Answer to Question #157445 in Optics for kris

Question

You need to use your cell phone, which broadcasts a signal at 900MHz

900MHz, but you are in an alley between two large buildings that have only a 5.0m

5.0m spacing between them. What is the angular width, in degrees, of the signal after it emerges from between the buildings?

Answer 1

Two radio antennas broadcasting in phase. $\nu$ = 900MHz

$dsin\theta = m\lambda \to m = 0, +1, -1, +2, -2...$ 1-equation

Since the resultant wave is detected at distances much greater than d = 5m we may use 1-equation to give the directions of the intensity maxima, the values of for which the path difference is zero or a whole number of wavelengths.

The wavelength is $\lambda = \large\frac{c}{\nu}$ .

From 1-equation with m = 0, and the intensity maxima are given by

$sin\theta = \large\frac{m\lambda}{d} = \large\frac{mc}{d\nu} = \frac{m*3*10^8}{5*9*10^8} = \frac{m}{15}$

$m = 0 \to \theta = 0, ... , m = 15 \to \theta = 90^o$