Question

If the intensity is increased by a factor of 20, by how many decibels is the intensity level increased:

Answer 1

Hint Intensity is defined as power per unit area carried out by a wave. We know the intensity level of a wave is given as \[10\log \left( {\dfrac{I}{{{I_o}}}} \right)\] Where ‘I’ is the intensity of wave & \[{I_o}\] is zero level intensity. This intensity level is also known as the loudness of the source and is measured in terms of decibel ( dB ).

Complete Step by step solution
Let ‘L’ be the intensity level of the wave
We know intensity level ‘L’ is given by \[10\log \left( {\dfrac{I}{{{I_o}}}} \right)\]
Therefore, \[L = 10\log \left( {\dfrac{I}{{{I_o}}}} \right)\] (1)
If the intensity is increased by a factor of 20, then we will get
 \[{I^f} = 20I\]
And intensity level \[{L^f} = 10\log \left( {\dfrac{{{I^f}}}{{{I_o}}}} \right) = 10\log \left( {\dfrac{{20I}}{{{I_o}}}} \right)\] (2)
From (1) and (2) we get
 \[{L^f} - L = 10\log \left( {\dfrac{{20I}}{{{I_o}}}} \right) - 10\log \left( {\dfrac{I}{{{I_o}}}} \right) = 10\log \left( {20} \right) = 13dB\]

Note Increasing the intensity of the sound wave increases the loudness of the sound. The intensity of a sound wave is increased by increasing the amplitude of the wave. In Fact the intensity is proportional to square of the amplitude.

Additional Information
If two waves pass through the same point, their amplitudes add up vectorially. Accordingly the intensity adds up.
 \[{I_{final}} \ne {I_1} + {I_2}\]
 \[{I_{final}} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \theta \]
Intuitively one would say that the intensities of the two waves get added up. But this is incorrect. As in the above equation, one can clearly see that there is this extra term \[2\sqrt {{I_1}{I_2}} \cos \theta \] .