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Prove that DeltaL=10log_(10)[(I_(1))/(I_(0))] decibel, from Weber-Fechner's law. |
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Answer» Solution :According to Weber-Fechner's law, ''loudness (L) is proportional to the logarithm of the actual intensity (I) measured with an accurate non-human instrument''. It is meant that `LproplnI` `L=klnI` The difference between two loudnesses, `L_(1)andL_(0)` MEASURES the relative loudness between two precisely measured intensities and is called as sound intensity level. Mathematically, sound intensity level is `DeltaL=L_(1)-L_(0)=klnI_(1)-klnI_(0)` `=KLN[(I_(1))/(I_(0))]` If k=1, then sound intensity level is measured in bel, in HONOUR of Alexander Graham Bell. `:.DeltaL=ln[(I_(1))/(I_(0))]`bel However, this is practically a bigger unit, so we use a convenient smaller unit, called DECIBEL. Thus, decibel `=(1)/(10)bel`. Hence, by multiplying and dividing by 10 we get `DeltaL=10(ln[(I_(1))/(I_(0))])(1)/(10)bel` `DeltaL=10ln[(I_(1))/(I_(0))]` decibel with k=10 For practical purposes, we use logarithm to base 10 instead of natural logarithm, `DeltaL=10log_(10)[(I_(1))/(I_(0))]` decibel. |
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