Q4E - University Physics with Modern Physics

Question

 Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point Q? (b) What is the longest wavelength for which there will be constructive interference at point Q?

Step-by-Step Solution

Answer

(a) The longest wavelength for which the interference at point Q is destructive is 

       $\mathrm{\lambda }=240 \mathrm{m}$

(b) The longest wavelength for which the interference at point Q is constructive is 

       $\mathrm{\lambda }=120 \mathrm{m}$

Step 1: (a) Determination of the longest wavelength for which the interference at point Q is destructive

The path difference given is ${\mathrm{r}}_2-{\mathrm{r}}_1=120\mathrm{m}$.

For destructive interference, the condition is,

${\mathrm{r}}_2-{\mathrm{r}}_1=\left(\mathrm{m}+\frac{1}{2}\right)\mathrm{\lambda }, \mathrm{m}=0,\pm 1,\pm 2,....$                                      ...(ii)    

So, here the wavelength is,

$$\begin{gathered} \frac{\mathrm{\lambda }}{2}=120\mathrm{m} \\ \mathrm{\lambda }=240\mathrm{m} \end{gathered}$$

Step 2: (b) Determination of the longest wavelength for which the interference at point Q is constructive

The path difference given is ${\mathrm{r}}_2-{\mathrm{r}}_1=120\mathrm{m}$.

For constructive interference, the condition is,

  ${\mathrm{r}}_2-{\mathrm{r}}_1=\mathrm{m\lambda } \mathrm{m}=0,\pm 1,\pm 2,...$                                     ...(ii)    

So, here the longest wavelength is,

$\mathrm{\lambda }=120\mathrm{m}$ for m = 1.