What is the decibel level of the radio with intensity 10−7 watts per square inch? Use a logarithmic model to solve.

Question

asked Jul 21, 2022 131k views

1 Answer

Aman Virk · Answer 1 · 2022-07-26T19:20:41+0000

Answer:

D = 50

Explanation:

Given

I = 10^(-7)W/m^2 --- Intensity

Required

Determine the decibel level (D)

This is calculated as:

D = 10 * log((I)/(I_n))

Where:

I_n = The threshold intensity

I_n =1 * 10^(-12)W/m^2

So, we have:

D = 10 * log((I)/(I_n))

This gives:

D = 10 * log((10^(-7)W/m^2)/(1 * 10^(-12)W/m^2))

D = 10 * log((10^(-7))/(10^(-12)))

Apply law of indices

D = 10 * log(10^(-7--12))

D = 10 * log(10^(5))

Apply law of logarithm

$loga^b = b\ log(a)$

So, we have:

D = 10 * 5 * log(10)

log(10) =1

So:

D = 10 * 5 *1

D = 50

Hence, the decibel level is 50