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Homework Help: Interference of Light Waves

  1. Oct 22, 2005 #1
    Equations dealing with the interference of light waves have a variable 'delta y'. My biggest problem is understanding what 'y' represents? Does it define the distance between two adjacent fringes? Or the distance of a fringe from the origin? this scepticism might have been the reason for my confusion in these questions.

    1.) Two antennas seperated by 300m simultaneously broadcast identical signals of = wavelength. a.) A car is travelling north and is at 400m north of the midpoint line between the two signals. What is the wavelength?

    My solution:
    y = 400m (from center/origin).

    I solved for wavelength and got an answer of 60m. the correct answer is supposedly 55.7m.

    Part 'b' of the question: How much further does the car have to travel to encounter the next minimum in reception? (Do not use small-angle approximation).

    I understand i must be solving for the dark fringes now. But the formula is confusing since 1.) i cant use small-angle approximation, and 2.) im still confused as to what 'y' represents.

    2.) Light with wavelength 442nm passes through a double-slit system that has a slit seperation of 0.400mm. Determine how far the screen must be placed away from the sources to create two fringes directly across from the slits and one frindge between those.

    I have wavelength, distance between slits, and i could potentially find y. I must ultimately solve for L (length between screen and slits). Fistly, i have to find where the two fringes which are directly across the slits really are. I assumed that the top fringe is 'd' away from the bottom fringe. Since there is a nother frindge in betwee, the distance between frindges would be 0.4mm/2. Once again, my problem with 'y' haunts me. Should 'y' be the 0.2mm or something else?

    Thank you.
  2. jcsd
  3. Oct 23, 2005 #2


    User Avatar
    Gold Member

    There are two equasions with y.
    One with y (not delta y) and 'n' which specifies which fringe you're talking about. This gives the distance of the n fringe from the middle at distance L:
    y = ((n - 0.5)*wavelength*L)/d
    The other one has delta y and no 'n'. This gives the distance between any two adjacent fringes at distance L. We got this equasion by using the first one and subtracting the distance of the n+1 fringe from the n fringe giving the distance between them:
    delta y = (wavelength*L)/d

    Also, if you can't use small angle approximation, then instead of saying that
    sin(angle) = (n-0.5)*wavelenght/d = y/L and finding L which is really only true for tan(angle), then just find the angle by
    sin(angle) = (n-0.5)*wavelenght/d, and find L by doing:
    tan(angle) = y/L
    Hope that helped!
  4. Oct 23, 2005 #3
    Can anyone verify the above, and also help on the other questions that i posted? Thank you.
  5. Oct 23, 2005 #4
    Revised Question

    Question number 2 was solved.

    Regarding Question number1:

    I use 300m for d (seperation of slits), 1000m for length between "screen" and slits (Sources). Should n = 2? And should y = 400m?

    Once again, i calculate 60m. But the correct answer should be 55.7m. Any ideas?
  6. Oct 25, 2005 #5
    ?? Anyone? .
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