Question

The ratio of intensities between two coherent sound sources is $9:1$. The difference of loudness in decibels $\left(\text{dB}\right)$ between maximum and minimum intensities when they interfere in space is

• A. $10\log 4$
• B. $10\log 9$
• C. $10\log 3$
• D. $\log 3$

Answer 1

The correct option is A $10\log 4$

Given that,
$\frac{I_1}{I_2}=\frac{9}{1}\Rightarrow \frac{\sqrt{I_1}}{\sqrt{I_2}}=3$
From formula
$\frac{I_{max}}{I_{min}}={\left[\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right]}^2$
i.e. $\frac{I_{max}}{I_{min}}={\left[\frac{\frac{\sqrt{I_1}}{\sqrt{I_2}}+1}{\frac{\sqrt{I_1}}{\sqrt{I_2}}-1}\right]}^2={\left[\frac{3+1}{3-1}\right]}^2={\left(\frac{4}{2}\right)}^2=4$

So, sound level for $I_{max}$, ${\beta }_{max}=10\log \left(\frac{I_{max}}{I_0}\right)$
and ${\beta }_{min}=10\log \left(\frac{I_{min}}{I_0}\right)$
So,
${\beta }_{max}-{\beta }_{min}=10\log \left(\frac{I_{max}}{I_0}\right)-10\log \left(\frac{I_{min}}{I_0}\right)$
$=10\log (\frac{I_{max}}{I_0}\times \frac{I_0}{I_{min}})$
$=10\log \left(\frac{I_{max}}{I_{min}}\right)=10\log 4$