Q53P - University Physics with Modern Physics

Question

Short-wave radio antennas A and B are connected to the same transmitter and emit coherent waves in phase and with the same frequency f . You must determine the value of f and the placement of the antennas that produce a maximum intensity through constructive interference at a receiving antenna that is located at point P, which is at the corner of your garage. First you place antenna A at a point 240.0 m due east of P. Next you place antenna B on the line that connects A and P, a distance x due east of P, where x < 240.0 m. Then you measure that a maximum in the total intensity from the two antennas occurs when x = 210.0 m, 216.0 m, and 222.0 m. You don’t investigate smaller or larger values of x. (Treat the antennas as point sources.) (a) What is the frequency f of the waves that are emitted by the antennas? (b) What is the greatest value of x, with x < 240.0 m, for which the interference at P is destructive?

Step-by-Step Solution

Answer

1. The frequency is 50Mhz
2. The maximum distance at which interference is destructive is 237m.

Step 1: Important Concepts

Constructive interference is at $\mathit{n}\mathit{\lambda }$

and destructive interference is at $\frac{\left(2n+1\right)\mathbf{\lambda }}{\mathbf{2}}$.

Where n is an integer.

Step 2: Find wavelength and frequency

We know that the interference at point P is a constructive interference.

This means that the path difference between the two waves of the two sources must be an integer number of X. We need to find the wavelength , which is the wavelength of the two sources. Noting that we are given three values of x in which source B where located to make the two waves interfere constructively at P. The difference between ant two locations of them is about 6.0 m.

 

                                         ${\mathrm{x}}_2-{\mathrm{x}}_1=216\mathrm{m}-210\mathrm{m}=6\mathrm{m}$

And 

                                         ${\mathrm{x}}_3-{\mathrm{x}}_2=222\mathrm{m}-216\mathrm{m}=6\mathrm{m}$

Hence, the wavelength is given by 

                     

                                            $\lambda =6.0m$       

Now we find the frequency using $v=f\lambda$

 

Since the source is light we get speed of light and then

                                                $f=\frac{c}{\lambda }$        

                                                 $$\begin{gathered} f=\frac{3.0\times {10}^8}{6.0} \\ f=50MHz \end{gathered}$$   

Step 3: Position of dark fringe

We know that the path difference should be 

                                                $∆x=(m+\frac{1}{2})\lambda$

In this case we are looking for 

                                              $240-x=(m+\frac{1}{2})\lambda$  

Solve for x   

                                                 $\mathrm{x}=240-(\mathrm{m}+\frac{1}{2})\mathrm{\lambda }$

From this equation above, the smallest vale of m will give the largest value of x

Hence we put m=0;

                                                $$\begin{gathered} \mathrm{x}=240-(0+\frac{1}{2})\mathrm{\lambda } \\ \mathrm{x}=240-\frac{1}{2}\mathrm{\lambda } \\ \mathrm{x}=240-\frac{1}{2}6.0 \\ \mathrm{x}=237\mathrm{m} \end{gathered}$$

Hence, the maximum distance at which interference is destructive is 237m