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)=(d)/(3)`

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

(a)

The intensity is given by,

`I=(P)/(\pi d^2)`

Where P is the power.

Thus the intensity after the distance is decreased is,

`\begin{gathered} I^(\prime)=(P)/(\pi(d^(\prime))^2) \\ =(P)/(\pi((d)/(3))^2) \\ =(P)/((\pi d^2)/(9)) \\ =(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*\log ((I)/(I_0))`

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

`\beta^(\prime)=10dB*\log ((I^(\prime))/(I_0))`

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

`\begin{gathered} \beta^(\prime)=10dB*\log ((9I)/(I_0)) \\ =10dB*\lbrack\log 9+\log ((I)/(I_0))\rbrack \\ =10dB*\log 9+10dB*\log ((I)/(I_0)) \\ =10dB*\log 9+\beta \\ =9.54+\beta \end{gathered}`

Thus the intensity level increase by an additive amount of 9.54