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In a double slit experiment, if monochromatic light is replaced by white light what effect does it have on the fringes. Answer given is below. Can someone please explain the answer.?

  1. Central fringe will be white
  2. Fringe closest on either side of the central white fringe is red and farthest will appear blue. After few fringes no clear fringe pattern is seen.
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I would describe the colors closest to the white central maximum as yellow (absence of blue) and magenta (absence of green). There is no "farthest" fringe, but next one sees a band of cyan (absence of red). These are complementary colors of the primary ones.

It has to do with the pigments in our eyes. Cameras and displays try to match these with their RGB channels. Here is an image I made, using a camera without a lens, using a $20\ \mu$m slit in the lens cap. I also decomposed the slit image in the separate color channels:

enter image description here https://commons.wikimedia.org/wiki/File:Diffraction_sunlight_-_color_channels.jpg

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    $\begingroup$ Can you comment in a bit more detail on how you took this picture? $\endgroup$ Aug 8 '18 at 10:29
  • $\begingroup$ @EmilioPisanty I added some information. A DSLR (digital single-lense reflex) camera, without a lens is perfect for optics experiments. $\endgroup$
    – user137289
    Aug 8 '18 at 10:36
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Its actually because of the fact that white light consists of seven colours and each of its components has its own wavelength due to difference in wavelength each components has different fringe width maximum interference for one may not be same for other except at the center where every component has max interference which again results to the formation of white light

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    $\begingroup$ White light doesn't consist of seven colours - it has a continuous spectrum. $\endgroup$ Aug 8 '18 at 10:31
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Well, the closest bright fringe next to white will be red. When you compare the wavelengths for visible spectrum, violet has the least, and you would expect violet, but that is not the truth, reddish/yellowish appearance is observed. Actually, here we need to find the closest destructive interference, or closest minima. As violet's maxima is closest, hence theoretically, violets minima is also closest to the white fringe. At violets minima, reddish/yellowish color will be observed (opposite color in the primary color schema). Hence the nearest fringe will be red in color. Central fringe will be white because all the wave fronts in YDSE, start from the central maxima, and then to First order, second order and so on. Hence when all the waves coincide at central maxima, white color is observed(as all those waves are counterparts of white only, achromatic source is used)

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If white light is used in the double-slit experiment, the different colours will be split up on the viewing screen according to their wavelengths. The violet end of the spectrum (with the shortest wavelengths) is closer to the central fringe, with the other colours being further away in order. For any component of white light having wavelength $λ$, the y-coordinate of nth maxima on the screen $= nλD/d$ (where $D$ = Distance between the screen and slits, $d$ = distance between the slits). Hence, for the 1st maxima, $y_1 = λD/d$. It's easy to see that y1 will be greater for greater wavelengths i.e. the colour having smallest wavelength will have its first maxima closest to the central bright $\Longrightarrow$ the fringe next to the white fringe will be violet.

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  • $\begingroup$ This is the wrong answer. As I showed above, the question did not describe the colours very well. $\endgroup$
    – user137289
    Dec 26 '19 at 18:34

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