Solution

Given,

d₁ = 25.0 m,
d₂ = 2.0 km = 2000 m,
I₁ = 45.0 dB,
I₂ = ?

Let, I₁ and I₂ be the intensities on the ground when the airplane is at an elevation of 200 m and 25 m respectively.

we have,

\(\frac{I_2}{I_1}=\frac{d_1^2}{d_2^2}=1.56\times10^{-4}\)

here, the intensity level at 25 m, is

\(\beta_1=10\times\log_{10}\left(\frac{I_1}{I_0}\right)\) -------- (i)

And the intensity level at 2 km is

\(\beta_2=10\times\log_{10}\left(\frac{I_2}{I_0}\right)\) ---------- (ii)

Now, subtracting equation (i) from equation (ii) we get,

\(\beta_2-\beta_1=10\times\log_{10}\left(\frac{I_2}{I_0}\right)-10\times\log_{10}\left(\frac{I_1}{I_0}\right)\)

∴ \(\beta_2=\beta_1+10\times\log_{10}\left(1.56\times10^{-4}\right)=6.94\) dB