# Waves & Interference

1. Oct 8, 2007

### TalonStriker

1. The problem statement, all variables and given/known data
There are 2 in-phase radio wave generators at x = 100 and x = 200. The frequency of each wave is 3.2 MHz. Find spot closest to x = 200 such that the interference at that point is maximum.

2. Relevant equations
$$\Delta\Phi = \frac{2\pi}{\lambda} \Delta x + \Phi_{0}$$
$$v = c = f\lambda$$ c = speed of light

3. The attempt at a solution

I assumed that point d was close to x = 300. So that
$$\Delta x = (100-d) - d$$
$$\Delta x = (100 - 2d)$$

$$2\pi m = \frac{2\pi}{c/f} (100 - 2d) + 0$$ for some integer m; \Phi_{0} = 0 since the waves are in-phase

$$m = \frac{1}{c/f} (100 - 2d)$$ cancel out 2 * pi

$$c/f * m = (100 - 2d)$$

$$(c/f * m ) - 100= 2d$$

$$d = \frac{(c/f * m) - 100}{2}$$

Ignoring d, I solved for m such that $$\frac{(c/f * m) - 100}{2}$$ was equal to 300. Using the m i found (rounding it to nearest integer), i plugged it back into $$d = \frac{(c/f * m) - 100}{2}$$ to find d.

Is the method I used correct? What did I do wrong? (Could you please provide me the correct answer & method...i have an exam in 5 hours).

Thanks!

Last edited: Oct 8, 2007