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Young's experiment

  1. Apr 9, 2005 #1
    In a double-slit experiment, the slit separation is 2.0mm, and two wavelengths, 750nm and 900nm, illuminate the slits. A screen is placed 2.0m from the slits. At what distance from the central maximum on the screen will a bright fringe from one pattern first coincide with the bright fringe from the other?

    what i know:
    d=2.0mm or 2.0e-3m
    lambda1=750nm or 750e-9m
    lambda2=900nm or 900e-9m
    y1=y2=Y <-- this is what we're looking for

    y=n lambda L / d
    d sin@ = n lambda

    by equating both Ys i get n1/n2 = lambda2/lambda1

    but I end up nowhere, any suggestions?
     
    Last edited: Apr 9, 2005
  2. jcsd
  3. Apr 9, 2005 #2
    anyone have an idea?
     
  4. Apr 10, 2005 #3
    still no takers?
     
  5. Apr 10, 2005 #4
    You need to work out the fringe difference, that is when the brightest parts are. Then you can find the point the two cross. Simple as that really.

    The Bob (2004 ©)
     
  6. Apr 10, 2005 #5
    do u mean trial and error?

    i don't know if this is correct, but if i use n<=(less than or equal)d/lambda
    there's going to be too many maximas on both wavelengths (thousands of them)
     
  7. Apr 10, 2005 #6
    You need the equation: [tex]\frac{\lambda}{d} = \frac{x}{L}[/tex]

    So for the wavelength of 750nm the fringes will be:

    [tex]\frac{750 \times 10 ^{-9}}{2.0 \times 10^{-3}} = \frac{x}{2.0}[/tex]

    [tex]\frac{750 \times 10 ^{-9}}{2.0 \times 10^{-3}} \times 2.0 = x = 7.5 \times 10^{-4} \ m[/tex]

    And for the wavelength of 900nm the fringes will be:

    [tex]\frac{900 \times 10 ^{-9}}{2.0 \times 10^{-3}} = \frac{x}{2.0}[/tex]

    [tex]\frac{900 \times 10 ^{-9}}{2.0 \times 10^{-3}} \times 2.0 = x = 9.0 \times 10^{-4} \ m[/tex]

    Then [tex]\frac{9.0 \times 10^{-4}}{7.5 \times 10^{-4}} = 1.2[/tex]

    1.2 needs to be multiplied by 5 times to get the first integer (which will be a bright fringe) which is 6. This means there will be 5 fringes of wavelength 900nm and 6 fringes of wavelength 750nm. This means the mixing fringe will be 4500nm or 4.5 x 10-6m from the central brightest fringe.

    No trial and error at all. Can you see anything wrong? (serious question. I like a second opinion)

    The Bob (2004 ©)
     
  8. Apr 10, 2005 #7
    thanks for the help, but i actually got it already from my original equation which i got n2/n1 = lambda1/lambda2

    I actually factored (700nm/900nm) smaller and got n2=n1 5/6

    which gives me n1=6 and n2=5

    which is the same as what you got :)
     
    Last edited: Apr 10, 2005
  9. Apr 10, 2005 #8
    how did u type in the lambda sign? so i dont have to type it all the time next time lol
     
  10. Apr 10, 2005 #9
    I was simply showing you another way of doing it because you said:

    The typing is using a system called LaTeX mathematical typesetting. This thread has more information but you simply type [tex ] /lambda [/ tex] but without the spaces in the tex bit.

    The Bob (2004 ©)
     
  11. Apr 10, 2005 #10
    For my interest, how did you factor down 700/900 and get 5/6??

    The Bob (2004 ©)
     
  12. Apr 10, 2005 #11
    oh sorry its 750/900
     
  13. Apr 10, 2005 #12
    don't get me wrong, i do appreciate (a lot) the help. it just amazes me how simple the problem is, but it took me so long to figure it out
     
  14. Apr 11, 2005 #13
    It's alright. I thought that there must have been something else to it as well because it didn't seem that bad but I have made silly posts in the past they have led other places and I have learnt other stuff from them. I guess I am saying don't worry. There are people that will mock you but we learn more from making mistakes than always getting the right answer. Questions are for the brave, cowards don't ask.

    Anyway, enough of the philsophy, I was happy to help and don't worry about how simple it was. :smile: In fact you have helped my revise for my test tomorrow. :smile:

    The Bob (2004 ©)
     
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