Question

The intensity level of two sounds are $100{\text{ }}dB{\text{ }}and{\text{ }}50{\text{ }}dB$. What is the ratio of their intensities?

A. ${10^1}$
B. ${10^3}$
C. ${10^{10}}$
D. ${10^5}$

Answer 1

Loudness is a parameter defined as the ten times the logarithmic value of intensity of the sound.
$L = 10{\log _{10}}(I)$
To find the ratio of two sound sources we can not simply take their ratio, instead we need to convert them to their corresponding loudness and then follow the required steps.

Complete step by step answer:
We are given with two sounds, first one is of $100dB$and second one is of $50{\text{ }}dB.$

$${L_1} = 100dB \\\ {L_2} = 50dB \\\$$

As we know that loudness is the parameter which needs to be compared, by definition, so we Convert above intensities to loudness, that is,

$${L_1} = 10{\log _{10}}({I_1}) \\\ \& \\\ {L_2} = 10{\log _{10}}({I_2}) \\\$$

Subtracting ${L_2}$ from ${L_1}$, we’ve,

$${L_1} - {L_2} = 10{\log _{10}}({I_1}) - 10{\log _{10}}({I_2}) = 10{\log _{10}}(\dfrac{{{I_1}}}{{{I_2}}}) \\\ 100 - 50 = 10{\log _{10}}(\dfrac{{{I_1}}}{{{I_2}}}) \\\ {\log _{10}}(\dfrac{{{I_1}}}{{{I_2}}}) = 5 \\\ \dfrac{{{I_1}}}{{{I_2}}} = {10^5} \\\$$

We, find that the ratio of their intensity is $10^5$, hence the correct option is $\left( D \right),$

Note:
Intensity and loudness are basically the same concept, but they are mathematically defined differently.
As silly as it may sound, there's an equivalent concept for Electromagnetic waves as well. Called as EM flux density.