Optics B/C

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Re: Optics B/C
We should start this back up...
Anyone want to ask a question?
Anyone want to ask a question?
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Re: Optics B/C
Can anyone draw a real diagram for this?
"Two thin converging lenses each with a focal length of 40 cm are positioned with 40 cm apart. An object is placed 60cm to the left of the first lens."
"Two thin converging lenses each with a focal length of 40 cm are positioned with 40 cm apart. An object is placed 60cm to the left of the first lens."
"Who's Fettywap?"

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Re: Optics B/C
I probably didn't draw this right (and I don't know how to attach an image to this post), but tell me if this link worksPettywap wrote:Can anyone draw a real diagram for this?
"Two thin converging lenses each with a focal length of 40 cm are positioned with 40 cm apart. An object is placed 60cm to the left of the first lens."
https://drive.google.com/file/d/0B24fq_ ... sp=sharing

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Re: Optics B/C
ngl I don't really know either but I thought that the very first line you drew was correct. I asked this question because I don't really know all of the rays.ClaireAndreasen wrote:I probably didn't draw this right (and I don't know how to attach an image to this post), but tell me if this link worksPettywap wrote:Can anyone draw a real diagram for this?
"Two thin converging lenses each with a focal length of 40 cm are positioned with 40 cm apart. An object is placed 60cm to the left of the first lens."
https://drive.google.com/file/d/0B24fq_ ... sp=sharing
"Who's Fettywap?"

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Re: Optics B/C
Hey I have a question:
Suppose the density of air in the atmosphere can be approximated by where d is density in kg/m^3 and h is altitude above sea level in km. The optical depth of a column of air at some location 15 km above sea level is taken to be 1.00. Find the opacity of that column of air in m^2/kg (assume opacity is constant).
And follow up question: If the sun shines with intensity 1.300 kw/m^2 at that same location 10 km above sea level, what will its intensity be when it reaches sea level?
Suppose the density of air in the atmosphere can be approximated by where d is density in kg/m^3 and h is altitude above sea level in km. The optical depth of a column of air at some location 15 km above sea level is taken to be 1.00. Find the opacity of that column of air in m^2/kg (assume opacity is constant).
And follow up question: If the sun shines with intensity 1.300 kw/m^2 at that same location 10 km above sea level, what will its intensity be when it reaches sea level?

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Re: Optics B/C
Tom_MS wrote:Hey I have a question:
Suppose the density of air in the atmosphere can be approximated by where d is density in kg/m^3 and h is altitude above sea level in km. The optical depth of a column of air at some location 15 km above sea level is taken to be 1.00. Find the opacity of that column of air in m^2/kg (assume opacity is constant).
And follow up question: If the sun shines with intensity 1.300 kw/m^2 at that same location 10 km above sea level, what will its intensity be when it reaches sea level?
Point of confusion: The equation given for density doesn't make sense dimensionally, so I'll modify the equation to log(d * 1 m^3/kg) = 3h/50 km. opacity (in m^2/kg) = attenuation coefficient (in m^1) / density (in kg/m^3) density (in kg/m^3) = 10^(3h/50 km) * 1 kg/m^3 attenuation coefficient (in m^1) = opacity (in m^2/kg) * 10^(3h/50 km) * 1 kg/m^3 1 = [math]\int_{0 km}^{15 km} \textrm{attenuation coefficient} (m^{1}) dh = \textrm{opacity} (\frac{m^2}{kg}) * 1 \frac{kg}{m^3} * \int_{0 km}^{15 km} 10^(\frac{3h}{50 km}) dh[/math] = according to WolframAlpha, 6.327 kgkm/m^3 * opacity (in m^2/kg) = 6327 kg/m^2 * opacity (in m^2/kg) 1 = 6327 kg/m^2 * opacity (in m^2/kg) opacity = 1/(6327 kg/m^2) = [b]0.000158 m^2/kg[/b] I = 1.300 kW/m^2 * exp(0.000158 m^2/kg * density * 10000 m). Let's use the value .542 kg/m^3 for density (mean of all of the densities between 0 km and 10 km) = 1.300 kW/m^2 * exp(0.000158 m^2/kg * .542 kg/m^3 * 10000 m) = [b]0.657 kW/m^2[/b] Not sure what I just did.

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Re: Optics B/C
Very well done! I started with the equation , then substituted in the equation for d, brought the over to the left and integrated. Optical depth would then be equal to the negative of the resulting on the left hand side. I then solved for opacity.
On the second part, I did a similar procedure except I solved for using the found value for opacity. You instead equivalently used the mean density over the column and multiplied it by the change in height. Very good! Unfortunately, I believe you miscalculated on your very last line the final answer for the second part.
Alright, your turn!
On the second part, I did a similar procedure except I solved for using the found value for opacity. You instead equivalently used the mean density over the column and multiplied it by the change in height. Very good! Unfortunately, I believe you miscalculated on your very last line the final answer for the second part.
Alright, your turn!

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