Q.74PE - College Physics (Urone)

Question

Find the sound intensity level in decibels of \[{\rm{2 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ W}} \cdot {{\rm{m}}^{{\rm{ - 2}}}}\] ultrasound used in medical diagnostics.

Step-by-Step Solution

Answer

The sound intensity level will be 103 dB that used for given intensity in medical diagnostics.

Step 1: Formula for sound intensity level in dB ( decibels)

The sound intensity level (SIL) or sound intensity level is the level (logarithmic quantity) of sound intensity relative to a reference value.

 

Now change this sound intensity level (dB) \[W \cdot {m^{ - 2}}\] by using the following formula.\[\beta  = 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\]

Here, \[\beta \] is the sound intensity level in dB, I is the intensity (in \[W \cdot {m^{ - 2}}\]) of ultrasound wave and \[{I_0}\] is the threshold intensity of hearing.

Step 2: Sound intensity level in dB of ultrasound of intensity

Consider the given data as below.

The intensity of ultrasound wave         

\[I = 2 \times {10^{ - 2}}{\rm{ }}W \cdot {m^{ - 2}}\]

Threshold intensity of hearing is,

\[{I_0} = {10^{ - 12}}{\rm{ }}W \cdot {m^{ - 2}}\]

 

Now the sound intensity level is given as

\[\beta  = 10{\log _{10}}\left( {\frac{I}{{{I_0}}}} \right)\]

 

By replacing values of I and \[{I_0}\] you will get following.

\begin{aligned}\beta &= 10{\log _{10}}\left( {\frac{{2 \times {{10}^{ - 2}}}}{{{{10}^{ - 12}}}}} \right)\\ &= 10{\log _{10}}\left( {2 \times {{10}^{10}}} \right){\rm{ }}\\ &= 10 \times 10.3\\ &= 103{\rm{ }}dB\end{aligned}

Hence, the sound intensity level will be \[\beta  = 103{\rm{ }}dB\] that used in medical diagnostics.