# Radio Antenna’s and Young’s Double split

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?

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