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!

Double Slit Experiment without the small angle approximation

  1. Jul 23, 2011 #1
    Two very narrow slits are spaced 1.8 [itex]\mu[/itex]m apart and are placed 35.0 cm from the screen. What is the distance from the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with [itex]\lambda[/itex] =550 nm

    r1-r2=m[itex]\lambda[/itex]
    dsin[itex]\theta[/itex]=m[itex]\lambda[/itex]
    Rtan[itex]\theta[/itex]=y

    Since Im only concerned with the distance i was thinking about just finding the distance between the bright fringes. The problem is I dont know how to get theta.

    Ive tried m[itex]\lambda[/itex]=dsin[itex]\theta[/itex] and solved for theta but that didnt work. Im not sure you can find r1 and r2 because the distance between the fringes is missing.
     
  2. jcsd
  3. Jul 23, 2011 #2
    You can use [itex]\theta = \arctan \frac{y}{R}[/itex].
     
  4. Jul 23, 2011 #3
    The problem is we dont know y that is what im trying to find, but im going to need[itex]\theta[/itex] first
     
  5. Jul 23, 2011 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Don't the "Dark Lines" correspond to destructive interference?
     
  6. Jul 24, 2011 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    SammyS is right, the problem asks about dark fringes. For them, the angle of deviation from the central line is related with d and λ as

    d sin([itex]\vartheta[/itex])=(2m+1)λ, with m=0, ±1, ±2 ....

    y is the distance of a dark spot from the centre of the interference pattern on the screen. If the screen is at D distance from the two slits, tan([itex]\vartheta[/itex])=y/D.
    You need the distance between the first and second dark lines. For the first one, m=0, and m=1 for the second one.

    Calculate [itex]\vartheta[/itex]-s with m=0 and m=1, calculate the corresponding y values and find their difference.

    ehild
     
  7. Jul 24, 2011 #6
    isnt the distance between the dark fringes the same as the distance between the bright fringes?

    so i would do

    dsin[itex]\theta[/itex]=[itex]\lambda[/itex]
    and dsin[itex]\theta[/itex]=3[itex]\lambda[/itex]

    solve for [itex]\theta[/itex] in terms of y and then solve for y
     
  8. Jul 24, 2011 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The title says "... without the small angle approximation"

    λ/d ≈ 3.5

    I wouldn't assume the lines in the diffraction pattern are equally spaced.
     
  9. Jul 24, 2011 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The values sin[itex]\theta[/itex] are equally spaced, but the position of the fringes are proportional to tan[itex]\theta[/itex].
    I am sorry, there is a mistake in my last post, I meant

    dsin[itex]\theta[/itex]=(2m+1)[itex]\lambda[/itex]/2

    So calculate the distances from
    dsin[itex]\theta[/itex]=[itex]\lambda[/itex]/2
    and dsin[itex]\theta[/itex]=3[itex]\lambda[/itex]/2.

    ehild
     
  10. Jul 24, 2011 #9
    Just to clear something up, If we were using the approximation then the bright and dark fringes would be the same length.

    Thank you so much for your help
     
  11. Jul 24, 2011 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, the approximate method yields equal distances but the approximation can be applied for small angles (less than 10°).

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




Similar Discussions: Double Slit Experiment without the small angle approximation
Loading...