Star light extinction and optical extinction

In summary: B.In summary, we can show that if the optical depth is proportional to frequency, then F(\lambda) = \frac{c_1}{\lambda} + c_2, where c_1 and c_2 are constants evaluated in terms of C, V, and B.
  • #1
TFM
1,026
0

Homework Statement



The normalized extinction is defined as

[tex] F(\lambda) = \frac{A_{\lambda}-A_v}{A_B - A_v} [/tex]

where [tex] A_{\lambda} [/tex] is the extinction (in magnitudes) at wavelength [tex] \lambda [/tex]. B and V are wavelength bands but for the purposes of this question we will
assume that they represent particular wavelengths with B = 420nm and V = 540 nm.

(a)

Show that if the optical depth is proportional to frequency (i.e. [tex] \tau = \frac{C}{\lambda} [/tex]), then [tex] F(\lambda) = \frac{c_1}{\lambda} + c_2 [/tex] where [tex]c_1[/tex] and [tex]c_2[/tex] are constants which should be evaluated in terms of C, V and B.

(b)

Hence sketch [tex] F(\lambda) [/tex] over the visible range, 300–800 nm.(c)

Find the asymptotic value of [tex] F(\lambda) as \lambda \leftharpoondown \infty [/tex] (i.e. the numerical value of [tex]c_2[/tex]).

Homework Equations



The equation of radiative transfer:

[tex] dI_{\nu} = (s_{\nu} - I_{\nu})d\tau_{\nu} [/tex]

[tex] s_{\nu} = j_{\nu}/\alpha_{\nu} [/tex]

[tex] \tau_{\nu} = \int \alpha_{\nu} ds [/tex]

The Attempt at a Solution



I am not quite sure where to start for part a). I know quite a few equations, which I have posted above. But could any tell me what is the best way to start this question?

Many Thanks,

TFM
 
Last edited:
Physics news on Phys.org
  • #2


Dear TFM,

To start, we can rewrite the equation for normalized extinction as:

F(\lambda) = \frac{A_{\lambda}-A_v}{A_B - A_v} = \frac{A_{\lambda}-A_{540}}{A_{420} - A_{540}}

We can then use the equation of radiative transfer to relate the extinction at different wavelengths to the optical depth:

A_{\lambda} = \int \alpha_{\lambda} ds = \int \alpha_{\lambda} \frac{ds}{d\tau_{\lambda}} d\tau_{\lambda} = \frac{1}{\alpha_{\lambda}} \int \alpha_{\lambda} d\tau_{\lambda}

Since we are assuming that the optical depth is proportional to frequency, we can write:

\alpha_{\lambda} = \frac{C}{\lambda}

Substituting this into the equation for extinction, we get:

A_{\lambda} = \frac{1}{\alpha_{\lambda}} \int \alpha_{\lambda} d\tau_{\lambda} = \frac{1}{C} \int C d\tau_{\lambda} = \frac{1}{C} \tau_{\lambda}

Therefore, we can rewrite the equation for normalized extinction as:

F(\lambda) = \frac{\frac{1}{C} \tau_{\lambda} - \frac{1}{C} \tau_{540}}{\frac{1}{C} \tau_{420} - \frac{1}{C} \tau_{540}} = \frac{\tau_{\lambda} - \tau_{540}}{\tau_{420} - \tau_{540}}

Since we are assuming that \tau \propto \lambda, we can write this as:

F(\lambda) = \frac{\frac{C}{\lambda} - \frac{C}{540}}{\frac{C}{420} - \frac{C}{540}} = \frac{C}{\lambda} \cdot \frac{540}{C} \cdot \frac{420}{540} = \frac{420}{\lambda}

Therefore, we can write F(\lambda) as:

F(\lambda) = \frac{420}{\lambda} + c_2

Where c_2 is a constant that can be evaluated in terms of C, V,
 
  • #3

To begin, we can use the equation of radiative transfer to relate the extinction (A_{\lambda}) to the optical depth (\tau_{\nu}). From the given information, we can assume that the wavelength bands B and V correspond to the blue and green parts of the visible spectrum, respectively. Therefore, we can rewrite the given equation as:

F(\lambda) = \frac{A_B-A_v}{A_B-A_v} = \frac{A_{420nm}-A_{540nm}}{A_{420nm}-A_{540nm}} = \frac{\tau_{420nm}-\tau_{540nm}}{\tau_{420nm}-\tau_{540nm}}

Next, we can use the given relationship between the optical depth and frequency (\tau = \frac{C}{\lambda}) to rewrite the equation as:

F(\lambda) = \frac{C/420nm - C/540nm}{C/420nm - C/540nm} = \frac{\frac{C}{420nm} - \frac{C}{540nm}}{\frac{C}{420nm} - \frac{C}{540nm}} = \frac{\frac{C}{420nm} - \frac{C}{540nm}}{\frac{C}{420nm} - \frac{C}{540nm}} = \frac{\frac{C}{420nm} - \frac{C}{540nm}}{\frac{C}{420nm} - \frac{C}{540nm}}

Now, we can factor out the constant C from the numerator and denominator, and we get:

F(\lambda) = \frac{C(\frac{1}{420nm} - \frac{1}{540nm})}{C(\frac{1}{420nm} - \frac{1}{540nm})} = \frac{\frac{1}{420nm} - \frac{1}{540nm}}{\frac{1}{420nm} - \frac{1}{540nm}}

Finally, we can simplify the fraction by combining the fractions in the numerator and denominator, and we get:

F(\lambda) = \frac{-\frac{120}{226800nm}}{-\frac{120}{226800nm}} = \frac{120}{120} = 1

Therefore, we have shown that if the optical depth is proportional to frequency, then F(\lambda) = 1, which
 

1. What is star light extinction?

Star light extinction refers to the decrease in brightness or intensity of light from a star as it passes through Earth's atmosphere or any other medium. This is due to various factors such as scattering, absorption, and diffraction of light by particles in the atmosphere.

2. How does star light extinction affect our view of the night sky?

Star light extinction can greatly affect our view of the night sky by making stars appear dimmer or even invisible. This is why astronomers often use specialized telescopes and instruments that can filter out the effects of extinction to get a clearer view of celestial objects.

3. What is optical extinction?

Optical extinction is a measure of how much light is lost as it passes through a medium, such as Earth's atmosphere. It is typically expressed in magnitudes, with higher values indicating more extinction. This is an important factor to consider when observing astronomical objects, as it can impact the accuracy of measurements and observations.

4. What causes optical extinction?

There are several factors that contribute to optical extinction, including atmospheric conditions, such as humidity, temperature, and air density. These can affect the scattering and absorption of light, leading to extinction. Other factors include the presence of particles, such as dust and pollutants, in the atmosphere.

5. How can we reduce the effects of star light extinction in astronomy?

To reduce the impact of star light extinction in astronomy, specialized instruments and techniques are used. These include adaptive optics, which can correct for atmospheric distortions in real-time, and using infrared or radio telescopes that are less affected by extinction. Astronomers also carefully choose observing locations with clear and dark skies to minimize the effects of extinction.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
922
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
792
  • Introductory Physics Homework Help
Replies
2
Views
945
  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
1K
Back
Top