About radiance equation in Radiometry

Thread starter dammage
Start date Jun 6, 2011
Tags

Radiance Radiometry

AI Thread Summary

The discussion centers on the radiance equation in radiometry, specifically the formula L = d2P / (dA*cos(a)*dw), where L represents radiance, P is power flux, A*cos(a) is the projected area, and w is the solid angle. A key point raised is the use of the second derivative of power, d2P, in the equation. The clarification provided indicates that radiance is defined as power per unit area per solid angle, which helps in understanding the necessity of the second derivative. The conversation concludes with an acknowledgment of the explanation, indicating a resolution to the initial confusion. Understanding the relationship between these variables is crucial for accurate calculations in radiometry.

Jun 6, 2011

dammage

Homework Statement

According to some pages on web, differential formula for calculating radiance L is :

L = d²P / ( dA*cos(a)*dw)

L : radiance
P : power flux
A*cos(a) : projected area
w : solid angle

Please tell me why does power P has derivaties two times d²P ?

Homework Equations

The Attempt at a Solution

Physics news on Phys.org

Jun 6, 2011

ideasrule

Homework Helper

Because radiance is power per unit area, per solid angle.

Jun 20, 2011

dammage

Ah, I understand now. Thank you !

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'

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 »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »