# Optics: Converging waves

1. Apr 24, 2015

### Wminus

Hi. A spherical wave $e^{i(kr-\omega t)}$ diverging from a single point $(x=0,y=0,z=-z_0)$ can be approximated as a parabolic wave in the paraxial case around the z-axis. I.e., $k r = k \sqrt{x^2+y^2+z^2} \simeq k (z +\frac{x^2+y^2}{2z})$.

OK, then let's say a lens is placed such that its optical axis coincides with the $z$ axis and its focus points are at $-z_0$ and $z_0$. In this case, the outgoing parabolic wave from $-z_0$ will be focused into the point $z_0$. My question is, how is this to be modeled mathematically? Intuitively I would guess that $k r \simeq k ( z - \frac{x^2+y^2}{2z})$, but what is $kr$ equal to in the accompanying case of a converging spherical wave? Something ala $e^{i(kr + \omega t)} e^{i \phi}$, where $\phi$ is some phase factor?

I would appreciate it if you guys could help me in clearing this stuff up :)

Thanks

Last edited: Apr 24, 2015
2. Apr 24, 2015

### blue_leaf77

If this is converging lens, then the rays will be focused at infinity, that is the outgoing rays are collimated.

3. Apr 25, 2015

### Wminus

crap, yeah you're right. I was thinking in terms of rays from the object plane being focused into the image plane, but I mixed it up. sorry.

But anyway, do you know the mathematical form of waves converging to a single point?

4. Apr 25, 2015

### blue_leaf77

converging spherical wave must be the inverse of the diverging one, the sign of the wavevector k must change (better change the sign of k rather than the sign of $\omega t$ as you did above, this will mean time reversal).