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Physics

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When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction peak that is 8.80 cm wide on a screen that is 2.75 m away. How wide is the slit?

  • Physics -

    Is there something wrong with this? http://www.jiskha.com/display.cgi?id=1335108759 The part poorly defined is the "peak that is ...wide". The formula I gave you is the distance to the first minimum.
    Some folks use the "width " of the peak to not the first minimum, but to the point where the peak has went down to the .707 RMS value.

  • Physics -

    I skipped over this question because I was not sure what was meant by width of the stripe. I think first minimum is most common.

  • Physics -

    The width of the central peak (max) is the distance between the first minima.
    The equation for diffraction minima is
    b•sinφ = k•λ.
    k =1,
    b = λ/sinφ.
    From the geometry of diffraction pattern
    tanφ =x1/L,
    where x1 =8.8/2 = 4.4 cm =0.044 m is the distance between the center of diffraction pattern and the first min,
    and L =2.75 m.
    tan φ = 0.044/2.75=0.016.
    sin φ = tan φ=0.016.
    b = λ/sinφ = 415•10^-9/0.016 =
    =2.59•10^-5 m =25.9 micrometers.

  • Physics -

    Thank you so much Elena! I forgot to divide 8.8 by two and, of course, kept getting double the answer which I knew was incorrect.

  • Physics -

    Anyone else from SF have to google this question, or was i the only one?

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