I'm working with fiber optics and I'd like to try to calculate the change of perceived brightness of visible light due to signal attenuation in a segment of fiber optic cable, based on its length.

fiber optic cables always have a value of signal attenuation $(\Delta P/d)$ measured in dB/m or dB/km.

Luminous flux (brightness as perceived by the eye), $(\Phi_V)$ , is measured in Lumens.

I am trying to get the two into one equation, using $\Phi_{V_1}$ and $\Delta P/d$ to solve for $\Phi_{V_2}$ based on some distance (d), and I'm unsure if I'm going in the right direction:

Many light sources list a luminous efficacy ($\eta$), that is, the amount of power required to create brightness, measured in lumens/watts.

If I have a value for luminous efficacy and luminous flux, I can calculate power in watts

$$P = \frac{\Phi_v}{\eta}$$

I have an equation for the difference between power levels in dB:

$$\Delta P = 10 \cdot \log(\frac{P_2}{P_1})$$

given $\Delta P / d = 0.2 dB/m$ and rearranging the above equation

I think

$$P_2=P_1 \cdot 10^{d \cdot 0.02}$$

can I combine this equation with the one relating power to luminous efficacy and luminous flux?

luminous efficacy should remain the same for both power equations, so

$$Φ_{V_2}=\Phi_{V_1} \cdot 10^{d \cdot 0.02}$$


Is this the proper way to calculate change in luminous flux due to signal attenuation in a fiber optic cable? am I smashing the wrong equations together?

  • $\begingroup$ Please use Mathjax for mathematics a it is the site standard and helps readability. $\endgroup$ – StephenG May 18 '18 at 18:30
  • 1
    $\begingroup$ updated with Mathjax $\endgroup$ – user2418372 May 18 '18 at 19:28

First, the efficacy $\eta$ refers to the process of convert input electrical input power into light. It doesn’t directly enter here.

But that’s ok, because your final ratio just depends on the loss ratio:

$P_{\rm{out}} = 10^{-\rm{dB}/10} P_{\rm{in}}$

Since your fiber loss is 0.2dB/m, for your distance $d$:

$P_{\rm{out}} = 10^{-0.2 d/10} P_{\rm{in}}$

which is what you had except for the minus sign.

Luminous intensity, if everything else remains the same, will go like the power in the light.


$\Phi_{\rm{out}} = 10^{-0.2 d/10} \Phi_{\rm{in}}$

  • $\begingroup$ Thanks for the response. I'm trying to arrive at a relationship between $\Phi_{out} $ and $\Phi_{in}$. Is there a way to relate this loss ratio to that? $\endgroup$ – user2418372 May 18 '18 at 19:32
  • $\begingroup$ @user2418372 Added another line to make Phi dependence explicit. $\endgroup$ – Bob Jacobsen May 18 '18 at 20:03

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