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

Question

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 $E_1 + E_2 = 1,25 + 0,19 = 1,44 W/m^2$.

Answer 1

The notation you are using is somewhat confusing to me but I think the angular term in the expression in your question comes from the fact that the intensity of radiation on a flat surface will depend on the angle the incoming rays make with respect to the normal direction to that surface. This is because rays that are incident at an angle with respect to the normal see a reduced projected area of the surface compared to the case where they are incident face-on. You might already have an intuitive grasp of this effect from daily life - it explains, for example, the seasonal variation of temperature of the northern and southern hemispheres due to the tilt of the earth's rotational axis with respect to its orbital plane.

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.