Final answer:
The ratio of the final sound intensity to the initial sound intensity when a door reduces the sound level by 36 dB is approximately 0.00025.
Step-by-step explanation:
When we talk about sound insulation and decibel (dB) reduction, we refer to a logarithmic scale used to measure the intensity of sound. In this case, a door reduces the sound level by 36 dB.
The relationship between sound intensity levels measured in decibels can be calculated using the formula:
\(L_2 = L_1 + 10 \log_{10}(I_2/I_1)\)
where \(L_2\) and \(L_1\) are the sound levels in dB, \(I_2\) and \(I_1\) are the respective sound intensities. Given that we're dealing with a reduction of 36 dB, we can rewrite the equation as:
\(-36 = 10 \log_{10}(I_2/I_1)\)
Dividing both sides by 10 and using the inverse of the logarithm to solve for the intensity ratio:
\(\log_{10}(I_2/I_1) = -3.6\)
\(I_2/I_1 = 10^{-3.6}\)
Which approximately equals \(0.00025119\). So the sound intensity behind the door is around 0.00025 times the intensity in front of it.