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free software acoustica cd-dvd label maker, acoustica cd label maker 3.40 registration code,.Q:

Light flux in the far field of a point source

I am trying to derive the formula for light flux in the far field of a point source in the image plane.
I have learned that the derived formula is as follows:
${\displaystyle \frac{d^2u_1}{{dw_1}^2}}=-{\displaystyle \frac{\cos\alpha}{r\sin\alpha}}{\displaystyle \frac{\sin u_1}{\sin\alpha}}-{\displaystyle \frac{\cos\alpha}{r\sin\alpha}}{\displaystyle \frac{dv_1}{dw_1}}$
Where $v_1$ and $u_1$ are the reflected and incident light vectors respectively, $\alpha$ is the angle between the normal and the vector $(u_1,v_1,w_1)$. Also, $r=\sqrt{u_1^2+v_1^2+w_1^2}$ is the reflection co-ordinate. I am trying to understand how the fact that $\cos\alpha$ shows up in this formula. Also, how can light undergo a reflection when it is point light source? Would it be a mistake in the book or can I just ignore this step? (The book uses the notation $\cos\alpha$ rather than $u_1\cos\alpha$)

A:

This is done by putting the point source in the origin.
When you do that, and you do it because you want to calculate the light flux in the far field, you get
\begin{equation}
\phi_r = \frac{L_0}{4\pi} = \frac{u_1\cos\alpha}{4\pi\sqrt{(u_1\cos\alpha)^2+(v_1\sin\alpha)^2+(w_1\sin\alpha)^2}}
\end{equation}
Afterwards it’s very clear that $\cos\alpha$ must be inside the numerator.
Also, the assumption of a point source is not invalidated, since you are not trying to calculate the light flux in the vicinity of the surface. The formula is only valid https://robag.ru/java-iobuffers-crack/

 

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