Question

36) The distance to a point source is decreased by a factor of three. (a) By what multiplicative factor does the intensity increase? (b) By what additive amount does the intensity level increase?

Answer 1

Given,

The distance of the point source is decreased by a factor of 3 times.

That is,

$d^{\prime}=\frac{d}{3}$

Where d' is the new distance and the d is the old distance.

(a)

The intensity is given by,

$I=\frac{P}{\pi d^2}$

Where P is the power.

Thus the intensity after the distance is decreased is,

$\begin{gathered} I^{\prime}=\frac{P}{\pi(d^{\prime})^2} \\ =\frac{P}{\pi(\frac{d}{3})^2} \\ =\frac{P}{\frac{\pi d^2}{9}} \\ =\frac{9P}{\pi d^2} \\ =9I \end{gathered}$

Therefore, the intensity is increased by a multiplicative factor of 9.

(b)

The intensity level when the distance is d is given by,

$\beta=10dB\times\log (\frac{I}{I_0})$

The intensity level when the distance id d' is given by,

$\beta^{\prime}=10dB\times\log (\frac{I^{\prime}}{I_0})$

On substituting I'=9I in the above equation,

$\begin{gathered} \beta^{\prime}=10dB\times\log (\frac{9I}{I_0}) \\ =10dB\times\lbrack\log 9+\log (\frac{I}{I_0})\rbrack \\ =10dB\times\log 9+10dB\times\log (\frac{I}{I_0}) \\ =10dB\times\log 9+\beta \\ =9.54+\beta \end{gathered}$

Thus the intensity level increase by an additive amount of 9.54