 19:03

• You’re welcome. If you think the answer is satisfactory please accept it by clicking the “tick” icon.
– Tausif Hossain
Nov 28 at 19:05

• This ans. is correct for standard arrangement. You might find it interesting that you can also shift the fringe pattern by more than you shift the envelope, by putting a thin piece of glass over the slits, slightly thicker at one slit than the other. In this way even numbers are possible (and c.f. “blazed grating”).
– Andrew Steane
Nov 28 at 19:21

3 Answers3

active

oldest

votes

3 Answers3

active

oldest

votes

active

oldest

votes

active

oldest

votes

In order for two maximum interference fringes to appear, it is necessary to break the reflection symmetry between the two slits. If this symmetry remains unbroken, the phases associated with paths through the top and bottom slits always match, and the amplitudes interfere constructively as mentioned by Tausif Hossain. There are many ways to break this symmetry, most of which involve doing so at a classical level (this is the case, for example, in Pieter’s answer pertaining to double slit interference of photons.)

In the case of double slit interference of electrons, it is possible to break the symmetry in a way that is intrinsically quantum mechanical (preserving the reflection symmetry at a classical level) using the Aharonov Bohm effect. If you insert a thin magnetic flux through and perpendicular to the electron beam and tune it to just the right value, you should be able to obtain an even number of maxima. (Whether you can get more than two maxima is another story, and probably depends more sensitively on how the magnetic flux is distributed.)

In order for two maximum interference fringes to appear, it is necessary to break the reflection symmetry between the two slits. If this symmetry remains unbroken, the phases associated with paths through the top and bottom slits always match, and the amplitudes interfere constructively as mentioned by Tausif Hossain. There are many ways to break this symmetry, most of which involve doing so at a classical level (this is the case, for example, in Pieter’s answer pertaining to double slit interference of photons.)

In the case of double slit interference of electrons, it is possible to break the symmetry in a way that is intrinsically quantum mechanical (preserving the reflection symmetry at a classical level) using the Aharonov Bohm effect. If you insert a thin magnetic flux through and perpendicular to the electron beam and tune it to just the right value, you should be able to obtain an even number of maxima. (Whether you can get more than two maxima is another story, and probably depends more sensitively on how the magnetic flux is distributed.)

In order for two maximum interference fringes to appear, it is necessary to break the reflection symmetry between the two slits. If this symmetry remains unbroken, the phases associated with paths through the top and bottom slits always match, and the amplitudes interfere constructively as mentioned by Tausif Hossain. There are many ways to break this symmetry, most of which involve doing so at a classical level (this is the case, for example, in Pieter’s answer pertaining to double slit interference of photons.)

In the case of double slit interference of electrons, it is possible to break the symmetry in a way that is intrinsically quantum mechanical (preserving the reflection symmetry at a classical level) using the Aharonov Bohm effect. If you insert a thin magnetic flux through and perpendicular to the electron beam and tune it to just the right value, you should be able to obtain an even number of maxima. (Whether you can get more than two maxima is another story, and probably depends more sensitively on how the magnetic flux is distributed.)

In order for two maximum interference fringes to appear, it is necessary to break the reflection symmetry between the two slits. If this symmetry remains unbroken, the phases associated with paths through the top and bottom slits always match, and the amplitudes interfere constructively as mentioned by Tausif Hossain. There are many ways to break this symmetry, most of which involve doing so at a classical level (this is the case, for example, in Pieter’s answer pertaining to double slit interference of photons.)

In the case of double slit interference of electrons, it is possible to break the symmetry in a way that is intrinsically quantum mechanical (preserving the reflection symmetry at a classical level) using the Aharonov Bohm effect. If you insert a thin magnetic flux through and perpendicular to the electron beam and tune it to just the right value, you should be able to obtain an even number of maxima. (Whether you can get more than two maxima is another story, and probably depends more sensitively on how the magnetic flux is distributed.)

edited Nov 28 at 20:05

answered Nov 28 at 19:57

fs137

2,540815

2,540815

One can get an even number of fringes by covering the pair of slits with a wedge of glass or plastic that gives half a wavelength more delay at one slit than at the other.

One can get an even number of fringes by covering the pair of slits with a wedge of glass or plastic that gives half a wavelength more delay at one slit than at the other.

One can get an even number of fringes by covering the pair of slits with a wedge of glass or plastic that gives half a wavelength more delay at one slit than at the other.

One can get an even number of fringes by covering the pair of slits with a wedge of glass or plastic that gives half a wavelength more delay at one slit than at the other.

answered Nov 28 at 19:41

Pieter

7,45231431

7,45231431

Notice that if it is strictly a double slit experiment with a coherent light source then there will always be a central bright fringe where the path difference between the waves is zero(which is that, there is a point in the screen which is equidistant from the slits which lies on the perpendicular bisector of the slits). Hence the constructive interference at the center and the bright fringe.

Thus, as long as that center fringe exists and also as there is symmetry ofcourse between the path differences on both sides of the central bright fringe. So the other bright fringes occur in pairs (like staring from 1 wavelength path difference from each side and so on).So, the number of fringes is always $$1+2n$$ (Where 1 is for the central bright fringe). Hence an odd number of bright fringes.

• I had doubts about it cause I read this question “In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maximum of the single slit pattern? (d=1mm)” The solution was like this Distance between the slits = d = 1 mm Path difference = a sinθ ≅ aθ = λ ⇒ θ = λ /a Width of central maximum of the single slit = 2 λ/a Width of the 20 maxima = 20 x fringe spacing = 20 x λ/d Width of central maximum of the single slit = Width of the 20 maxima of double slit 2 λ/a= 20 x λ/d a = d/10 = 0.1 mm So how is that possible?
– Learnero
Nov 28 at 18:57

• What I think the question meant is there are 20 Maxima apart from the central maxima. Hence you can say there are 10 Maxima on each side of the central fringe.
– Tausif Hossain
Nov 28 at 19:00

• Thank you Tausif
– Learnero
Nov 28 at 19:03

• You’re welcome. If you think the answer is satisfactory please accept it by clicking the “tick” icon.
– Tausif Hossain
Nov 28 at 19:05

• This ans. is correct for standard arrangement. You might find it interesting that you can also shift the fringe pattern by more than you shift the envelope, by putting a thin piece of glass over the slits, slightly thicker at one slit than the other. In this way even numbers are possible (and c.f. “blazed grating”).
– Andrew Steane
Nov 28 at 19:21

Notice that if it is strictly a double slit experiment with a coherent light source then there will always be a central bright fringe where the path difference between the waves is zero(which is that, there is a point in the screen which is equidistant from the slits which lies on the perpendicular bisector of the slits). Hence the constructive interference at the center and the bright fringe.

Thus, as long as that center fringe exists and also as there is symmetry ofcourse between the path differences on both sides of the central bright fringe. So the other bright fringes occur in pairs (like staring from 1 wavelength path difference from each side and so on).So, the number of fringes is always $$1+2n$$ (Where 1 is for the central bright fringe). Hence an odd number of bright fringes.

• I had doubts about it cause I read this question “In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maximum of the single slit pattern? (d=1mm)” The solution was like this Distance between the slits = d = 1 mm Path difference = a sinθ ≅ aθ = λ ⇒ θ = λ /a Width of central maximum of the single slit = 2 λ/a Width of the 20 maxima = 20 x fringe spacing = 20 x λ/d Width of central maximum of the single slit = Width of the 20 maxima of double slit 2 λ/a= 20 x λ/d a = d/10 = 0.1 mm So how is that possible?
– Learnero
Nov 28 at 18:57

• What I think the question meant is there are 20 Maxima apart from the central maxima. Hence you can say there are 10 Maxima on each side of the central fringe.
– Tausif Hossain
Nov 28 at 19:00

• Thank you Tausif
– Learnero
Nov 28 at 19:03

• You’re welcome. If you think the answer is satisfactory please accept it by clicking the “tick” icon.
– Tausif Hossain
Nov 28 at 19:05

• This ans. is correct for standard arrangement. You might find it interesting that you can also shift the fringe pattern by more than you shift the envelope, by putting a thin piece of glass over the slits, slightly thicker at one slit than the other. In this way even numbers are possible (and c.f. “blazed grating”).
– Andrew Steane
Nov 28 at 19:21

Notice that if it is strictly a double slit experiment with a coherent light source then there will always be a central bright fringe where the path difference between the waves is zero(which is that, there is a point in the screen which is equidistant from the slits which lies on the perpendicular bisector of the slits). Hence the constructive interference at the center and the bright fringe.

Thus, as long as that center fringe exists and also as there is symmetry ofcourse between the path differences on both sides of the central bright fringe. So the other bright fringes occur in pairs (like staring from 1 wavelength path difference from each side and so on).So, the number of fringes is always $$1+2n$$ (Where 1 is for the central bright fringe). Hence an odd number of bright fringes.

Notice that if it is strictly a double slit experiment with a coherent light source then there will always be a central bright fringe where the path difference between the waves is zero(which is that, there is a point in the screen which is equidistant from the slits which lies on the perpendicular bisector of the slits). Hence the constructive interference at the center and the bright fringe.

Thus, as long as that center fringe exists and also as there is symmetry ofcourse between the path differences on both sides of the central bright fringe. So the other bright fringes occur in pairs (like staring from 1 wavelength path difference from each side and so on).So, the number of fringes is always $$1+2n$$ (Where 1 is for the central bright fringe). Hence an odd number of bright fringes.

answered Nov 28 at 18:44

Tausif Hossain

2,6062618

2,6062618

• I had doubts about it cause I read this question “In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maximum of the single slit pattern? (d=1mm)” The solution was like this Distance between the slits = d = 1 mm Path difference = a sinθ ≅ aθ = λ ⇒ θ = λ /a Width of central maximum of the single slit = 2 λ/a Width of the 20 maxima = 20 x fringe spacing = 20 x λ/d Width of central maximum of the single slit = Width of the 20 maxima of double slit 2 λ/a= 20 x λ/d a = d/10 = 0.1 mm So how is that possible?
– Learnero
Nov 28 at 18:57

• What I think the question meant is there are 20 Maxima apart from the central maxima. Hence you can say there are 10 Maxima on each side of the central fringe.
– Tausif Hossain
Nov 28 at 19:00

• Thank you Tausif
– Learnero
Nov 28 at 19:03

• You’re welcome. If you think the answer is satisfactory please accept it by clicking the “tick” icon.
– Tausif Hossain
Nov 28 at 19:05

• This ans. is correct for standard arrangement. You might find it interesting that you can also shift the fringe pattern by more than you shift the envelope, by putting a thin piece of glass over the slits, slightly thicker at one slit than the other. In this way even numbers are possible (and c.f. “blazed grating”).
– Andrew Steane
Nov 28 at 19:21

• I had doubts about it cause I read this question “In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maximum of the single slit pattern? (d=1mm)” The solution was like this Distance between the slits = d = 1 mm Path difference = a sinθ ≅ aθ = λ ⇒ θ = λ /a Width of central maximum of the single slit = 2 λ/a Width of the 20 maxima = 20 x fringe spacing = 20 x λ/d Width of central maximum of the single slit = Width of the 20 maxima of double slit 2 λ/a= 20 x λ/d a = d/10 = 0.1 mm So how is that possible?
– Learnero
Nov 28 at 18:57

• What I think the question meant is there are 20 Maxima apart from the central maxima. Hence you can say there are 10 Maxima on each side of the central fringe.
– Tausif Hossain
Nov 28 at 19:00

• Thank you Tausif
– Learnero
Nov 28 at 19:03

• You’re welcome. If you think the answer is satisfactory please accept it by clicking the “tick” icon.
– Tausif Hossain
Nov 28 at 19:05

• This ans. is correct for standard arrangement. You might find it interesting that you can also shift the fringe pattern by more than you shift the envelope, by putting a thin piece of glass over the slits, slightly thicker at one slit than the other. In this way even numbers are possible (and c.f. “blazed grating”).
– Andrew Steane
Nov 28 at 19:21

I had doubts about it cause I read this question “In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maximum of the single slit pattern? (d=1mm)” The solution was like this Distance between the slits = d = 1 mm Path difference = a sinθ ≅ aθ = λ ⇒ θ = λ /a Width of central maximum of the single slit = 2 λ/a Width of the 20 maxima = 20 x fringe spacing = 20 x λ/d Width of central maximum of the single slit = Width of the 20 maxima of double slit 2 λ/a= 20 x λ/d a = d/10 = 0.1 mm So how is that possible?
– Learnero
Nov 28 at 18:57

I had doubts about it cause I read this question “In YDSE, what should be the width of each slit to obtain 20 maxima of the double slit pattern within the central maximum of the single slit pattern? (d=1mm)” The solution was like this Distance between the slits = d = 1 mm Path difference = a sinθ ≅ aθ = λ ⇒ θ = λ /a Width of central maximum of the single slit = 2 λ/a Width of the 20 maxima = 20 x fringe spacing = 20 x λ/d Width of central maximum of the single slit = Width of the 20 maxima of double slit 2 λ/a= 20 x λ/d a = d/10 = 0.1 mm So how is that possible?
– Learnero
Nov 28 at 18:57

What I think the question meant is there are 20 Maxima apart from the central maxima. Hence you can say there are 10 Maxima on each side of the central fringe.
– Tausif Hossain
Nov 28 at 19:00

What I think the question meant is there are 20 Maxima apart from the central maxima. Hence you can say there are 10 Maxima on each side of the central fringe.
– Tausif Hossain
Nov 28 at 19:00

Thank you Tausif
– Learnero
Nov 28 at 19:03

Thank you Tausif
– Learnero
Nov 28 at 19:03

You’re welcome. If you think the answer is satisfactory please accept it by clicking the “tick” icon.
– Tausif Hossain
Nov 28 at 19:05

You’re welcome. If you think the answer is satisfactory please accept it by clicking the “tick” icon.
– Tausif Hossain
Nov 28 at 19:05

3

This ans. is correct for standard arrangement. You might find it interesting that you can also shift the fringe pattern by more than you shift the envelope, by putting a thin piece of glass over the slits, slightly thicker at one slit than the other. In this way even numbers are possible (and c.f. “blazed grating”).
– Andrew Steane
Nov 28 at 19:21

This ans. is correct for standard arrangement. You might find it interesting that you can also shift the fringe pattern by more than you shift the envelope, by putting a thin piece of glass over the slits, slightly thicker at one slit than the other. In this way even numbers are possible (and c.f. “blazed grating”).
– Andrew Steane
Nov 28 at 19:21

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