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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP

up vote
3
down vote

favorite

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

enter image description here

Source of the pic: Hyperphysics

share|cite|improve this question

    up vote
    3
    down vote

    favorite

    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

    enter image description here

    Source of the pic: Hyperphysics

    share|cite|improve this question

      up vote
      3
      down vote

      favorite

      1

      up vote
      3
      down vote

      favorite

      1
      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

      enter image description here

      Source of the pic: Hyperphysics

      share|cite|improve this question

      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

      enter image description here

      Source of the pic: Hyperphysics

      optics waves double-slit-experiment

      share|cite|improve this question

      share|cite|improve this question

      share|cite|improve this question

      share|cite|improve this question

      edited Nov 28 at 22:32

      knzhou

      39.1k9109189

      39.1k9109189

      asked Nov 28 at 18:20

      Learnero

      232

      232

          3 Answers
          3

          active

          oldest

          votes

          up vote
          5
          down vote

          accepted

          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.)

          share|cite|improve this answer

            up vote
            10
            down vote

            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.

            share|cite|improve this answer

              up vote
              5
              down vote

              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.

              share|cite|improve this answer

              • 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

              • 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

              Your Answer

              StackExchange.ifUsing(“editor”, function () {
              return StackExchange.using(“mathjaxEditing”, function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [[“$”, “$”], [“\\(“,”\\)”]]);
              });
              });
              }, “mathjax-editing”);

              StackExchange.ready(function() {
              var channelOptions = {
              tags: “”.split(” “),
              id: “151”
              };
              initTagRenderer(“”.split(” “), “”.split(” “), channelOptions);

              StackExchange.using(“externalEditor”, function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using(“snippets”, function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: ‘answer’,
              convertImagesToLinks: false,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: “”,
              imageUploader: {
              brandingHtml: “Powered by u003ca class=”icon-imgur-white” href=”https://imgur.com/”u003eu003c/au003e”,
              contentPolicyHtml: “User contributions licensed under u003ca href=”https://creativecommons.org/licenses/by-sa/3.0/”u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href=”https://stackoverflow.com/legal/content-policy”u003e(content policy)u003c/au003e”,
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: “.discard-answer”
              ,immediatelyShowMarkdownHelp:true
              });

              }
              });

              draft saved
              draft discarded

              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin(‘.new-post-login’, ‘https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f443892%2fin-youngs-double-slit-experiment-can-there-be-an-even-number-of-maxima-in-the%23new-answer’, ‘question_page’);
              }
              );

              Post as a guest

              Required, but never shown

              3 Answers
              3

              active

              oldest

              votes

              3 Answers
              3

              active

              oldest

              votes

              active

              oldest

              votes

              active

              oldest

              votes

              up vote
              5
              down vote

              accepted

              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.)

              share|cite|improve this answer

                up vote
                5
                down vote

                accepted

                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.)

                share|cite|improve this answer

                  up vote
                  5
                  down vote

                  accepted

                  up vote
                  5
                  down vote

                  accepted

                  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.)

                  share|cite|improve this answer

                  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.)

                  share|cite|improve this answer

                  share|cite|improve this answer

                  share|cite|improve this answer

                  edited Nov 28 at 20:05

                  answered Nov 28 at 19:57

                  fs137

                  2,540815

                  2,540815

                      up vote
                      10
                      down vote

                      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.

                      share|cite|improve this answer

                        up vote
                        10
                        down vote

                        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.

                        share|cite|improve this answer

                          up vote
                          10
                          down vote

                          up vote
                          10
                          down vote

                          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.

                          share|cite|improve this answer

                          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.

                          share|cite|improve this answer

                          share|cite|improve this answer

                          share|cite|improve this answer

                          answered Nov 28 at 19:41

                          Pieter

                          7,45231431

                          7,45231431

                              up vote
                              5
                              down vote

                              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.

                              share|cite|improve this answer

                              • 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

                              • 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

                              up vote
                              5
                              down vote

                              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.

                              share|cite|improve this answer

                              • 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

                              • 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

                              up vote
                              5
                              down vote

                              up vote
                              5
                              down vote

                              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.

                              share|cite|improve this answer

                              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.

                              share|cite|improve this answer

                              share|cite|improve this answer

                              share|cite|improve this answer

                              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

                              • 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

                              • 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

                              • 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

                              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

                              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

                              draft saved
                              draft discarded

                              Thanks for contributing an answer to Physics Stack Exchange!

                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid

                              • Asking for help, clarification, or responding to other answers.
                              • Making statements based on opinion; back them up with references or personal experience.

                              Use MathJax to format equations. MathJax reference.

                              To learn more, see our tips on writing great answers.

                              Some of your past answers have not been well-received, and you’re in danger of being blocked from answering.

                              Please pay close attention to the following guidance:

                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid

                              • Asking for help, clarification, or responding to other answers.
                              • Making statements based on opinion; back them up with references or personal experience.

                              To learn more, see our tips on writing great answers.

                              draft saved

                              draft discarded

                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin(‘.new-post-login’, ‘https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f443892%2fin-youngs-double-slit-experiment-can-there-be-an-even-number-of-maxima-in-the%23new-answer’, ‘question_page’);
                              }
                              );

                              Post as a guest

                              Required, but never shown

                              Required, but never shown

                              Required, but never shown

                              Required, but never shown

                              Required, but never shown

                              Required, but never shown

                              Required, but never shown

                              Required, but never shown

                              Required, but never shown

                              Related Post

                              Leave a Reply

                              Your email address will not be published. Required fields are marked *