1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Young's double-slit experiment, fringe separation

  1. Sep 1, 2016 #1
    1. The problem statement, all variables and given/known data
    The distance between the 1st bright fringe and the 21st bright fringe in a Young's double-slit arrangement was found to be 2.7 mm. The slit separation was 1.0 mm and the distance from the slits to the plane of the fringes was 25 cm. What was the wavelength of the light?

    Answer: 5.4 * 10-7 m

    2. Relevant equations
    y = λD / a

    3. The attempt at a solution
    a = 1 mm = 10-3 m
    D = 25 cm = 0.25 m
    y (distance between 1st and 21st fringe) = 2.7 mm = 2.7 * 10-3 m
    m (number of slits) = 21
    λ = ?
    First of all I am not sure whether y and m are correct terms for the given data.

    y = (2.7 * 10-3) / (21-1) = 1.35 * 10-4 m
    Second of all I do not understand why we need to divide y by m and not even by 21 but by 20.

    λ = ya / D = 1.35 * 10-4 * 10-3 / 0.25 = 5.4 * 10-7 m

    The answer is correct, but I completely lack an understanding of what is being done and what is being found.

    Taking this image as an example: d is a = the separation of slits, through which the light from the source comes. This part I understand. Through the slits the light falls on the screen, which is L = D meters from the slits. This thing I also understand. As I suppose, y on the image is what I derive by dividing 2.7 mm by 20 slits. And this is where I am completely lost. What is 2.7 mm then? What are 21 or 20 slits? Why do we consider the y region at all?

    I see the whole experiment as on this http://h2physics.org/wp-content/uploads/2009/08/fringeseparation1.gif [Broken]. There are two slits through which light comes to a screen 4 m apart from the slits. Light from both slits comes in a form of a "triangle", both of triangles overlap and create a smaller triangle. Now what does the θ angle and that semi-triangle drawn in the image has anything to do with the experiment? The hypotenuse doesn't fit anywhere -- it's too far away at the beginning and too low at the end of the gif to the light-"triangles". But as I understand for some reason the opposite side of the θ angle is y, which we need to find, which looks like it doesn't represent anything at all.

    Honestly, I am very very lost in this topic and this looks like the basic idea without understanding of which I cannot continue. If anybody could explain these things in a simple or logical manner I would appreciate it very much. At this point I do not quite understand what is y, why we need to consider it and so on.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Sep 1, 2016 #2

    Charles Link

    User Avatar
    Homework Helper

    The formula you need is ## m \lambda=d \sin(\theta) ## where d is distance between the slits, and ## m ## is an integer. The formula gives the location (angle ## \theta ## of maximum (bright spot) number ## m ##, where ## m ## is an integer. ( ## \lambda ## is the wavelength. ) Try working it with this equation. You can use the approximation ## \sin(\theta)=\tan(\theta) ## for small angles in finding the angles where the maxima are located. Just a bit of background info on Young's experiment: You normally would expect to get a shadow pattern (of two bright lines) for the light passing through two slits, but instead it produces an interference pattern consisting of a series of many bright fringes that are nearly equally spaced. The wavelength of the light determines the spacing of the bright spots of the interference pattern. (Note: monochromatic light, i.e. one color and not a mixture of colors, needs to be used to get an interference pattern.) ## \\ ## And just a little additional info: The brightest spot occurs at the center of the pattern and is labelled m=0. The pattern is symmetric about ## m=0 ## and also has bright spots at (negative) angles ## \theta ## for each negative integer.
     
    Last edited: Sep 1, 2016
  4. Sep 6, 2016 #3
    Thank you for some hints. Did read more about the experiment in last the couple of days (that is why replying so late, sorry for that). I think if I'll use your given formula I'll need to have the required angle, right?

    If we take the original info which I have in the problem I see the situation as this:
    y is the distance between n / m = 0 (the adjacent line of the triangle) and the n / m = 1 which is the brightest slit aside from the center. And because we have n / m = 1 and n / m = 21 and we need the distance between m = 0 and m = 20, so we subtract 1 from 21. And in order to find the distance between 0 and 1 we just divide 2.7 mm by 20 and we find y. I think this should be correct.

    As said, y is the distance between m = 0 and m = 1, where m = 1 is the brightest slit after the center. And the hypothenuse points to m = 1.
     
    Last edited by a moderator: May 8, 2017
  5. Sep 6, 2016 #4

    Charles Link

    User Avatar
    Homework Helper

    The approximation that can be used is ## \sin(\theta)=\tan(\theta)=\theta ## provided ## \theta ## is measured in radians. With this approximation (which works for small angles, i.e. it works very well when the ##| \sin(\theta)|=1/10 ## or less and even somewhat good for ## | \sin(\theta)|<.25 ##), the bright spots are equally spaced for the interference pattern.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Young's double-slit experiment, fringe separation
Loading...