# Attenuation Coefficient in an optical fibre

#### says

1. The problem statement, all variables and given/known data
In a particular fibre, the attenuation coefficient for Rayleigh scattering is measured to be 0.30 dB/km at 1.20 μm.

(a) How large would it be expected to be at 0.80 μm?

(b) Ignoring other sources of attenuation, if a signal of a certain initial power was sent over a distance of 50 km of the fibre, what would be the power in the signal at the other end if the wavelength was 0.80 μm, as a fraction of the power that would be received at 1.20 μm?

2. Relevant equations
a = (10/L)log(Pin/Pout)

where L = distance
a= attenuation, units: dB/km

3. The attempt at a solution
a)

0.30 dB/km = (10/1.2*10-6)log(Pin/Pout)

0.30*10-6 / 10 = log(Pin/Pout)

3.0*10-8 = log(Pin/Pout)

a = (10/0.8*10-6)*3.0*10-8

a = 0.375 dB/km - this seems reasonable, as the distance is shorter, attenuation should be higher

b)
I don't really understand what b) is asking for though and am not really sure where to start.

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#### TSny

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Gold Member
The L in the formula represents the distance over which the signal travels. This will come into part (b).

I believe that 0.80 μm and 1.20 μm are wavelengths of the infrared light propagating in the fiber. To answer part (a), you need to know how the Rayleigh-scattering attenuation coefficient depends on wavelength. This link has some discussion

https://www.fiberoptics4sale.com/blogs/archive-posts/95048006-optical-fiber-loss-and-attenuation

Rayleigh scattering is discussed about halfway down the page. But, hopefully, this was covered in your textbook or lectures.

#### says

The loss due to Rayleigh scattering is proportional to λ-4 and obviously decreases rapidly with increase in wavelength

a(λ)=a00/λ)4

a0 = 0.30 dB/km
λ0 = 1.20 μm

a(λ)=0.30(1.2*10-6/0.8*10-6)4

a = 1.51875 dB/km - answer for part a)

#### TSny

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Gold Member
The loss due to Rayleigh scattering is proportional to λ-4 and obviously decreases rapidly with increase in wavelength

a(λ)=a00/λ)4

a0 = 0.30 dB/km
λ0 = 1.20 μm

a(λ)=0.30(1.2*10-6/0.8*10-6)4

a = 1.51875 dB/km - answer for part a)
That looks right to me.

#### says

That looks right to me.
Oh no, I will probably just round it to 3 sig figs with a= 1.52 db/km

For b)
a = (10/L)log(Pin/Pout)
a(λ)=a00/λ)4

I'm still not entirely sure what b) is asking.

If the signal input has power X, wavelength of 0.80 μm, and was sent through the fibre for 50km, what would the power be at the output Y.

If the signal input has power A, wavelength of 1.20 μm, and was sent through the fibre for 50km, what would the power be at the output B.

I think it's also asking for a ratio between the output power of both 0.80 μm and 1.20 μm wavelengths.

0.30 = (10/50,000)log(Pin/Pout)

1.52 = (10/50,000)log(Pin/Pout)

I can solve for log(Pin/Pout) for both 0.30 db/km and 1.52 db/km and then get a ratio of the two?

#### TSny

Homework Helper
Gold Member
Let P1, out be the output power of the 0.80 μm signal and let P2, out be the output power of the 1.20 μm signal.
Find the ratio of P2, out to P1, out assuming that the input power is the same for both wavelengths.

Since the attenuation coefficients are in units of dB/km, you will want to leave L in km.

#### says

0.30 = (10/50,000)log(Pin/Pout)

1.52 = (10/50,000)log(Pin/Pout)

(50,000*0.30)/10 = log(Pin/Pout) = 1500

(50,000*1.52)/10 = log(Pin/Pout) = 7600

ratio: 1500/7600?

#### TSny

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Gold Member
1.52 = (10/50,000)log(Pin/Pout)
It's a good idea to include units when you substitute in your numbers. You would then see that the units on the left side of the above equation don't match the units on the right side.

Once you fix the units, try to solve the equation for Pin/Pout.

#### says

(50*0.30)/10 = log(Pin/Pout) = 1.5
e^log(Pin/Pout) = Pin/Pout
e1.5 = 4.48
(50*1.52)/10 = log(Pin/Pout) = 7.6
e7.6 = 1998

#### TSny

Homework Helper
Gold Member
(50*0.30)/10 = log(Pin/Pout) = 1.5
e^log(Pin/Pout) = Pin/Pout
e1.5 = 4.48
(50*1.52)/10 = log(Pin/Pout) = 7.6
e7.6 = 1998
OK so far. Can you now find P2,out / P1,out if you assume Pin is the same for both signals?

#### says

If i divide both numbers by each other i'm getting (P2, in/P2, out)/(P1, in/P1, out)

4.48/1998=0.002

#### TSny

Homework Helper
Gold Member
If i divide both numbers by each other i'm getting (P2, in/P2, out)/(P1, in/P1, out)

4.48/1998=0.002
Looks good. But you have rounded the answer to only one significant figure.

#### says

Looks good. But you have rounded the answer to only one significant figure.
4.48/1998 = 0.00224224224

#### TSny

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Gold Member
4.48/1998 = 0.00224224224
How many significant figures would be appropriate?

#### says

3 sig figs... so 0.00224

#### TSny

Homework Helper
Gold Member
3 sig figs... so 0.00224
OK. 2 or 3 sig figs seems to me to be appropriate.

So, the question asks you to express P2, out as a fraction of P1, out.

#### says

Would that just be 1/0.00224?

#### TSny

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Not quite. You want to find a number c such that P2,out = c P1,out

#### says

I don't quite follow...

#### TSny

Homework Helper
Gold Member
You are asked to find P2, out as a fraction of P1, out.

Signal 2 has a larger attenuation constant. So, the output power of signal 2 will be less than the output power of signal 1. This can be expressed as saying that the output power of signal 2 is some fraction of the output power of signal 1.
That is, you could write P2, out = c P1, out for some number c. You just need to find c (which you have essentially done).

If c happened to equal .25, then you could say that the output power of signal 2 would be 1/4 the output power of signal 1.
You could write P2, out = 0.25 P1, out, or you could write P2, out = (1/4) P1, out
I think either way of writing it would be acceptable.

So, in your problem, what number should take the place of .25?

#### says

P2, out=0.00224P1, out

#### TSny

Homework Helper
Gold Member
Yes, that looks good.

"Attenuation Coefficient in an optical fibre"

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