First slide

Intensity of

Question

A two-fold increase in intensity of a wave implies an increase of (Given: $\log_{10} 2 = 0.3010$ .)

Moderate

A

2 dB
B

10 dB
C

3.01 dB
D

0.5 dB

Solution

$L = 10 \log_{10} \frac{I}{I_{0}} decibel$
$L^{'} = 10 \log_{10} \frac{2 I}{I_{0}} decibel$
$L^{'} - L = 10 [\log_{10} \frac{2 I}{I_{0}} - \log_{10} \frac{I}{I_{0}}]$
= 10 $\log_{10} 2 = 10 \times 0.3010 = 3.01 dB$

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