Hi! I need help with this question...
A double slit experiment, using light of wavelength 600 nm, results in fringes being produced on a screen. The fringe separation is found to be 1.0 mm.

When the distance between the double slits and the viewing screen is increased by 2.0 m, the fringe separation increases to 3.0 mm. What is the separation of the double slits producing the fringes?
Supposedly the answer is 0.6 mm, but I've been getting 1.2 mm. Where did I go wrong?

x=(wavelength*distance)/fringe separation

Eq 1 => 1*10^-3 = 6*10^-9 (D) /A
Eq 2 => 3*10^-3 = 6*10^-9 (D+2)/A

Help is much appreciated!!! :D

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asked by Leiah_
  1. Oh my gosh! I forgot nano is 10^-7. Oops! :C

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    posted by Leiah_
  2. 600 nm= 600E-9=.6E-9 meters=6E-7meters

    Is that your error?

  3. nano is -9....

  4. Hello Bob! Thanks for aswering :)
    Oh my gosh, nano is ^-9, what am I saying haha, so confused :C

    I think I've been using 6E-9 but I just can't seem to get it...

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    posted by Leiah_
  5. using light of wavelength 600 nm, " IS NOT 6E-9, it is 600 E-9, or 6E-7

  6. OH! Ok, I redid my working using 6E-7 this time :)

    Somehow, I managed to get a 6 in my answer haha, but not 0.6 mm.

    Eq 1 -> 3*10^-3 =6*10^-7(D+2)
    Eq 2 -> 1*10^-3 =6*10^-7(D)


    =6*10^-9 m
    =6*10^-6 mm

    Sorry for the messy working :C
    Still can't get it!!

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    posted by Leiah_
    General Certificate of Education
    Advanced Subsidiary Level and Advanced Level
    Paper 1 Multiple Choice
    October/November 2010
    1 hour
    Additional Materials: Multiple Choice Answer Sheet
    Soft clean eraser
    Soft pencil (type B or HB is recommended)

  8. I took a look at the paper but they only give the answer and they don't show the working method :C

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    posted by Leiah_

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