exponential decay HELP!!?

Question

The intesity of light falls exponentially with depth in water. Suppose that at a depth of 25 ft 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?

I've been working on it for about 5 hours and im clueless

yes, it is my homework, but i have also been working at it since about 11 o'clock last night and it is now 5 in the morning. that is why i resorted to this website. however, i FINALLY figured it out.

Answer 1

Exponential Functions



Examples of exponential growth and decay

Exponential Decay – atmospheric pressure



Atmospheric pressure as a function of altitude is also represented as an exponential decay function. For every 1000m increase of elevation, the atmospheric pressure is reduced by 11.5%.



Exponential Functions



Examples of exponential growth and decay

Exponential Decay – absorption of light in water



In lake and sea water, plant life can only exist in the top 10m or so, since daylight is gradually absorbed by the water. The light intensity as a function of depth of water through which the light must pass is modeled as exponential decay. (This is the Bouguer-Lambert law.) The rate of absorption depends on the purity of the water and the wavelength of the light beam.

There are several ways in which the law can be expressed:



where,



Here:

A is absorbance
I0 is the intensity of the incident light
I1 is the intensity after passing through the material
l is the distance that the light travels through the material (the path length)
c is the concentration of absorbing species in the material
α is the absorption coefficient or the molar absorptivity of the absorber
λ is the wavelength of the light
k is the extinction coefficient
In essence, the law states that there is a logarithmic dependence between the transmission of light through a substance and the concentration of the substance, and also between the transmission and the length of material that the light travels through. Thus if l and α are known, the concentration of a substance can be deduced from the amount of light transmitted by it.

The units of c and α depend on the way that the concentration of the absorber is being expressed. If the material is a liquid, it is usual to express the absorber concentration c as a mole fraction i.e. a dimensionless fraction. The units of α are thus reciprocal length (e.g. cm−1). In the case of a gas, c may be expressed as a density (units of reciprocal length cubed, e.g. cm−3), in which case α is an absorption cross-section and has units of length squared (e.g. cm2). If concentration c is expressed in moles per unit volume, α is a molar absorptivity (usually given the symbol ε) in units of mol−1 cm−2 or sometimes L mol−1 cm−1.

The value of the absorption coefficient α varies between different absorbing materials and also with wavelength for a particular material. It is usually determined by experiment.

In spectroscopy and spectrophotometry, the law is almost always defined in terms of common logarithms and powers of 10 as above. In general optics, the law is often defined in an alternate exponential form:



The values of α′ and A′ are approximately 2.3 (≈ln 10) times larger than the corresponding values of α and A defined in terms of base-10 functions. Therefore, care must be taken when interpreting data that the correct form of the law is used.

The law tends to break down at very high concentrations, especially if the material is highly scattering. If the light is especially intense, nonlinear optical processes can also cause variances.


[edit] Derivation
Assume that particles may be described as having an area, α, perpendicular to the path of light through a solution, such that a photon of light is absorbed if it strikes the particle, and is transmitted if it does not.

Define z as an axis parallel to the direction that photons of light are moving, and A and dz as the area and thickness (along the z axis) of a 3-dimensional slab of space through which light is passing. We assume that dz is sufficiently small that one particle in the slab cannot obscure another particle in the slab when viewed along the z direction. The concentration of particles in the slab is represented by c.

It follows that the fraction of photons absorbed when passing through this slab is equal to the total opaque area of the particles in the slab, αAc dz, divided by the area of the slab, or αc dz. Expressing the number of photons absorbed by the slab as dIz, and the total number of photons incident on the slab as Iz, the fraction of photons absorbed by the slab is given by


The solution to this simple differential equation is obtained by integrating both sides to obtain Iz as a function of z


For a slab of real thickness, ℓ, the difference in light intensity I0 at z = 0, and I1 at z = ℓ, is given by


or



It is instructive to consider the consequences of error in the assumption that one particle in a slab cannot obscure another particle in the slab. Implicit in the integration step is an extension of this assumption, namely that one particle cannot obscure another particle in any other slab. This assumption can only approach accuracy, of course, in very dilute solutions, and it becomes increasingly inaccurate with increasingly concentrated solutions. In practice, the accuracy of the assumption is better than the accuracy of most spectroscopic measurements up to an absorbance of 1 (or : .

Go to wikipedia and enter Beer-Lambert Law to see the equations

Answer 2

Ok, first of all well done for asking the question with a clear style of English, many of the people on here who ask Physics questions have a lot to learn from you....lol.
I=I0exp(-px) thats the physical law for what you are describing, in which x is the distance below the surface, p is a constant for the medium in question and is measured in inverse distance units, lets say ft^-1 so to make it easier, actually doesn't matter. I0 is the intensity of the light striking the surface, i.e. the noon day Sun and I is the intensity at a distance of x ft. The exponential shows that the relation between the two intensities is an exponential decay and the minus sign demonstrates that its a decay process. We don't need to know the absolute value of the intensity of the noon-day Sun, since intensities are given with respect to it, i.e as a fraction or ratio much like the 15% you give in the question for a depth of 25ft.
Therefore the equation becomes:
I/I0=exp(-px).......we need to determine p from the initial situation that you specify in the question, i.e. a 15% intensity drop after a depth of 25ft has been reached.
=> putting in the constants, 0.85=exp(-p(25ft))
=>-ln[0.85]=p(25ft) we are going to express p in inverse feet
=>0.1625=p(25ft)=>p=0.1625/25 ft^-1=0.006501ft^-1
Now we have p, we should proceed to calculate the depth x at which I/I0 is 0.000001,
=> 0.000001=exp(-0.006501x)
=> ln(0.000001)/[-0.006501]=x
=> x=2125.1362ft
HOPE THAT HELPS!!!!

Answer 3

It is quite easy really.

Let the number of ft required be x
If the original light intensity is 1, after 25 ft it becomes
(1- 0.15) = 0.85

At depth x ft the intensity is
0.85 ^ x/25 = 1/ 300,000

Taking logs,
x/25 log 0.85 = log1 - log 300,000 = - log 300,000
(Note that log 1 is zero).
So, x = - 25 log 300,000/ log 0.85
= 1940 ft

Answer 4

If the intensity falls exponentially with depth then

light intensity = intensity at surface * exp (-gamma * depth)

Also

fraction of maximum intensity = exp (-gamma * depth)

if 15% is absorbed at 25 ft

then

0.85 = exp (-gamma *25)

so gamma = ln(0.85) / (-25)

once you have gamma you can work out the depth for the second part.

1/300,000 = exp(-gamma * depth)

put in numbers in your calculator and you should get the answer

Answer 5

Exponential decay is expressed as :
y (light remaining) = exp(constant x depth)
so if at 25ft, 85% is left. 0.85 = exp(c * 25) gives c=-0.0065
y = 1/300000 occurs when 1/300000 = exp(-0.0065 * x)
x = 1940 feet

Answer 6

Cant help but think we are doing your homework lol