To solve the problem of finding the intensity at specified distances from the source, let's use the formula for intensity $I$ of a wave emitted isotropically by a point source:

$I = \frac{P}{4\pi r^2}$

Where:

• $I$ is the intensity (in W/m²),
• $P$ is the power of the source (1.7 W in this case),
• $r$ is the distance from the source (1.0 m for part (a) and 2.8 m for part (b)).

Solution:

Part (a): At $r = 1.0 \, \text{m}$

$I = \frac{1.7}{4 \pi (1.0)^2}$ $I = \frac{1.7}{4 \pi} \approx \frac{1.7}{12.566} \approx 0.1353 \, \text{W/m}^2$

Part (b): At $r = 2.8 \, \text{m}$

$I = \frac{1.7}{4 \pi (2.8)^2}$ $I = \frac{1.7}{4 \pi (7.84)} \approx \frac{1.7}{98.491} \approx 0.0173 \, \text{W/m}^2$

Final Answers:

(a) $I = 0.1353 \, \text{W/m}^2$
(b) $I = 0.0173 \, \text{W/m}^2$

Let me know if you need further details or explanations!

Additional Questions for Practice:

1. How does the intensity change if the power of the source is doubled?
2. What is the intensity at a distance of 5.0 m from the source?
3. What is the total energy passing through a sphere of radius 2.0 m per second?
4. How does the intensity relate to the inverse-square law?
5. If the medium absorbs some energy, how does the intensity at a point change?

Tip:

Always ensure to square the distance $r$ in intensity calculations, as small errors in this step can significantly affect the result.