# Double-slit experiment with radio towers!

1. Aug 20, 2008

### gigli

1. The problem statement, all variables and given/known data
Two shortwave radio antennas broadcast identical, in-phase signals at the same frequency. The transmitters are 176.0 m north, and 176.0 m south of Western Ave, respectively, as shown (that is, they are separated by 352.0 m). Western Ave is 452.0 m long. Starting at the end of that avenue, a car drives north along Negundo Street, which lies parallel to the line joining the two radio antennas. The car first encounters a minimum in reception after it travels 124.0 m. What is the wavelength of the radio waves? Assume that the car and the transmitters are all at the same altitude.

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2. Relevant equations
2-slit interference for dark fringes (minimums)
sin$$\theta$$=[(m+.5)$$\lambda$$]/d

3. The attempt at a solution
First encounters a minimum of reception means the first dark spot from the midpoint. Therefore m, the fringe number=0.
Solving for the angle of separation between the first dark spot at 124m and the midpoint. We have two sides of a triangle so arctan(124/452)= 15.3408908 degrees

sin(15.3408908)=[(.5)$$\lambda$$]/352
2*sin(15.3408908)*352=$$\lambda$$=186.251202 m

However this is incorrect! Any help you can give me would be excellent!

http://www.instantimagehosting.com/storage/Untitled_6.jpg [Broken]

Last edited by a moderator: May 3, 2017
2. Aug 20, 2008

### alphysicist

The condition for destructive interference here is that the path length difference be related to the wavelength by:

$$\Delta L= \left(m+\frac{1}{2}\right)\lambda$$
where $\Delta L=L_1-L_2$. So on the left hand side you find the distance from one source to the receiver, find the distance from the other source to the receiver, and then find the difference between those two paths.

In cases where the distance between the sources is much smaller than the distance to the receiver (like in Young's experiment), there is an approximation that can be used:

$$\Delta L = d \sin\theta$$

This is what you were using; however, for this to be valid, you would need the distance between the transmitters to be much smaller than the length of Western Avenue. Since this is not the case, I believe you would need to use the original formulation.

Last edited by a moderator: May 3, 2017