Re: Optics B/C
Posted: January 3rd, 2018, 5:55 am
We should start this back up...
Anyone want to ask a question?
Anyone want to ask a question?
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."
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
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.
Good, your turn.Pettywap wrote:violet