Gravitational lensing in Newtonian physics Famously, when Eddington attempted to measure gravitational lensing during the Eclipse, it was the measured magnitude of the lensing that gave gravity [pun, obviously, intended] to General Relativity - not the measurement of lensing itself. That is, Newtonian physics also predicted a lensing deflection, but only half of the deflection predicted by GR. 

Question is: Why? I've read a lot about this, and I can see how when one integrates the Newtonian acceleration along the photon path with constant |v|=c this gives the 'correct' Newtonian value - but intuitively, I can't wrap my head around it. Why should a massless photon be affected by gravity at all in Newtonian gravity?
A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical corpuscle of light--appropriately enough, a theory of light advanced by Newton himself.

There are many problems with the Newtonian conception of light, but that's an issue for electromagnetism rather than gravity.

Critically, for a test particle, the trajectory depends on the initial velocity only, and not the mass. So we don't even have to address the mass of the light corpuscle at all to talk about its trajectory, as the acceleration is the gradient of the gravitational potential and if considering force (or gravitational potential energy), the mass cancels out anyway. If you wish to bring in the mass explicitly, we can still think of the trajectory of a massless particle as a limit of trajectories of particles with masses tending to zero but having the same velocity &mdash; a trivial limit because they're all the same trajectory in the Newtonian theory. On the other hand, if we recognize that light carries momentum, a Newtonian light corpuscle shouldn't have zero mass, so the question of what to do with a genuinely massless particle evaporates.

When people talk about the Newtonian deflection of light, they are typically considering a hyperbolic trajectory of test particle at the speed of light under Newtonian gravity. If the angle between the asymptotes is $\theta$, then $\theta = \pi$ represents a completely straight trajectory unaffected by gravity, and the eccentricity is
$$e = 1 + \frac{v_\infty^2R_p}{GM} = \frac{1}{\cos\left(\frac{\theta}{2}\right)} = \frac{1}{\sin\psi}\text{,}$$
where $R_p$ is the periapsis distance and $2\psi = \pi-\theta$ is the measure of deflection. On the scale of the solar system, it doesn't matter whether we set $v_\infty = c$ or anywhere else along the trajectory. For example, if the velocity at peripasis is $c$ instead, then $e\mapsto e-2$, and so is negligible for light deflection due to the Sun, $e &gt; 10^{5}$. The total deflection is approximately
$$2\psi \approx e^{-1} \approx \frac{2GM}{c^2R_p}\text{,}$$
which is half the correct general-relativistic prediction.

Note that here we did not assume that the speed is constant along the hyperbolic orbit. That would not be consistent with Newtonian gravity. Rather, what we have is a situation where if $v = c$ anywhere along the trajectory, then the speed along any other point is so close to $c$ that it doesn't practically matter for considering light deflection.