# I Regarding the wavevector spread.

1. Jul 14, 2016

### otaKu

So I was reading THIS book on nano optics. It says that the maximum possible spread in the wavevector component k(The spread can occur for instance when the light field converges towards a focus, e.g. behind a lens.) is the total length of the free space wavevector k=2π/λ.
Can anyone please explain this to me?! Thanks.

2. Jul 14, 2016

### otaKu

Here's the screenshot of the text if the google link doesn't work.

3. Jul 14, 2016

### Andy Resnick

What are you struggling with? Do you understand how to get eqn 1.2? The text seems to simply substitute 'k' into 'Δk', which is questionable, but if you like you should be able to determine Δk (start with writing down dk/dλ) and go from there.

4. Jul 14, 2016

### otaKu

I understand equation 1.2. What I don't understand is the reason why the author substituted k into Δk.

5. Jul 15, 2016

### Andy Resnick

Does footnote 1 say anything relevant? That specific sentence seems to be the crux of the matter, and I don't quite understand it either.

6. Jul 17, 2016

### otaKu

No it doesn't. It says "for real lens this must be corrected by the numerical aperture.'

7. Jul 18, 2016

### Andy Resnick

Blech. The rest of the book seems to be better- chapter 2 was straightforward, chapter 3 is also reasonable (although I objected to a few things here and there).

8. Jul 19, 2016

### nasu

The component along a direction cannot be larger than the magnitude of the vector. And cannot be less than zero. So the interval of values for any component of the vector k is from zero to k (magnitude). This is the maximum possible "spread" he is talking about.
The actual spread will be less that this maximum value and will depend on the angle of the cone of light. Which is given by the numerical aperture of the lens.

9. Jul 25, 2016

Thanks!