Respuesta :
Answer:
The intensity of the beam 16 feet below the surface is 0.06% of the initial intensity of the incident beam.
Step-by-step explanation:
"When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional to I(t), where t represents the thickness of the medium (in feet).
This means that the intersity can be modeled by the following differential equation:
[tex]\frac{dI(t)}{dt} = -kt[/tex]
In which k is the decrease rate.
This differential equation leads to the following solution:
[tex]I(t) = I(0)e^{-kt}[/tex]
In which I(0) is the initial intensity.
The intensity 3 feet below the surface is 25% of the initial intensity I0 of the incident beam.
This means that I(3) = 0.25I(0). We use this to find k. So
[tex]I(t) = I(0)e^{-kt}[/tex]
[tex]0.25I(0) = I(0)e^{-3k}[/tex]
[tex]e^{-3k} = 0.25[/tex]
[tex]\ln{e^{-3k}} = \ln{0.25}[/tex]
[tex]-3k = \ln{0.25}[/tex]
[tex]k = -\frac{\ln{0.25}}{3}[/tex]
[tex]k = 0.4621[/tex]
So
[tex]I(t) = I(0)e^{-0.4621t}[/tex]
What is the intensity of the beam 16 feet below the surface
This is I(16). So
[tex]I(16) = I(0)e^{-0.4621*16} = 0.0006I(0)[/tex]
The intensity of the beam 16 feet below the surface is 0.06% of the initial intensity of the incident beam.
