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Hair in between two glass plates. Decreasing the distance between the maxima

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  1. Oct 2, 2014 #1
    1. The problem statement, all variables and given/known data
    wfwXk.jpg
    A thin hair is placed between two microscope slides. When laser light is shined down onto the slides, periodic intensity maxima are seen to span the slides. Which of the following changes will decrease the distance between the maxima?

    I. Decrease the wavelength of the laser light.
    II. Fill the air gap between the slides with carbon disulfide (n=1.6)
    III. Move the hair to the left, while keeping the slides in contact with each other at their left edges.

    The correct answer is I and III.


    2. Relevant equations

    2t = (m+1/2) * (wavelength / n) for constructive interference.

    3. The attempt at a solution

    i'm really unsure as to how to approach this problem. I initially saw it a a diffraction problem, in which a phase change occurs at the first interface because the glass has a higher n than the air. Then in the second interface because you're going from air to glass

    Then there is another reflection as you go from glass to air but there is no phase change in this reflection.

    Finally, in the third transition, you go from air to glass so there is a phase change.

    However that does not tell me what i qualitatively need to know to answer this question. The answer key in the book I am using classifies this as a double slit diffraction problem to explain the answer.

    Decreasing the wavelength of light would decrease the distance between maxima. Apparently this is because decreasing the wavelength would create more wavelets which would thus decrease the space between maxima. What I dont understand is the connection between wavelength and # of wavelts and distance between maxima.
     
  2. jcsd
  3. Oct 3, 2014 #2

    haruspex

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    You have two reflections which interfere, one from each glass slide. The interference arises from the difference in path length, which will be twice the distance between the slides at that point. From there on its just geometry.
    If the wave length is shorter, you don't have to go as far along the slides, horizontally, to find the next place where a whole number of waves occupy the airspace.
     
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