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Waves and Light

  1. Dec 22, 2003 #1
    1. A double-slit experiment is performed using a monochromatic light source, two slits spaced 0.10 mm apart, and a screen located 150 cm away. The bright fringes are located 0.30 cm apart. If the screen distance was changed to 3.0 m from the sources, what would the average distance between bright fringes become?

    2. Again a double-slit experiment is performed using a monochromatic light source with a wavelength of 5.00 ´ 10–7 m. The pattern appears on a screen 150 cm away and the bright fringes are 0.40 cm apart. If the wavelength of the light used is changed to 7.50 ´ 10–7 m, what would the average distance between bright
  2. jcsd
  3. Dec 22, 2003 #2
    I don't think that i have the right formula to use for this question, but i know that for question 1
    X= 0.004m

    Whats the formula? I think that if i have the formula for question 1 I can solve for 2.
  4. Dec 22, 2003 #3
    Another question
    Two speakers producing exactly the same frequency and always in phase are located on an east–west line. The sound is going north. The speakers are 50 cm apart and a microphone is located 80 cm due north of one speaker. Which of the following wavelengths would produce sound with the least intensity at the location of the microphone?

    I drew a diagram and i understand how the speakers and mic are set up but i think that all i need is the formula for destructive interference. i have the formula sin(theta)=(n - 1/2)ë/d, but i don't think that its the right one b/c I don't have an angle of a nodal line #. What is the formula that i'm looking for?
  5. Dec 22, 2003 #4
    Just so you know, there's an "Edit" button on your post. Instead of double posting and wasting space, simply press the edit button, and put your questions in your original post. Thanks.
  6. Dec 23, 2003 #5


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    For the first one you have to use an equation that looks like

    [tex]n\lambda = dSin(\theta)[/tex]

    NOTE: those terms may not be correct, check your formula sheet

    the n is the fringe number, lambda is wavelength, d is the distance between the lines in the diffraction grating (the inverse of the lines/m rating the grating has) and theta is the angle made between light going straight and going to one of the fringes.

    To find theta, you have to use the inverse tangent of the distance between fringes divided by the distance between the diffraction grating and the screen.
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