# Two-Antenna Interference Maxima Calculation ## Problem Summary - **Separation between antennas:** \( d = 1.60\,\text{m} \) - **Frequency:** \( f = 750\,\text{MHz} = 7.50 \times 10^8\,\text{Hz} \) - **Circle radius:** \( R = 20.\,\text{m} \) - **Both antennas emit **identical waves** (in phase, same frequency). - **Question:** As you walk around the circle, **how many maxima** (constructive interference points) do you detect? --- ## Step 1: Wavelength Calculation \[ c = 3.00 \times 10^8\,\text{m/s} \\ \lambda = \frac{c}{f} = \frac{3.00 \times 10^8}{7.50 \times 10^8} = .400\,\text{m} \] --- ## Step 2: Path Difference for Maxima For constructive interference (maxima), the path difference must be an integer multiple of the wavelength: \[ \Delta L = d \sin \theta = m \lambda \quad (m = , \pm1, \pm2, \ldots) \] Where \(\theta\) is the angle from the perpendicular bisector of the line joining the antennas. --- ## Step 3: Maximum Value of \(m\) The maximum possible \(|\sin\theta|\) is 1, so: \[ |m| \leq \frac{d}{\lambda} \] \[ \frac{d}{\lambda} = \frac{1.60}{.400} = 4 \] So \( m = -4, -3, -2, -1, , 1, 2, 3, 4 \): **9 possible integer values**. --- ## Step 4: Number of Maxima Around the Circle - For each \(m\) (excluding the endpoints where \(\sin\theta = \pm 1\)), there are **two directions** (one on either side). - For \(m = \pm 4\), \(\sin\theta = \pm 1\) occurs **once** each (at \(\theta = 90^\circ, 270^\circ\)). - For \(m = \), the central maximum occurs once (at \(\theta = ^\circ, 180^\circ\)). **But on a full circle, each value of \(m\) (except the extreme values) gives two maxima (one on each side of the axis).** So, for \(m = , \pm1, \pm2, \pm3, \pm4\): - \(m = \): 2 maxima (front and back) - \(m = \pm1, \pm2, \pm3\): 2 each side \(\to\) 2 x 3 x 2 = 12 - \(m = \pm4\): 2 maxima (left and right) But actually, for a **full circle**, each value of \(m\) gives **two directions** (except for the extreme values), but since the circle is continuous, these extreme cases are included. **Total maxima:** \[ \text{Number of maxima} = 2 \times \left(\frac{d}{\lambda}\right) + 1 = 2 \times 4 + 1 = 9 \] --- ## Final Answer \[ \boxed{9} \] You will detect **9 maxima** (points of constructive interference) as you walk around the circle. --- ### Diagram (optional)  *Antennas separated by \(d\), maxima directions indicated on a circle of radius \(R\).*