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"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

User Chizou
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Answer:

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.

User Vit Kos
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