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    The light from a `special LED' consists of two colours of light with wavelengths of 530nm and 630nm respectively. The light is shone through a diffraction grating with 500lines/mm, and the two colours need to be separated by at least 5.0∘. What is the minimum order of interference needed in order to do this?

    How do I do this one?
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    For the diffraction grating (not youngs slits) you ahve an equation

    nλ = d sin theta

    The n has to be the same for both, d is the diffraction grating spacing and you have two wavelengths.

    At the moment I can't see any other way to do it that start from n=1 (n=0 will give angle zero for both) and solve for theta until you find 5º of difference for the two wavelengths. (ie, start with n=1 and if the difference isn't enough, try n=2 and so on)

    Someone else more inspired might come up with a shortcut.
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    (Original post by joyoustele)
    The light from a `special LED' consists of two colours of light with wavelengths of 530nm and 630nm respectively. The light is shone through a diffraction grating with 500lines/mm, and the two colours need to be separated by at least 5.0∘. What is the minimum order of interference needed in order to do this?
    How do I do this one?

    (Original post by phys981)
    For the diffraction grating (not youngs slits) you have an equation

    nλ = d sin θ

    The n has to be the same for both, d is the diffraction grating spacing and you have two wavelengths.

    At the moment I can't see any other way to do it that start from n=1 (n=0 will give angle zero for both) and solve for theta until you find 5º of difference for the two wavelengths. (ie, start with n=1 and if the difference isn't enough, try n=2 and so on)

    Someone else more inspired might come up with a shortcut.

    I would not say the following is a shortcut but some learning moments for the students after they have successfully answered the question.

    For such question, I would usually have some extension questions or discussions depending on the level of students.

    If this is the first time that the students are attempting such question, I would get them to use the brute force method outlined by phys981.

    Then I would have a discussion using a spreadsheet calculation based on the brute force method or guiding questions to explore for a “shortcut”.

    Let θSn and θLn be the angles for the wavelength of 530nm and 630nm respectively based on the diffraction grating formula (d sin θ = nλ)

    Compute the ratio λ /d for the wavelength of 530nm and 630nm respectively and then find the difference (which is 0.05).


    Using the brute force method, the following can be “deduced” for "small angle":



     \theta_{Ln} - \theta_{Sn} \approx \dfrac{n \lambda_{Ln}}{d} - \dfrac{n \lambda_{Sn}}{d}


    Note that the angle in radians NOT degree.

    So for 5°, convert it to radian which is about 0.087 rad. Substitute 0.087 rad to the above approximated equation to solve for n.






    Spoiler:
    Show



     \theta_{Ln} - \theta_{Sn} \approx n \left( \dfrac{\lambda_{Ln}}{d} - \dfrac{\lambda_{Sn}}{d} \right)

    0.087 ≈ n(0.05)

    n ≈ 1.7


    So the order is 2.





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    (Original post by Eimmanuel)
    I would not say the following is a shortcut but some learning moments for the students after they have successfully answered the question.

    For such question, I would usually have some extension questions or discussions depending on the level of students.

    If this is the first time that the students are attempting such question, I would get them to use the brute force method outlined by phys981.

    Then I would have a discussion using a spreadsheet calculation based on the brute force method or guiding questions to explore for a “shortcut”.

    Let θSn and θLn be the angles for the wavelength of 530nm and 630nm respectively based on the diffraction grating formula (d sin θ = nλ)

    Compute the ratio λ /d for the wavelength of 530nm and 630nm respectively and then find the difference (which is 0.05).


    Using the brute force method, the following can be “deduced” for "small angle":




     \theta_{Ln} - \theta_{Sn} \approx \dfrac{n \lambda_{Ln}}{d} - \dfrac{n \lambda_{Sn}}{d}



    Note that the angle in radians NOT degree.

    So for 5°, convert it to radian which is about 0.087 rad. Substitute 0.087 rad to the above approximated equation to solve for n.








    Spoiler:
    Show




     \theta_{Ln} - \theta_{Sn} \approx n \left( \dfrac{\lambda_{Ln}}{d} - \dfrac{\lambda_{Sn}}{d} \right)

    0.087 ≈ n(0.05)

    n ≈ 1.7



    So the order is 2.






    I just didnt understand the question well. Thanks a lot, Very much appreciate it.
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    (Original post by joyoustele)
    I just didnt understand the question well. Thanks a lot, Very much appreciate it.
    I hope you understand it now.
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    (Original post by Eimmanuel)
    I hope you understand it now.
    I do, it was very simple. Thank you very very much.
 
 
 
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