1. Not finding help here? Sign up for a free 30min 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!

Wavelength of light

  1. Apr 23, 2007 #1
    1. The problem statement, all variables and given/known data

    The interference pattern on a screen 1.2 m behind a 710 line/mm diffraction grating is shown in Figure P22.45 (http://i137.photobucket.com/albums/q208/infinitbelt/p22-45alt.gif ), in which y1 = 52.3 cm and y2 = 107.6 cm. What is the wavelength of the light?


    2. The attempt at a solution

    I believe that is a single slit diffraction grating. Thus the equation is a*sin(theta) = m(lambda), right? I then have sin(theta) = y/L Where do I go from there?

    I tried having two equations (one for y1 and one for y2) and setting them equal to one another by setting them individually equal to a. But that did not get me the right answer.


    Thanks!
     
  2. jcsd
  3. Apr 23, 2007 #2

    hage567

    User Avatar
    Homework Helper

    Can you show more of your calculation?
     
    Last edited: Apr 23, 2007
  4. Apr 23, 2007 #3
    I had:

    a*(y1 / L) = m1*lambda
    a*(y2 / L) = m2*lambda

    (lambda*L) / y1 = (2*lambda*L) / y2

    lambda = [(y2)(L)] / [(y1)(L)]


    Is that right?


    Thanks!
     
  5. Apr 23, 2007 #4

    hage567

    User Avatar
    Homework Helper

    y2 is not the total distance to the second maximum from the centre, it is the distance between the first order and second order maxima. Also, be careful of the approximation of sin(theta) = y/L. That doesn't work when the angle gets too large. So that is something to consider.
     
  6. Apr 23, 2007 #5
    So should y2 be y2 + y1? I am still a bit confused.

    Thanks!
     
  7. Apr 23, 2007 #6

    mezarashi

    User Avatar
    Homework Helper

    This experiment uses a diffraction grating, and you need to use the equations for that. You should focus on the first (closest to center) maxima. As hage567 has mentioned, the equations are only approximations based on the fact that (sin x = x) for small values of x. If you are using large angles, then your equations will not be accurate.

    This page has a good geometric explanation for the equation you mentioned.
    http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratcal.html
     
  8. Apr 23, 2007 #7
    Thanks! I got the right answer but am still not sure why. Here's what I did:

    y = ((lambda)*(L)) / d

    ((0.523 * 1.408E-6) / 1.2) - (1.076E-7 - 0.523E-7) = 558.5 nm.


    Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Wavelength of light
  1. Wavelength of light (Replies: 4)

Loading...