The intensity of light $I ( d )$ at a depth d below the surface of a body of water changes at a rate proportional to I.If the intensity at a depth of $d = 2.4$ feet is half of the surface intensity $I _ { 0 }$ to the nearest tenth of a foot,at what depth is the intensity $15 \%$ of $I _ { 0 } ?$
A)6.6 feet
B)9.9 feet
C)13.2 feet
D)7.9 feet

Question

Quizplus · Accepted Answer

The intensity of light $I(d)$ decreases exponentially with depth, following the formula $I(d) = I_0 e^{-kd}$, where $k$ is a constant. Given that at $d = 2.4$ feet, $I(d) = 0.5I_0$, we can find $k$ by solving $0.5 = e^{-2.4k}$. To find the depth at which the intensity is $15\%$ of $I_0$, we solve $0.15 = e^{-kd}$ using the same $k$. Solving these equations leads to a depth of approximately 6.6 feet for the intensity to be $15\%$ of the surface intensity.

The Intensity of Light I(d)I ( d )I(d) At a Depth D Below the Surface of a Body

The Intensity of Light $I ( d )$ At a Depth D Below the Surface of a Body