Question

"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). In clear seawater, the intensity 3 feet below the surface is 25% of the initial intensity I0 of the incident beam. What is the intensity of the beam 16 feet below the surface

Answer 1

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:

$\frac{dI(t)}{dt} = -kt$

In which k is the decrease rate.

This differential equation leads to the following solution:

$I(t) = I(0)e^{-kt}$

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

$I(t) = I(0)e^{-kt}$

$0.25I(0) = I(0)e^{-3k}$

$e^{-3k} = 0.25$

$\ln{e^{-3k}} = \ln{0.25}$

$-3k = \ln{0.25}$

$k = -\frac{\ln{0.25}}{3}$

$k = 0.4621$

So

$I(t) = I(0)e^{-0.4621t}$

What is the intensity of the beam 16 feet below the surface

This is I(16). So

$I(16) = I(0)e^{-0.4621*16} = 0.0006I(0)$

The intensity of the beam 16 feet below the surface is 0.06% of the initial intensity of the incident beam.