0
$\begingroup$

Background:

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

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

Question:

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

Then

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.