✔ 最佳答案
Given b = 10 log ( I / I o) where Io is a constant & b1 = 105, b2 = 80
Le the intensity of b1 = Ib1, the intensity of b2 = Ib2 , so that:
105 = 10 log ( Ib1 / Io ) .............(1)
80 = 10 log ( Ib2 / Io ) ...............(2)
(1) - (2):
105 - 80 = 10 log ( Ib1 / Io ) - 10 log ( Ib2 / Io )
25 = 10 [log ( Ib1 / Io ) - log ( Ib2 / Io )]
25/10 = log [( Ib1 / Io ) / ( Ib2 / Io )]
5/2 = log ( Ib1 / Ib2 )
Ib1 / Ib2 = 105/2
Ib1 / Ib2
= 102 + (1/2) = 102101/2 =100√(10)
~ 316.23 ( corrected to 2 decimal places)
Therefore, the intensity of b1 = 105dB ( factory ) is ~ 316.23 times greater than the intensity of b2 = 80dB ( busy street ).