Maximum Number of reflections in optical fibre

[physicoo]

Question:

An optical fibre is 1.0 metre long and has a diameter of 20micro-meter. Its ends are perpendicular to its axis. Its index of refraction is 1.30. What is the maximum number of reflections a light ray entering one end will make before it emerges from the other end?

A) 18000
B) 22000
C) 34000
D) 42000

I have calculated the number of reflections to be 41533 reflections. However, does anyone know if it's supposed to be C or D? I suppose it can't exceed 41533 reflections, hence the answer should be C?

 

[NobodySpecial]

You have the data rounded to 2 sig figures (the length and ND) - how many figures do you round the answer to ?

 

[physicoo]

The data is basically given as what the question has provided:

Length of fibre - 1.0m
Diameter of fibre - 20micro-m
Index of refraction - 1.3

This was what I did:

Sin(theta) = n(air)/n(fibre)
theta = sin^-1(1/1.3) = 50.28486277 degrees

From the internal reflection (and Pythogoras theorem):

tan(50.28486277) = length of reflection / 20 micro meters
-> Length of Reflection = 24.07717062 micro-m = 0.000024077m

Number of Reflections = 1m / 0.000024077m = 41533.11931 ~= 41533 reflections.

 

[NobodySpecial]

theta = sin^-1(1/1.3) = 50.28486277 degrees
No this is wrong.
If you only know n to 2 figures how can you calculate an angle to 10?

 

[physicoo]

theta = sin^-1(1/1.3) = 50.28486277 degrees
No this is wrong.
If you only know n to 2 figures how can you calculate an angle to 10?

Sorry, could you explain to me what you meant by "calculate an angle to 10?" ?

 

[NobodySpecial]

To 10 "significant figures".

In physics it's important to give an answer to the correct accuracy.
You only know the N to 2 sig figures, a 1% accuracy, so you can only quote the answer to the same accuracy - in this case around 50.3 deg.
It's slightly trickier with sin - so you are probably justified in adding another decimal place but you can't claim you know the answer to an accuracy of a part per 100,billion like you quoted

 

[physicoo]

Rounding off the values for each calculating to 2sf would result in my end result as 42000.

However, I think there's a need to calculate accurately (keeping all the sfs/decimals) to determine the exact number of reflections (41533 in this case) since it determines the 'maximum' number of reflections in the optical fiber, does it not?

Rounding my values all to 2sf, subsequently resulting in a 42000 value doesn't seem feasible since the exact number of maximum reflections is 41533.

Would appreciate any help! :)

 

[Dadface]

As NobodySpecial said you need to consider the correct number of significant figures to use and this means rounding off your final answer.When you look at the information given in your question it can be incorrectly implied that the data is exactly correct,for example that the diameter is exactly 20mum all along the length of the fibre.All data is subject to experimental errors and to be shown in greater detail it can be presented as a figure along with the uncertainty (eg...width=20mum plus or minus a certain percentage).Because of the uncertainties in the given data there will be uncertainties in any quantities you calculate with that data and hence the need for rounding off.

 

[physicoo]

So based on the advice, I suppose the correct answer would be D - 42000 (rounded off due to uncertainties) 'even though' the exact calculations is 41533 then?

 

[NobodySpecial]

So based on the advice, I suppose the correct answer would be D - 42000 (rounded off due to uncertainties) 'even though' the exact calculations is 41533 then?

This is important - the exact answer is 42000 not 41533.
The exact answer would be 41533 if the diameter of the fibre was 20.0000um and1.00000m long and the ND was given as 1.300000.

In experimental physics understanding accuracy and error is as important as the answer.

Suppose you were asked how many steps is it for you to walk to work.
You are told it's 2.0km and your stride is about 60cm. So the answer is 3333.3 ?
No because it might actually be between 1.95 and 2.05km to work and your stride might be between 59 and 61cm to the accuracy you are given the data.
The range of answers could be between 2.05k/0.59=3474 and 1.95K/0.61=3196
So to two sig fig you would say it's 3300 steps to work.