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!

Uncertainty in single slit diffraction

  1. Sep 9, 2016 #1

    Titan97

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    An electron is moving in a parallel beam along the x-direction with momentum, p=mv. It encounters a slit of width w. Assuming that the electron gets diffracted somewhere within the central maximum of small angular magnitude Δθ, estimate the uncertainty Δp in its momentum component transverse to the direction of motion. Check that uncertainty principle is satisfied in this experiment.

    2. Relevant equations
    ΔyΔp = ħ/2

    3. The attempt at a solution
    I don't know where to apply the above equation. What should be the Δy here? Is it the width w or the width Δθ?

    What is the uncertainty in momentum here? If Δx is the width w, then I have to consider the Δp at the slit. (is that correct?)
     
  2. jcsd
  3. Sep 9, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi,
    You need some more under 2. What about the width of the central maximum ? Any expressions ?
     
  4. Sep 9, 2016 #3

    Titan97

    User Avatar
    Gold Member

    the width of central maxima is obtained from ##n\lambda=w\sin\frac{\Delta\theta}{2}##

    Here, n=1.
     
  5. Sep 9, 2016 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##\lambda## ?
    Yes. And you measure Δp by looking at the width of the diffraction pattern.

    (you use y and x for the same direction ?)
     
    Last edited: Sep 9, 2016
  6. Sep 9, 2016 #5

    Titan97

    User Avatar
    Gold Member

    what is delta P? the change in momentum?
     
  7. Sep 9, 2016 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes/No. You don't have to worry about things 'before the slit'.
    Assume x is horizontal and the slit is horizontal too (hyperphysics picture:)
    sinslit.gif .
    His a is your w.

    At the slit:

    The uncertainty in position (in the y direction) is equal to the width of the slit.
    And the uncertainty of the momentum (in the y direction) there (at the slit) appears as the width of the central maximum (on the screen).

    [edit] picture hicks up. It's here: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html


    ##\lambda## ?
     
  8. Sep 9, 2016 #7

    Titan97

    User Avatar
    Gold Member

    Can you explain this?

    $$\lambda=\frac{h}{mv}$$
     
  9. Sep 9, 2016 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's called the de Broglie wavelength, which google.
    [edit] it is indeed the wavelength you need here. And it may seem strange, but you need the p = mv in the x-direction to determine it.
     
    Last edited: Sep 9, 2016
  10. Sep 9, 2016 #9

    Titan97

    User Avatar
    Gold Member

    @BvU i am not asking about de broglie wavelength. I am asking about this

     
  11. Sep 9, 2016 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, sorry.
    Well, if the momentum uncertainty (in the y direction) at the slit would be zero (the average momentum itself in that direction is zero) then there would be no diffraction pattern, just a bright line with the same height as the slit.
     
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: Uncertainty in single slit diffraction
Loading...