When the intensity level (IL) is doubled, by how many decibels does the level increase?

When the intensity level (IL) is doubled, by how many decibels does the level increase?

• A. 1 dB
• B. 3 dB
• C. 6 dB
• D. 10 dB

Answer: (B)

When discussing sound intensity and its relationship to decibels, it is essential to understand how the decibel scale operates. The decibel scale is logarithmic, which means that an increase in intensity is not linear. Specifically, sound intensity increases logarithmically based on a reference intensity. When the intensity level is doubled, the increase in the decibel level is calculated using the formula for decibels, which states that a difference in intensity in decibels (dB) can be expressed as: \[ \text{dB} = 10 \log_{10} \left( \frac{I_2}{I_1} \right) \] Where \( I_2 \) is the new intensity and \( I_1 \) is the reference intensity. If \( I_2 \) is twice \( I_1 \) (i.e., when the intensity level is doubled), the calculation becomes: \[ \text{dB} = 10 \log_{10}(2) \] Calculating \( \log_{10}(2) \) gives approximately 0.301, so: \[ \text{dB} = 10 \times 0.301 = 3.01 \

When discussing sound intensity and its relationship to decibels, it is essential to understand how the decibel scale operates. The decibel scale is logarithmic, which means that an increase in intensity is not linear. Specifically, sound intensity increases logarithmically based on a reference intensity.

When the intensity level is doubled, the increase in the decibel level is calculated using the formula for decibels, which states that a difference in intensity in decibels (dB) can be expressed as:

[ \text{dB} = 10 \log_{10} \left( \frac{I_2}{I_1} \right) ]

Where ( I_2 ) is the new intensity and ( I_1 ) is the reference intensity. If ( I_2 ) is twice ( I_1 ) (i.e., when the intensity level is doubled), the calculation becomes:

[ \text{dB} = 10 \log_{10}(2) ]

Calculating ( \log_{10}(2) ) gives approximately 0.301, so:

[ \text{dB} = 10 \times 0.301 = 3.01 \