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

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

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:

`(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 = -(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.