Question

Two loudspeakers are being compared, and one is perceived to be 32 times louder than the other.
What will be the difference in intensity levels between the two?
(A) $3dB$
(B) $10dB$
(C) $25dB$
(D) $50dB$​

Answer 1

Loudness is a measure of the response of the ear to the sound. Its amplitude defines the loudness of a sound. Loudness is that sensation of hearing where the sound ranges from quiet to loud.
Sound intensity is the sum of sound energy that travels into our ears per second. These terms are often used interchangeably, so students need to be very cautious while solving such questions.

Formulae used:
$\beta = 10{\log _2}\left( {\dfrac{I}{{{I_0}}}} \right)$ where $\beta$ is the intensity of the sound, $I$ is the loudness of the sound and ${I_0}$ is a constant reference intensity ${10^{ - 12}}W{m^{ - 2}}$

Complete step by step solution:
According to the given question,
Let the intensity of the first loudspeaker be ${\beta _1}$
The intensity of the second loudspeaker becomes ${\beta _2}$
Let the loudness of the first loudspeaker be ${I_1} = I$
Since the second one is $32$ times louder than first one,
the loudness of the second loudspeaker becomes ${I_2} = 32I$
Thereafter, using the formula co relating loudness and frequency, we get
$\Rightarrow {\beta _1} = 10{\log _2}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right) \\\ \\\$ _________________(i)

Now, doing the same for the second loudspeaker, we have
$\Rightarrow {\beta _2} = 10{\log _2}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right) \\\ \\\$ ____________________(ii)
Now, subtracting equation (i) from equation (ii), we get
$\Rightarrow {\beta _2} - {\beta _1} = 10\left[ {{{\log }_2}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right) - {{\log }_2}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right)} \right] \\\ \Rightarrow \Delta \beta = 10{\log _2}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right) \\\$
Substituting the respective values in their respective places, we get
$\Rightarrow \Delta \beta = 10{\log _2}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right) \\\ \Rightarrow \Delta \beta = 10{\log _2}\left( {\dfrac{{32I}}{I}} \right) \\\ \Rightarrow \Delta \beta = 10{\log _2}\left( {32} \right) \\\ \Rightarrow \Delta \beta = 10{\log _2}\left( {{2^5}} \right) \\\ \Rightarrow \Delta \beta = 10 \times 5dB \\\ \Rightarrow \Delta \beta = 50dB \\\$
Hence, the intensity increases by $50dB$.

Note: In many places, people confuse loudness with intensity and often interchange those terms. To remember them correctly, we need to remember that $\beta$ is measured in $dB$ and $I$ is just perceived in factors. $\dfrac{I}{{{I_0}}}$ is a ratio and has no unit, whereas $\beta$ has a unit.