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Why does diffracted white light have a white central fringe?

  1. Sep 12, 2016 #1
    Hey everybody,

    1. The problem statement, all variables and given/known data:

    I came across a question discussing Young's Double Slit interference and was wondering how come when we diffract white light through the slits it gives us a central antinodal fringe that is also white?

    The question itself came in parts firstly considering monochromatic light diffracting and interfering through double slits. The section I couldn't answer states "The monochromatic illumination is replaced by sunlight. Explain how this will assist the experimenter to determine the position of the new central maximum bright fringe"

    2. Relevant equations:

    The small angle approximation and the diffraction grating formula

    3. My attempt:

    Obviously when you do the experiment with monochromatic light all the fringes will be the same colour however, my understanding is that white light actually consists of all the different frequencies of visible light and hence these diffract at different angles to produce a spectra of colour. But when I googled an image for the interference pattern of white light through a diffraction grating all the pictures show the central fringe is white flanked by overlapping coloured fringes (Which must answer the physical question stated in the book about how the experimenter would be able to determine the position of the central maximum - by spotting the white fringe). But the answer doesn't go on to explain why this fringe occurs and I can't understand or find a reason or an equation that explains why this central fringe isn't a spectrum of colours as well?

    I assume this must have something to do with path difference of the light waves. At the central antinode there is no path difference so does that mean that the waves of the different frequencies add to give white light. That would be my guess but the answer in the book is very vague and doesn't confirm this at all.
    Last edited by a moderator: Sep 13, 2016
  2. jcsd
  3. Sep 13, 2016 #2


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    Hello Gigi777. Welcome to PF !

    As the moderators will tell you,
    1. You should use the template which is provided. Don't erase it.
    2. You need to make an attempt at a solution.
    You need to understand how Young's Double Slit works for monochromatic light before you can understand the white light case.
  4. Sep 13, 2016 #3
    Hi Sam,

    Thanks for the reply I'll be sure to keep that in mind for next time - Should I go edit the post to follow the template?
    Also I do believe I understand how it works for light in that the bright coloured fringes occur when the diffracted light meets in phase to constructively interfere and produce an antinode (seen as the bright light) and vice versa for the dark fringes. My attempt at a solution (since this is a question where you have to explain the situation) was I researching what occurs when white light diffracts - hence I learnt about the spectra of light produced but I don't understand why this doesn't apply for the central fringe.
  5. Sep 13, 2016 #4


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    I think this may satisfy the mentors regarding an attempt to at least understand the problem. As a newbie, probably your apology and willingness to edit your post to include the template will satisfy the Mentors. They are very reasonable.

    As for the stated problem:

    For monochromatic light: Do you understand how the spacing of the bright fringes depends upon the wavelength of the light ?
  6. Sep 13, 2016 #5
    I hope so! Went and edited it so hopefully it follows the required template now. I'm honestly not to sure I do understand the relationship between those two - is it that a greater wavelength leads to greater diffraction and thus more spacing? Perhaps I'm completely off but that would be my guess :&
  7. Sep 13, 2016 #6


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    The central spot is a maximum for all colors at the same time - and all colors on top of each other give white light if your source is white light.
  8. Sep 13, 2016 #7
    Ah should have known it was as simple as that - thank you so much for the help! :)
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