Question

For the same signal emitted by a radio antenna, Observer A measures its intensity to be $16$ times the intensity measured by Observer B. The distance of Observer A from the radio antenna is what fraction of the distance of Observer B from the radio antenna?

• A. $\frac{1}{4}$
• B. $\frac{1}{16}$
• C. $\frac{1}{64}$
• D. $\frac{1}{256}$

Answer 1

The correct option is A $\frac{1}{4}$

Given for both the Observers, signal emitted by the radio antenna is the same. Hence, $P$ is same for both $A$ and $B$
Now, for observer $A$ ; $I_A=\frac{P}{4\pi {r_A}^2}$
And for observer $B$ ; $I_B=\frac{P}{4\pi {r_B}^2}$
Given, $I_A=16I_B$
$\Rightarrow \frac{P}{4\pi {r_A}^2}=16\left(\frac{P}{4\pi {r_B}^2}\right)$
$\Rightarrow {r_B}^2=16{r_A}^2$
$\Rightarrow r_B=4r_A$
$\Rightarrow r_A=\frac{1}{4}{r}_B$