Solving for Magnitude of Sirius (-1.5) Through Triple Window

  • Thread starter Thread starter Jussi Lundahl
  • Start date Start date
  • Tags Tags
    Magnitude
AI Thread Summary
The discussion revolves around calculating the apparent magnitude of Sirius (-1.5) as observed through a triple window that reflects away 15% of the incident light. The intensity of Sirius is reduced to 0.614 after passing through the three windows, calculated using the formula (0.85)^3. The difference in apparent magnitudes is determined using the equation m1 - m2 = -2.5 * log10(F1/F2), leading to a final apparent magnitude of approximately -0.97. Participants emphasize the importance of understanding the variables in the equation and the implications of light reflection. The conversation highlights the need for adherence to forum rules regarding homework assistance.
Jussi Lundahl
Messages
6
Reaction score
0

Homework Statement


Stars are observed through a triple window.
Each surface reflects away 15% of the incident light.
a) What is the magnitude of Sirius (m = -1.5) seen
through the window?

And I know that the solution is -0,97.

Homework Equations


m1-m2=-2,5*log10*(F1/F2)

The Attempt at a Solution


I have no idea what to do...
 
Physics news on Phys.org
Ok, start by telling us what are the F1 and F2 in the equation.
 
If 15% is deflected each time, then sirius's intensity, which let's say is 1 becomes 0.614 since (0.85)^3=0.614

Then, log(1/0.614)*2.5=0.5295

0.5295 is hence the difference between the apparnt magnitudes of the star after it has passed through the three windows. -1.5+0.5295=-0.9704

There's the answer.
 
  • Like
Likes Jussi Lundahl
Bandersnatch said:
Ok, start by telling us what are the F1 and F2 in the equation.

I think those are called flux density in English. $$F=\frac{L}{4 \pi r^2}$$, where the L is luminosity and r is radius. $$ L= \omega r^2$$ where $$\omega$$ is solid angle?? and r is radius.
 
Voltageisntreal said:
If 15% is deflected each time, then sirius's intensity, which let's say is 1 becomes 0.614 since (0.85)^3=0.614

Then, log(1/0.614)*2.5=0.5295

0.5295 is hence the difference between the apparnt magnitudes of the star after it has passed through the three windows. -1.5+0.5295=0.9704

There's the answer.

Thank you! You saved my day :)
 
Voltageisntreal said:
There's the answer.
Please, do not provide full answers to questions posted in the homework forum. Spoon-feeding is against the rules.
 
Member warned not to use text-speak in posts.
Bandersnatch said:
Please, do not provide full answers to questions posted in the homework forum. Spoon-feeding is against the rules.

;o, mb didn't kno rip
 
Voltageisntreal said:
;o, mb didn't kno rip
Text speak is also not permitted!
 
Back
Top