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    Like how am i supposed to know if a certain colour has a greater wavelength than another colour, like wavelength of red > wavelength of white?
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    (Original post by Ayaz789)
    Like how am i supposed to know if a certain colour has a greater wavelength than another colour, like wavelength of red > wavelength of white?
    First of all, white light doesn't have a wavelength, it's a combination of all the wavelengths of visible light. Secondly, you just have to learn it. Memorise this diagram:
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    (Original post by Plagioclase)
    First of all, white light doesn't have a wavelength, it's a combination of all the wavelengths of visible light. Secondly, you just have to learn it. Memorise this diagram:
    Okay ill learn that diagram thanks!
    So therefore according to that the wavelength of blue> wavelength of red
    & the wavelength of any colour> wavelength of white?
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    (Original post by Ayaz789)
    Okay ill learn that diagram thanks!
    So therefore according to that the wavelength of blue> wavelength of red
    & the wavelength of any colour> wavelength of white?
    No the wavelength of red (~700nm) is longer than the wavelength of blue (~450nm). White light doesn't have a wavelength, the whole point is that it's a combination of different wavelengths of light. It makes no sense to say that something has a longer wavelength than white light because white light doesn't have a wavelength, it's a continuum of wavelengths.
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    (Original post by Ayaz789)
    Okay ill learn that diagram thanks!
    So therefore according to that the wavelength of blue> wavelength of red
    & the wavelength of any colour> wavelength of white?
    Small wavelengths are at the top of the diagram. An angstrom is smaller than a kilometre.
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    (Original post by Plagioclase)
    No the wavelength of red (~700nm) is longer than the wavelength of blue (~450nm). White light doesn't have a wavelength, the whole point is that it's a combination of different wavelengths of light. It makes no sense to say that something has a longer wavelength than white light because white light doesn't have a wavelength, it's a continuum of wavelengths.
    http://www.egsphysics.co.uk/files/a_...W-MS-JAN09.PDF
    So why in this ms does it say that in q7b?
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    (Original post by Ayaz789)
    http://www.egsphysics.co.uk/files/a_...W-MS-JAN09.PDF
    So why in this ms does it say that in q7b?
    That's the average wavelength of the white light. If the mark scheme was talking about the wavelength of the white light then it wouldn't make sense because there are infinitely many different wavelengths for the white light.
    White light has a continuous distribution of wavelengths, but you can still take the average of them(in every case as far as I know). If you've studied calculus then if the allowed values of \lambda in the white light are in I then \bar{       \lambda} = \dfrac{\displaystyle\int_I \lambda P(\lambda)\mathrm{d}\lambda}{        \displaystyle\int_I P(\lambda)\mathrm{d}\lambda}. If not then imagine taking the mean of a finite number of (weighted) values then extending this to infinitely many.
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    (Original post by morgan8002)
    That's the average wavelength of the white light. If the mark scheme was talking about the wavelength of the white light then it wouldn't make sense because there are infinitely many different wavelengths for the white light.
    White light has a continuous distribution of wavelengths, but you can still take the average of them(in every case as far as I know). If you've studied calculus then if the allowed values of \lambda in the white light are in I then \bar{       \lambda} = \dfrac{\displaystyle\int_I \lambda P(\lambda)\mathrm{d}\lambda}{        \displaystyle\int_I P(\lambda)\mathrm{d}\lambda}. If not then imagine taking the mean of a finite number of (weighted) values then extending this to infinitely many.
    I have studied calculus but that does look difficult haha Okay so what is the wavelength of average white?
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    (Original post by Ayaz789)
    http://www.egsphysics.co.uk/files/a_...W-MS-JAN09.PDF
    So why in this ms does it say that in q7b?
    It says average... red is at one of the extreme ends of the visible spectrum the middle of a fringe made with white light would be in a different place to the middle of a fringe made with red light.

    fwiw A couple of mnemonics for the order of colours by decreasing wavelength

    Richard Of York Gave Battle In Vain
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    (Original post by Joinedup)
    It says average... red is at one of the extreme ends of the visible spectrum the middle of a fringe made with white light would be in a different place to the middle of a fringe made with red light.

    fwiw A couple of mnemonics for the order of colours by decreasing wavelength

    Richard Of York Gave Battle In Vain
    Rock On You Great Big Indian Vole
    Okay thanks ill remember them!
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    (Original post by Joinedup)
    It says average... red is at one of the extreme ends of the visible spectrum the middle of a fringe made with white light would be in a different place to the middle of a fringe made with red light.

    fwiw A couple of mnemonics for the order of colours by decreasing wavelength

    Richard Of York Gave Battle In Vain
    Rock On You Great Big Indian Vole
    By decreasing wavelength you mean wavelength of red is greater than wavelength of blue?
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    (Original post by Ayaz789)
    I have studied calculus but that does look difficult haha Okay so what is the wavelength of average white?
    It depends on the specific distribution of the light and is given by the equation I wrote above. It'll be somewhere below red and above blue.
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    (Original post by morgan8002)
    It depends on the specific distribution of the light and is given by the equation I wrote above. It'll be somewhere below red and above blue.
    Okay np
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    (Original post by Ayaz789)
    By decreasing wavelength you mean wavelength of red is greater than wavelength of blue?
    yes



    (Original post by Ayaz789)
    Okay so what is the wavelength of average white?
    I'll stick my neck out and say it's about 540 or 550 nm for practical purposes... which is about the colour of a green LED
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    (Original post by Joinedup)
    yes





    I'll stick my neck out and say it's about 540 or 550 nm for practical purposes... which is about the colour of a green LED
    Thanks haha
 
 
 
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