Question about the derivation of the intensity formula of a diffraction grating












4












$begingroup$


In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:



$$I = I_0
left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
left( frac{sin(kasintheta)}{kasintheta} right)^2
$$



They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?






waves diffraction







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$endgroup$

















    4












    $begingroup$


    In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:



    $$I = I_0
    left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
    left( frac{sin(kasintheta)}{kasintheta} right)^2
    $$



    They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?






    waves diffraction







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:



      $$I = I_0
      left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
      left( frac{sin(kasintheta)}{kasintheta} right)^2
      $$



      They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?






      waves diffraction







      share|cite|improve this question











      $endgroup$




      In the notes I have, they have a diffraction grating with $2N + 1$ slits, a slit width of $2a$ and a slit spacing of $d$. They then say that the equation for the diffraction intensity pattern is given by:



      $$I = I_0
      left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)^2
      left( frac{sin(kasintheta)}{kasintheta} right)^2
      $$



      They don't, however, give any proof or reason why this is the formula. I have been looking online for a way to justify this formula but I cant find anything. Anybody have a nice proof for this?






      waves diffraction




      waves diffraction






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 13 hours ago









      Thomas Fritsch

      1,546515




      1,546515










      asked 15 hours ago









      A. PavlenkoA. Pavlenko

      344




      344






















          1 Answer
          1






          active

          oldest

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          5












          $begingroup$

          diffraction grating

          (image from Antonine education)



          The light amplitude $E(theta)$ into direction $theta$ can be calculated
          straight-forward by summing the contributions




          • of all the slits ($n$ from $-N$ to $+N$)

          • and of the parts of each individual slit ($x$ from $-a$ to $+a$)


          The path difference of each contributing ray
          (compared to the path length of the ray originating from the center of the grating)
          is $(nd+x)sintheta$.
          And hence its phase is $k(nd+x)sintheta$.



          Summing these contributions you get
          $$
          begin{align}
          E(theta)
          &= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
          &= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
          left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
          &= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
          left( 2afrac{sin(kasintheta)}{kasintheta} right)
          end{align}
          $$



          And finally you get the intensity by taking the absolute square of the amplitude
          $$I(theta) = |E(theta)|^2$$






          share|cite|improve this answer











          $endgroup$














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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            diffraction grating

            (image from Antonine education)



            The light amplitude $E(theta)$ into direction $theta$ can be calculated
            straight-forward by summing the contributions




            • of all the slits ($n$ from $-N$ to $+N$)

            • and of the parts of each individual slit ($x$ from $-a$ to $+a$)


            The path difference of each contributing ray
            (compared to the path length of the ray originating from the center of the grating)
            is $(nd+x)sintheta$.
            And hence its phase is $k(nd+x)sintheta$.



            Summing these contributions you get
            $$
            begin{align}
            E(theta)
            &= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
            &= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
            left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
            &= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
            left( 2afrac{sin(kasintheta)}{kasintheta} right)
            end{align}
            $$



            And finally you get the intensity by taking the absolute square of the amplitude
            $$I(theta) = |E(theta)|^2$$






            share|cite|improve this answer











            $endgroup$


















              5












              $begingroup$

              diffraction grating

              (image from Antonine education)



              The light amplitude $E(theta)$ into direction $theta$ can be calculated
              straight-forward by summing the contributions




              • of all the slits ($n$ from $-N$ to $+N$)

              • and of the parts of each individual slit ($x$ from $-a$ to $+a$)


              The path difference of each contributing ray
              (compared to the path length of the ray originating from the center of the grating)
              is $(nd+x)sintheta$.
              And hence its phase is $k(nd+x)sintheta$.



              Summing these contributions you get
              $$
              begin{align}
              E(theta)
              &= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
              &= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
              left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
              &= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
              left( 2afrac{sin(kasintheta)}{kasintheta} right)
              end{align}
              $$



              And finally you get the intensity by taking the absolute square of the amplitude
              $$I(theta) = |E(theta)|^2$$






              share|cite|improve this answer











              $endgroup$
















                5












                5








                5





                $begingroup$

                diffraction grating

                (image from Antonine education)



                The light amplitude $E(theta)$ into direction $theta$ can be calculated
                straight-forward by summing the contributions




                • of all the slits ($n$ from $-N$ to $+N$)

                • and of the parts of each individual slit ($x$ from $-a$ to $+a$)


                The path difference of each contributing ray
                (compared to the path length of the ray originating from the center of the grating)
                is $(nd+x)sintheta$.
                And hence its phase is $k(nd+x)sintheta$.



                Summing these contributions you get
                $$
                begin{align}
                E(theta)
                &= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
                &= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
                left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
                &= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
                left( 2afrac{sin(kasintheta)}{kasintheta} right)
                end{align}
                $$



                And finally you get the intensity by taking the absolute square of the amplitude
                $$I(theta) = |E(theta)|^2$$






                share|cite|improve this answer











                $endgroup$



                diffraction grating

                (image from Antonine education)



                The light amplitude $E(theta)$ into direction $theta$ can be calculated
                straight-forward by summing the contributions




                • of all the slits ($n$ from $-N$ to $+N$)

                • and of the parts of each individual slit ($x$ from $-a$ to $+a$)


                The path difference of each contributing ray
                (compared to the path length of the ray originating from the center of the grating)
                is $(nd+x)sintheta$.
                And hence its phase is $k(nd+x)sintheta$.



                Summing these contributions you get
                $$
                begin{align}
                E(theta)
                &= E_0 sum_{n=-N}^{+N} int_{-a}^{+a} e^{ik(nd+x)sintheta} text{d}x \
                &= E_0 left( sum_{n=-N}^{+N} e^{ikndsintheta}right)
                left( int_{-a}^{+a} e^{ikxsintheta} text{d}x right) \
                &= E_0 left( frac{sin((N+frac{1}{2})kdsintheta)}{sin(frac{1}{2}kdsintheta)} right)
                left( 2afrac{sin(kasintheta)}{kasintheta} right)
                end{align}
                $$



                And finally you get the intensity by taking the absolute square of the amplitude
                $$I(theta) = |E(theta)|^2$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 9 hours ago

























                answered 14 hours ago









                Thomas FritschThomas Fritsch

                1,546515




                1,546515






























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