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The Numerical Aperture (NA) (for fiber optics) is usually used to denote the acceptance cone for a multi-mode fiber.

Does NA also describe the expansion of light emitted from the end of a fiber?

I have a 1 mm core 0.22 NA PMMA fiber, I would like to collimate the light once it's reached a 1.5" diameter.

$$ n\sin(\theta)=NA \\ \theta = \arcsin(\frac{.22}{1.4914}) \\ \frac{.5 \cdot 38 [mm]}{ \tan(\theta)}=\left \{ \text{focal length} \right \} $$

So, I should need a ~130 mm focusing lens?

edited in response to answer below

Is n the index of refraction of the fiber core (1.4914), or free space (1)?

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Yes, but.

The quality of the collimated beam may not be fantastic. And it depends on how the light is launched into the fiber. If the incident light has a narrow angular dispersion, and the fiber is short and/or of very good quality with few bends, then the light coming out will have a low angular dispersion. But if the incident light is converging at an angle close to the NA of the fiber, then your analysis should be ok.

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I'm sorry, but your formulas do not seem correct.

First, the NA is defined in free space, not inside the fiber. So, NA = 0.22 means that is the NA in air. Therefore, the angle is $sin(\theta) = 0.22 \rightarrow \theta = 12.7 deg$. [no need to divide by the index of refraction]

Next, NA refers to the half angle, not the full angle (see: http://en.wikipedia.org/wiki/Numerical_aperture ). So: $ tan(\theta) = \frac{D/2}{f} = \frac{38 mm/2}{f}$ which gives you $f = \frac{17.5 mm}{tan(\theta)} = 77.6 mm$ .

Finally, please appreciate that collimation is identical to imaging at infinity. What this means in practice is two-fold

  1. If you look on a wall that is much more than 77 mm from your lens, you'll see an image of the fiber tip (defects included!)
  2. The divergence angle of the collimated light is non-zero. Therefore, it does not stay the same diameter as it propagates. As soon as the light exits the lens, it will begin to diverge at a rate of: $\theta \approx \frac{D_{fiber}/2}{f} = \frac{0.5 mm}{77.6 mm} = 6.4 mrad = 0.37 deg $
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  • $\begingroup$ No need to apologize. Yes you're right. $\theta$ is only the half angle! I missed that. It does look like n should be the index of refraction in air too. However I ended up using a 150 mm lens which got me close, and that matches fairly closely if you use n=1.4914 in the equation (this could be due to the effects @garyp noted). Good point about the divergence angle being non-zero. However, in the case of (true) collimation an image should never be formed! The light rays will remain parallel out to infinity $\endgroup$ – Ben Aug 29 '14 at 5:38

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