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PLZ help: diffraction off a CD ROM question ... Watch

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    CD ROM is what they used before digital download kids! :P

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    I got the first question part right and it's approx 720µ metres for the spacing ... okay ... so I just did pretty much the same equation, but using slightly different angles.

    I think I did something like 720µm * sin(60) = lamda, and 60 is because it is 55 (previous angle which is right)+5 ... which felt wrong and I wasn't sure why ... & it turns out IT IS wrong anyway,

    So how do I do it? Any pointers please?
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    (Original post by DrSebWilkes)
    CD ROM is what they used before digital download kids! :P

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Views: 6
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    I got the first question part right and it's approx 720µ metres for the spacing ... okay ... so I just did pretty much the same equation, but using slightly different angles.

    I think I did something like 720µm * sin(60) = lamda, and 60 is because it is 55 (previous angle which is right)+5 ... which felt wrong and I wasn't sure why ... & it turns out IT IS wrong anyway,

    So how do I do it? Any pointers please?
    Diffraction grating spacing:

    dsin(\theta) = n\lambda

    first order refraction angle (n=1)

    \theta = 90 - tan^{-1}(\frac{0.35}{0.5})

    \theta = 55^{o}

    d = \frac{(590 \textrm {x10}^{-9})}{sin(55)} = 720x10-9 m

    d = 0.720 \mum
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    If the CD rom is tilted through 5 degrees then would the grating element ‘d’ (value in part one) itself become ‘dcos(5)’? Then you could just use the formula as before?
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    (Original post by uberteknik)
    snip got the first 4 marks no problem
    (Original post by physcamy)
    If the CD rom is tilted through 5 degrees then would the grating element ‘d’ (value in part one) itself become ‘dcos(5)’? Then you could just use the formula as before?
    Hmm I think I see where you are coming from; the effective "area" (if you like) goes down with rotation ... seems kinda obvious when you think about it.

    Here is the markscheme answer but I honestly don't understand it.
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    ANY IDEAS ANYONE? plz
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    (Original post by DrSebWilkes)
    Hmm I think I see where you are coming from; the effective "area" (if you like) goes down with rotation ... seems kinda obvious when you think about it.

    Here is the markscheme answer but I honestly don't understand it.
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Views: 6
Size:  97.9 KB

    ANY IDEAS ANYONE? plz
    It just about makes sense but it’s nothing like what I suggested - probably because I was naive in making the assumption that the ‘slits’ are always perpendicular to the incident light rays, which clearly isn’t the case. Essentially they’re finding the path difference and are saying that whichever wavelength fits into that path difference exactly must be the wavelength observed. It’s a nice question but let’s hope it doesn’t appear on the PAT tomorrow 😉
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