# Where do you get $I_p=\frac{P}{4 \pi r^2} sin \alpha$?

This problem is about color science and it is short but tricky: Where do you get $I_p=\frac{P}{4 \pi r^2} sin \alpha$, where $I_p$ is radiant intensity(color science: radiometric quantities). Look at the link( exercise 2 a)): http://cs.joensuu.fi/~ot14/vo/IMG.jpg I know that radiant intensity is $I=\frac{\Phi}{\Omega} = \frac{\Phi}{4\pi}$ if source radiates uniformly to all directions as they are according to problem.

The right solution is: $I=\frac{\Phi}{\Omega} = \frac{\Phi}{4\pi}$ as stated previously( radiant intensity), because source(s) radiate uniformly to all directions. Then you use that $E=\frac{I}{d^2} \cos \theta$. So for source 1( lefthand side), $\Omega=P = 100 W$, $I=\frac{\Omega}{4 \pi}=\frac{100}{4 \pi}= 7,96 W/sr$, $d=\sqrt{1,5^2+1,5^2}= 2,12 m$, $\alpha = 45^{\circ}$, $\theta = 45^{\circ}$ and $E_p = \frac{I}{d^2}cos \theta = 1,25 W/m^2$

And for source 2( righthand side), $\Omega=P = 40 W$, $I=\frac{\Omega}{4 \pi}=\frac{40}{4 \pi}= 3,18 W/sr$, $d=\sqrt{2,5^2+1,5^2}= 2,92 m$, $\alpha = 30,96^{\circ}$, $\theta = 59,04^{\circ}$ and $E_p = \frac{I}{d^2}cos \theta = 0,19 W/m^2$

So answer to the problem 2 a) is: Irradiance at point a is $E_1 + E_2 = 1,25 + 0,19 = 1,44 W/m^2$.

-
It is not clear what is confusing you. Are you solely concerned with where the $\sin \alpha$ factor comes from in the formula in your question? Or are you asking that we check the solution you've written? –  kleingordon Jun 14 '12 at 22:54
well just check that solution is right. And of course if you can tell where $I_p=\frac{P}{4 \pi r^2}sin \alpha$ comes from then it would be better. –  laovultai Jun 15 '12 at 6:29
This is not appropriate use of the site, but you got a good answer. –  Ron Maimon Jul 17 '12 at 4:05

Incidentally, I am used to seeing this angular dependence expressed as a $\cos\theta$ factor where $\theta$ is the angle with respect to the normal. So, a surface that is rotated to a completely edge-on orientation ($\theta = \pi/2$) will absorb zero intensity, whereas a surface absorbing the radiation face-on ($\theta = 0$) gets to absorb the full power.
The $1/(4 \pi r^2)$ factor is the familiar inverse square law that accounts for geometric dilution of flux lines - it is the same effect that is responsible for the $1/r^2$ dependence of the laws of gravity and Coulomb attraction. To put it simply in this case for radiation, if a source is emitting energy at a constant power $P$ and if the radiation is not getting attenuated anywhere up to a distance $r$ away, then the power passing through the sphere of radius $r$ centered on the source must also be $P$. So then the power per unit area passing through a patch of this sphere, which is by definition a radiant intensity, must be $P / (4 \pi r^2)$. If you place a surface to absorb the energy at a point on this sphere, then you must also account for the angle the radiation makes with the surface as discussed in the first paragraph.