Can Initial Intensity Differ from Incident Intensity?

[vivekfan]

Homework Statement

In the equation Beta=10log (I/I (initial)

is the I inital always the incident intensity 10^-12? Or if you are moving from one intensity to another, then can the initial intensity be the one you were initially at? For example, if you were moving from an intensity of 5 to 10, then could the initial be 5 instead of the incident intensity? If so, why?

Homework Equations

Given above.

The Attempt at a Solution

Well, I actually solved a problem in which something went from an initial intensity to a final, and plugged in the initial intensity and got the right answer, but I am confused on why you don't always use incident intensity?

 

[CompuChip]

In principle you should always use the reference intensity of I0 = 10^(-12).
However, suppose you want to calculate the difference in sound level when going from intensity Ia to Ib. Then mathematics tells you that

10log( Ia / I0 ) - 10log( Ib / I0 ) = 10log( (Ia / I0) / (Ib / I0) ) = 10log( Ia / Ib )

 

[vivekfan]

In principle you should always use the reference intensity of I0 = 10^(-12).
However, suppose you want to calculate the difference in sound level when going from intensity Ia to Ib. Then mathematics tells you that

10log( Ia / I0 ) - 10log( Ib / I0 ) = 10log( (Ia / I0) / (Ib / I0) ) = 10log( Ia / Ib )

Well, could you help me work through this problem...

A speaker is producing a total of 5 W of sound, and you hear 10 dB. Someone turns up the
power to 50 W. What level of sound do you hear?
(a) 10 dB, (b) 15 dB, (c) 100 dB, (d) 40 dB, (e) 20 dB

I don't know why I plugged in 50/5 for I and I initial after I used the math function you gave. Please help.