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Diffraction grating of white light

  1. Mar 23, 2004 #1
    White light containing wavelengths from 405 nm to 760 nm falls on a grating with 7510 lines/ cm. How wide is the first-order spectrum on a screen 2.29 m away?

    I'm not sure the way this question is worked out, but what I have so far is

    sin(theta)=[m*Lambda+ m*Lambda]/d
    d=1/7510cm=1.33e-6m
    sin(theta)=[(407e-9m)+(705e-9m)]/1.33e-6m= 0.876 theta=61 degrees
    width=2x(tan theta)=(2*2.29)*(tan 61)=8.32m

    I have the wrong answer so I must have done something wrong. Can someone help?
     
  2. jcsd
  3. Mar 24, 2004 #2

    ZapperZ

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    n*lambda = d*sin(theta)

    n=1 in this case, and you know d.

    Find theta for lambda=405 nm, and then find the location of this 1st order on the screen. Next, find theta again for lambda=760 nm, and again find the location on the screen. The continuous spectrum will be in between these two locations.

    The mistake that you made was thinking that delta(sin(theta)) is the same as delta(theta) or sin(delta(theta)).

    Zz.
     
  4. Mar 25, 2004 #3
    Is this correct

    sin theta= (1)*(4.05e-7m)/(1.33e-6m)= 0.305 theta=17.8
    location= 2x(tan 17.8)= (2*2.29)*0.321=1.47m

    sin theta= (1)*(7.60e-7m)/(1.33e-6m)= 0.571 theta=34.8
    location= 2x(tan 34.8)= (2*2.29)*0.695= 3.18m

    so then should the answer to the question be
    3.18m+1.47m/2= 2.33m
     
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