Answer:
(a) [tex]I = 3.7\times 10^{-2}\ W/m^{2}[/tex]
(b) [tex]I = 1.9\times 10^{-2}\ W/m^{2}[/tex]
Explanation:
Given:
Point source [tex]P_{b}=1.2\ W[/tex]
Distance from source Part a [tex]L = 1.6\ m[/tex]
Distance from source Part b [tex]L = 2.2\ m[/tex]
Solution:
Part (a)
Using intensity formula at distance L from an isotropic point.
[tex]I = \frac{P_{b}}{4\pi(L)^{2}}[/tex] -------------------(1)
Substitute [tex]L = 1.6\ m[/tex] and [tex]P_{b}=1.2\ W[/tex] in equation 1..
[tex]I = \frac{1.2}{4\times 3.14(1.6)^{2}}[/tex]
[tex]I = \frac{1.2}{32.15}[/tex]
[tex]I = 0.037\ W/m^{2}[/tex]
[tex]I = 3.7\times 10^{-2}\ W/m^{2}[/tex]
Part (b)
Substitute [tex]L = 2.2\ m[/tex] and [tex]P_{b}=1.2\ W[/tex] in equation 1.
[tex]I = \frac{1.2}{4\times 3.14(2.2)^{2}}[/tex]
[tex]I = \frac{1.2}{60.79}[/tex]
[tex]I = 0.019\ W/m^{2}[/tex]
[tex]I = 1.9\times 10^{-2}\ W/m^{2}[/tex]
Therefore, the intensity at distance 1.6 m from the source:
[tex]I = 1.9\times 10^{-2}\ W/m^{2}[/tex].
And, the intensity at distance 2.2 m from the source:
[tex]I = 1.9\times 10^{-2}\ W/m^{2}[/tex].
