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The Nature of Light

  1. Apr 28, 2007 #1
    a) A light with a wavelength of 500 nm illuminates two narrow slits that are 0.10 mm apart. If the viewing screen is 1.20 m from the slits calculate the distance between each maximum.

    b)how far would the fourth maximum be from the central maximum?



    a) Given: lambda = 500 x 10e-9 m
    d= 0.1 x 10e-3 m
    L= 1.2 m

    delta x = lambda x L/d
    =(500 x 10e-9 m) (1.2 m)/(0.1 x 10 e-3 m)
    =0.0006 m
    =6.0 x 10 e-3

    Therefore, the distance between each maximum is 6.0 x 10 e-3 m

    b) Since the distance between each maximum is 6.0 x 10 e-3 m the fourth maximum would be 4 x (6.0 x 10 e-3 m)

    Therefore, 4 x (6.0 x 10 e-3 m) = 2.4 x 10 e-2

    Therefore, the fourth maximum would be 2.4 x 10 e-2 m away from the central maximum.


    Does this look like I'm doing this question right to anybody?
     
  2. jcsd
  3. Apr 28, 2007 #2

    hage567

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    Homework Helper

    Yes, you've got it right.
     
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