When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional toI(t), where t represents the thickness of the medium (infeet). In clear seawater, the intensity 3 feet below thesurface is 25% of the initial intensity I0 of theincident beam.

What is the intensity of the beam 15 feet below the surface?

The intensity of the beam at 15feet below the surface is 0.098%Io

Step-by-step explanation:

The differential equation for the intensity of beam of light is given by: dI/dt= KI

dI/I = Kdt

Integrating both sides

Integral dI/I = integral Kdt

Ln/I/ = Kt + c1

I = e^(Kt + c1)

I= e^ it e^c1

I = Ce^it

At I(0) = Io

Io = C = I = Io e^ kt

At 3ft below surface

0.25 Io = Io e^3k

0.25 = e^3k

3k = Ln(0.25)

K = Ln(0.25)/3

K= 0.4621

I = Ioe^(-0.4621)

At 15 ft

I = Ioe^(-0.4621 ×15)

I= Io e^-0.69315

I = 0.00098Io

I= 0.098% Io

i dont know and i dont care for math bc it is a stupied subject

step-by-step explanation:

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1