LED light flux over a bacterial plate surface

Question

I am carrying out an experiment, in which I radiate a bacterial plate with an LED above the plate. In the LED datasheet, I have a graph giving me the relative intensity $I_r$ as a function of the angle $\phi$ to the normal of the LED surface. As I understand this, if I measure the flux on a spherical surface with radius $R$ away from the LED, then I will observe the flux behave like $I_r$($\phi$).

The relative flux at 0$^{\circ}$ from the normal to the LED surface is 100% and the relative flux at 10$^{\circ}$ from the normal is 50% etc. All the flux from the LED is within $\pm$45$^{\circ}$ angle from the normal to the LED surface and I know the total radiant flux (power) $P$ from the LED.

I want to calculate the radiant intensity (flux per unit area) at the bacterial plate, which is located distance $h$ from the LED surface, the LED being located at the center of the bacterial plate horizontally. I know I have to use surface integrals but my calculus classes took place long ago, so I would appreciate any help!

Answer 1

First of all, to avoid misunderstandings, let us set some definitions according to SI terms (re. e.g. to this Wikipedia article)

• Radiant Flux (or Power) ($W$): radiant energy per unit time
• Radiant Intensity ($W/{sr}$): radiant flux per solid angle
• Irradiance (Flux density) ($W/m^2$): radiant flux per unit area

For a point-like source, the power irradiated within a cone with vertex in the source will be constant, and the Irradiance therefore will decrease with the square of the distance from the source. Or equivalently, the power emitted within a steradiant (Intensity) is constant.
Now if you have a diagram (normally given in polar form) of the relative Intensity, that shall mean that the total power irradiated by the led is given by the following integral for $\phi$ ranging from $0$ to $\pi /2$. $$ \begin{gathered} P = \int_{\;0}^{\,\pi /2} {I_0 (R)I_r (\phi )\,2\pi R\sin \phi \,R\,d\phi } = \hfill \\ = I_0 (R)\,2\pi R^{\,2} \int_{\;0}^{\,\pi /2} {I_r (\phi )\,\sin \phi \,d\phi } = \hfill \\ = P_0 \;\int_{\;0}^{\,\pi /2} {I_r (\phi )\,\sin \phi \,d\phi } \hfill \\ \end{gathered} $$ where $P_0$ is the power of the led if it was radiating with a constant flux equal to that in the vertical, i.e. with $I_r (\phi)=1$. Otherwise, knowing $P$, then $P_0$ will represent a scale constant.

That premised, the sketch above shows clearly that the power emitted inside the solid conic angle of aperture $d\alpha$, which is $P_0 \;I_r (\phi )\,\sin \phi \,d\phi $, on the plane at distance $h$ will distribute over a surface of $2\,\pi \,\rho (\phi )\,\,d\rho (\phi )$. Since we have the following relations: $$ \left\{ \begin{gathered} \rho = h\;\tan \phi \hfill \\ d\rho = \frac{h} {{\cos ^{\,2} \phi }}\;d\phi \hfill \\ \end{gathered} \right.\quad \Leftrightarrow \quad \left\{ \begin{gathered} \phi = \arctan \left( {\rho /h} \right) \hfill \\ d\phi = \frac{{\cos ^{\,2} \phi }} {h}d\rho = \frac{h} {{\left( {h^{\,2} + \rho ^{\,2} } \right)}}d\rho \hfill \\ \sin \phi = \frac{\rho } {{\sqrt {h^{\,2} + \rho ^{\,2} } }} \hfill \\ \; \hfill \\ \end{gathered} \right. $$ the relative Irradiance (irradiance per 1 totally emitted W) on the plane surface will be:

• in terms of $\phi$ and $h$

  $$ Q (\phi ,h) = I_r (\phi )\,\frac{{\sin \phi \,d\phi }} {{2\,\pi \,\rho (\phi )\,\,d\rho (\phi )}} = I_r (\phi )\,\frac{{\cos ^{\,3} \phi }} {{2\,\pi \,h^{\,2} }} $$

• in terms of $\rho$ and $h$

  $$ Q (\rho ,h) = I_r \left( {\arctan \left( {\rho /h} \right)} \right)\,\frac{{\frac{\rho } {{\sqrt {h^{\,2} + \rho ^{\,2} } }}\,\frac{h} {{\left( {h^{\,2} + \rho ^{\,2} } \right)}}d\rho }} {{2\,\pi \,\rho \,\,d\rho }} = I_r \left( {\arctan \left( {\rho /h} \right)} \right)\frac{1} {{2\,\pi h^{\,2} \left( {1 + \left( {\rho /h} \right)^{\,2} } \right)^{\,3/2} }}\, $$

The above relative Irradiance shall be then multiplicated by $P_0=I_0(R) 2 \pi R^2$ in order to get the value in $W/m^2$. Note that $I_0(R)=Q(0,R)$ as should be expected.