Does the inverse-square law apply to linearly polarized light?












3












$begingroup$


It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?






polarization







share|cite|improve this question











$endgroup$












  • $begingroup$
    What exactly is M in your question ?
    $endgroup$
    – Lelouch
    Dec 30 '18 at 7:34
















3












$begingroup$


It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?






polarization







share|cite|improve this question











$endgroup$












  • $begingroup$
    What exactly is M in your question ?
    $endgroup$
    – Lelouch
    Dec 30 '18 at 7:34














3












3








3





$begingroup$


It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?






polarization







share|cite|improve this question











$endgroup$




It's a stupid question but: We did and experiment using linearly polarized microwave radiation generators and receivers. Our teacher asked to check experimentally if the receiver measurements are proportional to the intensity of the radiation I or to the intensity of the electric field $E$. To check that we took the measurements M of the receiver to different distances R between the receiver and the generator. She said that if the receiver goes with I the diagram of $M=f(frac{1}{R^2})$ will fit better to linear fitting than the $M=f(frac{1}{R})$ diagram. If the reverse happens the receiver goes with $E$. So that's why I am confused. Shouldn't the $frac{1}{R^2}$ relationship indicate that the receiver's measurements are proportional to $E$?






polarization




polarization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 9:41









Wasserwaage

3591416




3591416










asked Dec 30 '18 at 2:14









JimmyNikJimmyNik

183




183












  • $begingroup$
    What exactly is M in your question ?
    $endgroup$
    – Lelouch
    Dec 30 '18 at 7:34


















  • $begingroup$
    What exactly is M in your question ?
    $endgroup$
    – Lelouch
    Dec 30 '18 at 7:34
















$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34




$begingroup$
What exactly is M in your question ?
$endgroup$
– Lelouch
Dec 30 '18 at 7:34










2 Answers
2






active

oldest

votes


















1












$begingroup$

Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.



A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:



$$P=4pi r^2 I(r).$$



And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by



$$I(r)=frac{P}{4pi}frac{1}{r^2}.$$



Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as



$$I=kappa|phi|^2,$$



With $kappa$ being a constant. As such, we can then say that



$$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$



Which falls off as $1/r$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).



    If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok I think I got it but does it have to do with linear polarization?
      $endgroup$
      – JimmyNik
      Dec 30 '18 at 14:08










    • $begingroup$
      @JimmyNik No, these properties are the same whatever the polarization state.
      $endgroup$
      – Buzz
      Dec 30 '18 at 14:21











    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',
    autoActivateHeartbeat: false,
    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%2f451100%2fdoes-the-inverse-square-law-apply-to-linearly-polarized-light%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.



    A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:



    $$P=4pi r^2 I(r).$$



    And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by



    $$I(r)=frac{P}{4pi}frac{1}{r^2}.$$



    Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as



    $$I=kappa|phi|^2,$$



    With $kappa$ being a constant. As such, we can then say that



    $$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$



    Which falls off as $1/r$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.



      A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:



      $$P=4pi r^2 I(r).$$



      And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by



      $$I(r)=frac{P}{4pi}frac{1}{r^2}.$$



      Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as



      $$I=kappa|phi|^2,$$



      With $kappa$ being a constant. As such, we can then say that



      $$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$



      Which falls off as $1/r$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.



        A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:



        $$P=4pi r^2 I(r).$$



        And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by



        $$I(r)=frac{P}{4pi}frac{1}{r^2}.$$



        Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as



        $$I=kappa|phi|^2,$$



        With $kappa$ being a constant. As such, we can then say that



        $$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$



        Which falls off as $1/r$.






        share|cite|improve this answer











        $endgroup$



        Nope. Point sources of waves have fields falling off as $1/r$ while intensities fall off as $1/r^2$. To see why, think about conservation of energy.



        A point source is spherically symmetric, and so the intensity should only be a function of the distance $I(r)$. Intensity is defined as the area flux density of energy due to the wave. Consider a sphere of radius $r$ around the point source. The total power going through this sphere should be the product of the area and intensity:



        $$P=4pi r^2 I(r).$$



        And since the power leaving the sphere should be the same at any distance (all the energy has to leave the sphere in order to be conserved) then the intensity is given by



        $$I(r)=frac{P}{4pi}frac{1}{r^2}.$$



        Which clearly falls off as $1/r^2$. To see why the field (electric field, in your case) goes as $1/r$, remember that intensity and field $phi$ are related as



        $$I=kappa|phi|^2,$$



        With $kappa$ being a constant. As such, we can then say that



        $$|phi|=sqrtfrac{P}{4pikappa}frac{1}{r},$$



        Which falls off as $1/r$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 4:15

























        answered Dec 30 '18 at 3:53









        Gabriel GolfettiGabriel Golfetti

        1,3221713




        1,3221713























            1












            $begingroup$

            Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).



            If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ok I think I got it but does it have to do with linear polarization?
              $endgroup$
              – JimmyNik
              Dec 30 '18 at 14:08










            • $begingroup$
              @JimmyNik No, these properties are the same whatever the polarization state.
              $endgroup$
              – Buzz
              Dec 30 '18 at 14:21
















            1












            $begingroup$

            Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).



            If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ok I think I got it but does it have to do with linear polarization?
              $endgroup$
              – JimmyNik
              Dec 30 '18 at 14:08










            • $begingroup$
              @JimmyNik No, these properties are the same whatever the polarization state.
              $endgroup$
              – Buzz
              Dec 30 '18 at 14:21














            1












            1








            1





            $begingroup$

            Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).



            If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.






            share|cite|improve this answer









            $endgroup$



            Static electric fields fall off as $1/R^{2}$, but the ${bf E}$ field associated with a spherical wave does not. The electric field of a radiating spherical wave falls off as $1/R$. In finding the solutions of problems with accelerating charges, the difference in behavior can often be used to separate space into a "quasi-static zone" (close to the sources where the ${bf E}$ field falls off as $1/R^{2}$) and a "radiation zone" (farther away, where ${bf E}$ falls as $1/R$).



            If you are measuring a wave's intensity as a function of distance from the source, your should observe the $1/R^{2}$ behavior. The reasons for this is that the amount of energy carried by an outgoing wave is proportional to ${bf E}^{2}$, and thus falls off as $1/R^{2}$ even though ${bf E}$ itself goes down as $1/R$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 30 '18 at 3:48









            BuzzBuzz

            3,54431726




            3,54431726












            • $begingroup$
              Ok I think I got it but does it have to do with linear polarization?
              $endgroup$
              – JimmyNik
              Dec 30 '18 at 14:08










            • $begingroup$
              @JimmyNik No, these properties are the same whatever the polarization state.
              $endgroup$
              – Buzz
              Dec 30 '18 at 14:21


















            • $begingroup$
              Ok I think I got it but does it have to do with linear polarization?
              $endgroup$
              – JimmyNik
              Dec 30 '18 at 14:08










            • $begingroup$
              @JimmyNik No, these properties are the same whatever the polarization state.
              $endgroup$
              – Buzz
              Dec 30 '18 at 14:21
















            $begingroup$
            Ok I think I got it but does it have to do with linear polarization?
            $endgroup$
            – JimmyNik
            Dec 30 '18 at 14:08




            $begingroup$
            Ok I think I got it but does it have to do with linear polarization?
            $endgroup$
            – JimmyNik
            Dec 30 '18 at 14:08












            $begingroup$
            @JimmyNik No, these properties are the same whatever the polarization state.
            $endgroup$
            – Buzz
            Dec 30 '18 at 14:21




            $begingroup$
            @JimmyNik No, these properties are the same whatever the polarization state.
            $endgroup$
            – Buzz
            Dec 30 '18 at 14:21


















            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f451100%2fdoes-the-inverse-square-law-apply-to-linearly-polarized-light%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







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more