Respuesta :
Given,
The distance of the point source is decreased by a factor of 3 times.
That is,
[tex]d^{\prime}=\frac{d}{3}[/tex]Where d' is the new distance and the d is the old distance.
(a)
The intensity is given by,
[tex]I=\frac{P}{\pi d^2}[/tex]Where P is the power.
Thus the intensity after the distance is decreased is,
[tex]\begin{gathered} I^{\prime}=\frac{P}{\pi(d^{\prime})^2} \\ =\frac{P}{\pi(\frac{d}{3})^2} \\ =\frac{P}{\frac{\pi d^2}{9}} \\ =\frac{9P}{\pi d^2} \\ =9I \end{gathered}[/tex]Therefore, the intensity is increased by a multiplicative factor of 9.
(b)
The intensity level when the distance is d is given by,
[tex]\beta=10dB\times\log (\frac{I}{I_0})[/tex]The intensity level when the distance id d' is given by,
[tex]\beta^{\prime}=10dB\times\log (\frac{I^{\prime}}{I_0})[/tex]On substituting I'=9I in the above equation,
[tex]\begin{gathered} \beta^{\prime}=10dB\times\log (\frac{9I}{I_0}) \\ =10dB\times\lbrack\log 9+\log (\frac{I}{I_0})\rbrack \\ =10dB\times\log 9+10dB\times\log (\frac{I}{I_0}) \\ =10dB\times\log 9+\beta \\ =9.54+\beta \end{gathered}[/tex]Thus the intensity level increase by an additive amount of 9.54
