Light scattering homework question (rayleign scattering)

AI Thread Summary
Infrared telescopes can observe deeper into star-forming regions due to less scattering of infrared light compared to visible light. The homework question involves calculating the wavelength of light where scattering is only 2% of that at 490nm. The correct approach involves using the formula I(490)/IR=(lambda/490nm)^4 and equating it to 0.02 to solve for lambda. A misunderstanding in calculations led to an incorrect result, highlighting the importance of proper equation manipulation. The discussion emphasizes the significance of infrared wavelengths, which range from 700nm to 1mm.
haitianstudent
Messages
3
Reaction score
0

Homework Statement


Infrared telescopes, which use special infrared detectors, are able to peer farther into star-forming regions of the galaxy because infrared light is not scattered as strongly as is visible light by the tenuous clouds of hydrogen gas from which new stars are created.


Homework Equations



For what wavelength of light is the scattering only 2.00% that of light with a visible wavelength of 490nm ?

The Attempt at a Solution


the formula is I(490)/IR=(lamda/490nm)^4.
I'm really not sure what I am doing wrong here
 
Physics news on Phys.org
Did you solve for lamda? What did you do?
 
I first took lamda and 490 to the 4th power and I got (lamda)^4/5.77*10^10. Then I crossed multiply by 200, I then recived 1.15E13...and I took the fourth root of that and received 1841 but its apparently wrong
 
haitianstudent said:
the formula is I(490)/IR=(lamda/490nm)^4.
2% is not = 1 / 200

The formula is correct. Equate it to 0.02 and solve for the sole unknown, viz., λ

Are you told the correct answer?
 
Note infrared is 700nm to 1mm.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Light scattering (rayleigh scattering)
Replies
5
Views
14K
The thickness of glass and the longest wavelength
Replies
1
Views
4K
Help Understanding Light/Atom Thoeries
Replies
3
Views
2K
The masking effect of the sky scattering of visible light
Replies
3
Views
2K
Time for light to reach earth homework
Replies
3
Views
4K
B Questions about the Eddington Limit and Super - Eddington black holes
Replies
0
Views
1K
B Infrared Detectors & The 2nd Law of Thermodynamics
4
Replies
152
Views
9K
Spectrum of Light from Hydrogen
Replies
1
Views
1K
How does atmospheric scattering explain the faintness of starlight?
Replies
16
Views
3K
Question regarding temperature of a star and its solar system/planets
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top