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Q.
If the intensity level of a sound wave is increased by factor 20 the corresponding change in decibels level of the sound would be
VMMC MedicalVMMC Medical 2015
Solution:
As we know that, $ \beta =10\log \left( \frac{{{I}_{2}}}{{{I}_{1}}} \right) $ If the actual intensity of sound is $ {{I}_{0}}. $ Then, when the intensity is I $ {{\beta }_{1}}=10\log \left( \frac{I}{{{I}_{0}}} \right) $ When the intensity is increased by a factor 20 Then, $ {{\beta }_{2}}=10\log \left( \frac{20I}{{{I}_{0}}} \right) $ Thus, change in decibels of sound $ {{\beta }_{2}}-{{\beta }_{1}}=10\log \left( \frac{20I}{{{I}_{0}}} \right)-10\log \left( \frac{I}{{{I}_{0}}} \right) $ $ =10\log \left( \frac{20I}{I} \right)=10\log 20 $ $ =13dB. $