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Radio Antennas and Young's Double slit

  1. Mar 26, 2014 #1

    STJ

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    1. The problem statement, all variables and given/known data

    Two radio antennas separated by d = 294 m as shown in the figure below simultaneously broadcast identical signals at the same wavelength. A car travels due north along a straight line at position x = 1 290 m from the center point between the antennas, and its radio receives the signals. Note: Do not use the small-angle approximation in this problem.

    (a) If the car is at the position of the second maximum after that at point O when it has traveled a distance y = 400 m northward, what is the wavelength of the signals? ANS: 43m (Correct)

    (b) How much farther must the car travel from this position to encounter the next minimum in reception? ANS: 399.8m (Incorrect)

    2. Relevant equations

    I just used trig, this does resemble double split though but I didn't see how any double split formulas would help when trig should be enough.


    3. The attempt at a solution

    I determined the wavelength(w) to be 43m, which is correct. I started by knowing that for the car to reach the next minimum it must travel some distance x past the 400m mark. S1 = ((400 + x + 147)^2 + 1290^2)^.5 m and S2 = ((400 + x - 147)^2 + 1290^2)^.5 m. The difference between the top radio and bottom radio (DS) = 235200+588x. This must equal 2.5w as the next minimum is 1 half wavelength away from the 2nd maximum. Plugging in w and solving for x I get x = -399.8. Through common sense the next minimum can't be backwards, it has to be forwards so I make it positive instead. I'm being told I am wrong though. Some hints perhaps?
     
  2. jcsd
  3. Mar 26, 2014 #2
    Seems like you forgot about the ^.5 outside the parenthesis.

    Say double slit, not double split three times in a row.
     
    Last edited: Mar 26, 2014
  4. Mar 26, 2014 #3

    STJ

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    It seems I did forget about that. After looking at this for a minute, how is it possible to solve

    (1963309 + 1094x + x^2)^.5 - (1728109 + 506 + x^2)^.5 = 2.5w

    for x? I don't see any way of solving that for x, so I guess I have to use some formulas or something?

    EDIT: would Ydark = Ltan(theta dark) be the proper formula for this? If so is theta equal to the angular difference between the "lines" to the car from the antennas? I'm not sure where to go from here.
     
  5. Mar 26, 2014 #4
    Square the whole equation, isolate the one remaining term that has a square root in it, square the equation again, you should find a quadratic for x. Solve it.
     
    Last edited: Mar 26, 2014
  6. Mar 26, 2014 #5

    STJ

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    There has got to be a faster way to do that problem. Doing this during a 1 hour exam would take most of my time. There is no simpler way to solve this problem?
     
  7. Mar 26, 2014 #6

    rude man

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    EDIT: remember the admonition not to use small-angle approximations.

    Thus, the rays from both transmitters at the 1st minimum are not parallel and you have to deal with a scalene triangle. So the difference between the two beams must be 1/2 wavelength. The rest is trig. Draw the picture and find the required distance form the starting point (midpoint between the stations projected perpendicularly to the road.)

    I'm a bit surprised you got the right answer for part (a) and not for part (b).

    Would help if you posted the diagram.
     
    Last edited: Mar 26, 2014
  8. Mar 26, 2014 #7
    It's a bit of work, but not way too bad. Solve the problem algebraically and only plug in the data at the very last step. Using vector notation simplifies the algebra somewhat
     
  9. Mar 26, 2014 #8
    You mean the difference between the two beams must be 3/2 wavelength (3rd minimum). Part (b) is much more challenging than part (a).
     
  10. Mar 26, 2014 #9

    rude man

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    Right. Sorry. But it's the second minimum, not the 3rd: 1st max (center), 1st min, 2nd max, 2nd min. At least that's my reading.

    No matter: you have a scalene triangle with r2 - r1 = 3(lambda)/2 and you need to calculate the distance from the center road position to that point, then subtract 400m.

    PS still no diagram ...
     
    Last edited: Mar 26, 2014
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