Hard calculus problem?

Question

The fact that sunlight is absorbed by water is well known to any diver who has dived to a depth of 100 feet. It is also true that the intensity of light falls exponentially with depth. Suppose that at a depth of 25 feet the water absorbs 15% of the light that strikes the surface. At what depth would the light at noon be as bright as a full moon, which is one three-hundred-thousandth as bright as the noonday sun?


Depth in feet

Pascal you are right!!!! 5 stars

Answer 1

According to the problem, at a depth of 25 feet, 85% of the light remains. Since the light level falls exponentially, at x feet the light level is only (.85)^(x/25). We wish to find the point at which the light level is 1/300000 of the original. So:

(.85)^(x/25) = 1/300000
(x/25) log (.85) = log (1/300000)
x/25 = log (1/300000)/log (.85)
x=25 log (1/300000)/log (.85)
x ≈ 1940 ft

Edit: yeah, latter poster's right. Water absorbs way more light than that. Perhaps your book meant to say that the water absorbs all but 15% of the light that strikes the surface. In this case, we would have the equation (.15)^(x/25) = 1/300000, so x=25 log (1/300000)/log (.15) ≈ 166.2 ft.

Answer 2

Need to know your assumptions about the real world. The problem totally ignores scattering. Actually, at 10 meters (30 ft), 85% of the surface light has been absorbed or scattered. (not 15%) See http://www.punaridge.org/doc/factoids/Light/Default.htm

Answer 3

Exponential decay function

Let Io (I subscript nought) be the intensity of sunlight at the surface.
s = depth in feet

I(s) = Io e^(-ks)

I(25) = 0.15 Io = Io e^(-25k)
0.15 Io = Io e^(-25k)
0.15 = e^(-25k)
ln 0.15 = -25k
k = (ln 0.15) / (-25)

Using k,

I(x) = (1/300,000) Io = Io e^(-kx)
(1/300,000) Io = Io e^(-kx)
(1/300,000) = e^(-kx)
ln(1/300,000) = -kx
x = ln(1/300,000) / (-k)

Plug in you solution for k and you have your depth in feet.