## Optics B/C

Froggie
Member
Posts: 302
Joined: June 19th, 2017, 2:12 pm
Division: B
State: PA
Location: See above.

### Re: Optics B/C

We should start this back up...
Anyone want to ask a question?
"A lot of people have quotes in their signature. Maybe I should have a quote in my signature. "
- Froggie

Pettywap
Member
Posts: 109
Joined: November 4th, 2017, 11:15 am
Division: C
State: PA

### 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."
"Who's Fettywap?"

ClaireAndreasen
Member
Posts: 2
Joined: December 30th, 2017, 7:13 pm
Division: B
State: DE

### 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."
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 works

Pettywap
Member
Posts: 109
Joined: November 4th, 2017, 11:15 am
Division: C
State: PA

### 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."
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 works
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.
"Who's Fettywap?"

Tom_MS
Member
Posts: 41
Joined: April 28th, 2015, 11:08 am
State: PA

### Re: Optics B/C

Hey I have a question:
Suppose the density of air in the atmosphere can be approximated by $log(d)=-3h/50$ 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?

UTF-8 U+6211 U+662F
Exalted Member
Posts: 1518
Joined: January 18th, 2015, 7:42 am
Division: C
State: PA

### Re: Optics B/C

Hey I have a question:
Suppose the density of air in the atmosphere can be approximated by $log(d)=-3h/50$ 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 = $\int_{0 km}^{15 km} \textrm{attenuation coefficient} (m^{-1}) dh = \textrm{opacity} (\frac{m^2}{kg}) * 1 \frac{kg}{m^3} * \int_{0 km}^{15 km} 10^(\frac{-3h}{50 km}) dh$ = 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.

Tom_MS
Member
Posts: 41
Joined: April 28th, 2015, 11:08 am
State: PA

### Re: Optics B/C

Very well done! I started with the equation $dI = k*d*I*dh$, then substituted in the equation for d, brought the $I$ over to the left and integrated. Optical depth would then be equal to the negative of the resulting $ln(If/I0)$ on the left hand side. I then solved for opacity.
On the second part, I did a similar procedure except I solved for $If$ 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.

UTF-8 U+6211 U+662F
Exalted Member
Posts: 1518
Joined: January 18th, 2015, 7:42 am
Division: C
State: PA

### Re: Optics B/C

What color has the highest deviation in a prism?

Pettywap
Member
Posts: 109
Joined: November 4th, 2017, 11:15 am
Division: C
State: PA

### Re: Optics B/C

violet
"Who's Fettywap?"

UTF-8 U+6211 U+662F
Exalted Member
Posts: 1518
Joined: January 18th, 2015, 7:42 am
Division: C
State: PA

violet