A child sets off the firecracker at a distance of 100 m from the family house. what is the sound intensity β100 at the house?

To solve this problem, we use the formula:

I100 / I1 = [P / 4π (100m) ^2] / [P / 4π (1m) ^2]

I100 / I1 = 1 / 100^2

I100 / I1 = 10^-4

Therefore the change in intensity from 1m to 100m in decibels is:

B100 - B1 = 10 log (10^-4) dB = - 40 dB

So the intensity at 100m is calculated as:

B100 = B1 - 40 dB = 140 dB - 40 dB = 100 dB

Answer:

100 dB