# Question on how much intensity of light has been scattered

In summary: Unfortunately, I don't remember the equation for scattering at long distances, so I can't give you a precise answer. However, if you reduce the probability of scattering by the ratio (5/7)^4, you will get the correct answer of 167 nW.
Homework Statement
The intensity of light coming from a distant star is measured using two identical instruments A and B, where A is placed in a satellite outside the Earth's atmosphere and B is placed on the
Earth's surface. The results are as follows:

For green (500nm wavelength), intensity of light at A and B (in nW) is 100 and 50 respectively.
For red (700nm wavelength), intensity of light at A and B (in nW) is 200 and x respectively.

Assuming that there is scattering, but no absorption of light in the Earth's atmosphere at these wavelengths, the value of x can be estimated as:

Options:
(a) 177
(b) 167
(c) 157
(d) 147
(e) 137
Relevant Equations
(1) ##I = e\sigma T^4##
(2) ## \lambda T = constant = k##
I actually am not sure what equations are relevant here but I thought these are the relevant ones.

My Approach:
By Stefan-Boltzman Law, the intensity absorbed by the Earth is given as ## I = e \sigma T^4## where e is the emissivity of Earth, ##\sigma## is Stefan-Boltzman constant and T is the temperature of the Earth. The values of ##I## for green and red then are ## 50-100 = -50## and ##x-200## respectively.

Now this is where I am stuck. I am not sure if what I do next is valid: By using Wien's displacement law, ##I## may be given as ##I = \frac{e\sigma k }{\lambda^4} ##. Since the wavelengths are given, I simply divide the respective intensity and obtain

\frac{x-200}{-50} =\frac{500^4}{700^4}

Solving for ##x##, I get ##x \approx 187\ nW##. But this is not the right answer. The correct answer is 167 ##nW##.

Can someone please explain where I am going wrong?

.
Assuming that there is scattering, but no absorption of light in the Earth's atmosphere at these wavelengths, ...
.
Relevant Equations:: (1) ##I = e\sigma T^4##
(2) ## \lambda T = constant = k##
Can someone please explain where I am going wrong?
Since no one has replied yet, I will make a few comments.

The question has nothing to do with black-body radiation. So the formulae you are quoting are irrelevant. Also, the temperatures (T in your two equations (1) and (2)) are different temperatures - so it makes no sense to combine the two equations.

The question is about the scattering of light as it passes through the atmosphere; this reduces the intensity reaching ground-level. Neither the atmosphere nor the Earth act as a black-body absorber in this question.

It looks like the question is about the wavelength-dependence in Rayleigh scattering. You need to read-up about Rayleigh scattering before you attempt the question.

In addition, it would be worth revising-black-body radiation so that you understand the meanings of your equations (1) and (2) in case you ever have a question which is actually about black-body radiation!

Delta2 and BvU
As @Steve4Physics points out, the question has nothing to do with Stefan-Boltzmann. By coincidence, though, Rayleigh scattering also involves λ4 and so leads to (5/7)4.
But I could not understand your logic with the 50, 100 and 200. What fraction of the green light was scattered? How is that fraction adjusted by the (5/7)4? What fraction does that lead to for the red light?

That said, I don't get 167 either, so maybe it is not Rayleigh scattering, or perhaps some subtlety I am missing.

Steve4Physics and Delta2
haruspex said:
That said, I don't get 167 either, so maybe it is not Rayleigh scattering, or perhaps some subtlety I am missing.

You can't just multiply the light lost by (5/7)^4. The cross-section for scattering is reduced by (5/7)^4. The intensity falls off as $e^{-\alpha x}$ where x is the distance and α is the probablilty of being scattered. If you reduce α by the ratio (5/7)^4, you in fact get the answer of 167.

Steve4Physics, rsk and BvU
phyzguy said:
You can't just multiply the light lost by (5/7)^4. The cross-section for scattering is reduced by (5/7)^4. The intensity falls off as $e^{-\alpha x}$ where x is the distance and α is the probablilty of being scattered. If you reduce α by the ratio (5/7)^4, you in fact get the answer of 167.
Ah, yes of course. I should have figured that out.

## 1. How is the intensity of light scattered measured?

The intensity of light scattered can be measured using a variety of techniques, such as spectrophotometry, radiometry, or photometry. These methods involve measuring the intensity of the scattered light at different wavelengths or angles to determine the amount of light that has been scattered.

## 2. What factors affect the intensity of light scattered?

The intensity of light scattered can be affected by several factors, including the size, shape, and composition of the scattering particles, the wavelength of the incident light, and the concentration of the scattering particles in the medium.

## 3. How does the intensity of light scattered relate to the concentration of scattering particles?

The intensity of light scattered is directly proportional to the concentration of scattering particles in the medium. This means that as the concentration of particles increases, the intensity of scattered light also increases.

## 4. Can the intensity of light scattered be used to determine the size of scattering particles?

Yes, the intensity of light scattered can be used to estimate the size of scattering particles. This is because the intensity of scattered light is inversely proportional to the size of the particles, meaning that larger particles will scatter light with lower intensity compared to smaller particles.

## 5. How is the intensity of light scattered used in scientific research?

The intensity of light scattered is used in a variety of scientific research fields, such as atmospheric science, material science, and biology. It can provide valuable information about the properties of scattering particles and their interactions with light, which can help researchers understand various phenomena and develop new technologies.

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