Firstly, is that correct?

Yes your intuitive understanding for this part of the Coriolis effect is correct.

The second part, that is, why wind in the East direction is deflected South, is a bit trickier, and involves the use of centripetal force. this is given by the equation:

$F = \frac{mv^2}{r}$

If we re-arrange the above equation, we can find $r$ in terms of $v$, and we arrive at:

$r = \frac{mv^2}{F}$

This tells us that as velocity increases, the radius required to maintain the orbit also increases.

Now let's apply this concept to winds on the Earth. If we feel no wind on the Earth, then the air in the atmosphere is travelling at the same velocity as the Earth. The Earth is naturally spinning towards the East.

In the case of an additional Eastward wind felt on the Earth, this wind has effectively increased its velocity, and therefore the above equation tells us that the radius of orbit must increase as well. Radius in this case is the distance, measured perpendicularly of the Earth's axis, between the axis and the wind.

In order for the radius to increase, the wind moves southwards, where the radius is larger.

Similarly, wind moving in the West direction, is moving in the direction opposite of that to the Earth, and therefore its velocity is decreased. Consequently this wind moves towards the North, where the radius is less.

The above image shows what happens. The wind moving East begins to expand its radius, thus moving outwards. Gravity pulls it back, and the wind moves South, in order to maintain the larger radius required for its increased velocity.

I will assume that you are asking about laser beams in the fundamental, diffraction limited Gaussian mode. The standard expression for the divergence angle of a Gaussian beam in the far field is (see the Wikipedia page on Gaussian beams)
$$
\theta=\frac{\lambda}{\pi\omega_0}
$$
where $\omega_0$ is the so-called waist size of the Gaussian beam. From here you can calculate the solid angle subtended by the beam which is given, in the small $\theta$ limit, as
$$
\Theta\simeq\pi\theta^2=\frac{\lambda^2}{\pi\omega_0^2}=\frac{\lambda^2}{A},
$$
where $A$ is the area of the beam's waist.

If you are looking for a derivatation which starts at a more fundamental level than that, then you should pick up any textbook on lasers. Any textbook you can find will cover the derivation of the Gaussian modes of a laser beam from the wave/Helmholtz equation. You can also look at section 2.1 of my thesis where I sketch out the derivation from the Helmholtz equation although I stop slightly short of deriving the divergence angle.

## Best Answer

Coriolis force is not an actual force, but rather an effect observed in rotating frame of reference. The light path is not actually bent, so it doesn't matter that the photon has no mass, the Earth's rotation will have an affect on the photon's

apparentpath.This does not contradict your calculation of $F_{Coriolis}=0$, because you have to put this force in $F=ma$, where $m$ is, again, zero. However, you can take the limit of both sides for $m\rightarrow 0$ and calculate $a$.