# What is the path of the bended light beam in a mirage?

Is it like this: Or like this: If the second diagram is the right one, could you explain why the light beam "keeps curving" after arriving horizontal to the ground? I would rather say that the light beam keeps going horizontal, because there is no more change of the refractive index at the same height.

Both diagrams are correct. I should like to address your insightful question:

could you explain why the light beam "keeps curving" after arriving horizontal to the ground?

which I read as follows. You imagine the sky to be discrete horizontal layers of optically homogeneous material, but with the different layers having different refractive indices. You also correctly state that if a ray found itself moving parallel to the layers, there would be no way for it to be bent sideways back towards the ground.

This is an astute question and the answer somewhat subtle:

1. If the real system were in fact a system of discrete layers, there would be no way for a ray to become horizontal and confined in a layer. The only way a refracted ray can emerge from an interface parallel to the interface is at the onset of total internal reflexion. Imagine the ray approaching an interface at a shallower and shallower angle; when the angle is very shallow, most power is reflected from the interface. Thus, in a discretized layer model, the ray is never horizontal and the whole path is a polygon; the highest point of this path is a vertex at an interface where the ray reflects symmetrically (i.e. according to the law of reflexion) from the highest reached interface;

2. In a continuous model, the ray path must go through a point where the ray direction is horizontal, but at the point where it is horizontal, you do not have a plane wave in a homogeneous medium; you have a plane wave "straddling" an inhomogeneous medium with the index varying in the transverse direction. Such a wave always has an effective "force" bending it towards the region of higher index. This is what happens in an optical fiber. Don't forget that a ray is always represents a plane wave of small, but still many-wavelength (as a condition for the Eikonal equation to hold), sideways breadth, so the inhomogeneity always acts in this way. Indeed, Hamilton's equations of motion for a ray in this system read as follows and you can see the downwards "force" acting on the ray when it is horizontal in equation (3) below:

$$H(p_x,\,p_y,\,p_z,\,x,\,y,\,z,\,\tau) = \frac{1}{2}\frac{p_x^2+p_y^2+p_z^2}{n(z)^2}\tag{1}$$ $$\frac{\mathrm{d}}{\mathrm{d}\tau} p_x = -\frac{\partial H}{\partial x} =\frac{\mathrm{d}}{\mathrm{d}\tau} p_y=-\frac{\partial H}{\partial y}=0\tag{2}$$ $$\frac{\mathrm{d}}{\mathrm{d}\tau} p_z = -\frac{\partial H}{\partial z}= \frac{p_x^2+p_y^2+p_z^2}{n(z)^3}\frac{\mathrm{d}n(z)}{\mathrm{d}z}\tag{3}$$ $$\frac{\mathrm{d}}{\mathrm{d}\tau}x = \frac{p_x}{n^2}\tag{4}$$ $$\frac{\mathrm{d}}{\mathrm{d}\tau}y = \frac{p_y}{n^2}\tag{5}$$ $$\frac{\mathrm{d}}{\mathrm{d}\tau}z = \frac{p_z}{n^2}\tag{6}$$

where the path parameter $\tau$ is the optical pathlength and the momentums are the direction cosines of the ray direction multiplied by the refractive index.