Question on how much intensity of light has been scattered

[Nirmal Padwal]

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

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?

 

[Steve4Physics]

.
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!

 

[haruspex]

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.

 

[phyzguy]

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.

 

[haruspex]

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.