Photon Escape Angle From Black Hole Consider a photon source emitting photons near the surface of a Schwarzschild black hole.
What angle, as a function of the source's radius from the event horizon, must the photons be emitted at such that they can escape to an observer at infinity?
 A: At the Schwarzschild radius, a photon must be emitted exactly normal to the surface in order to escape. As you travel outwards, the angle of emission decreases such that just above 1.5 times the Schwarzschild radius (i.e. the Photon Sphere) the photon can be emitted parallel to the tangent of the horizon and still escape.
According to this source (which is also a good source for all fun things regarding Schwarzschild black holes), there is a critical emission angle for photons from a stationary source some radius, $R$, from the black hole. Note that this equation technically should work for radii less than the Schwarxschild radius (the event horizon radius, $r_s=\frac{2GM}{c^2}$), but it'll give you negative angles because photons can't escape. Also note that all angles are given relative to the radial direction. $\theta=0$ means directed radially outwards and $\theta=\pi$ is radially inwards.
Inside the photon sphere, $R\le{3\over2}r_s$, the angles at which photons can escape are given by:
$$\theta\le\arcsin\left[\frac{\sqrt{27}r_s}{2R}\sqrt{1-\frac{r_s}{R}}\right]$$
Outside the photon sphere, $R\ge{3\over2}r_s$, the escape angles are:
$$\theta\le\pi-\arcsin\left[\frac{\sqrt{27}r_s}{2R}\sqrt{1-\frac{r_s}{R}}\right]$$
To get correct angles, just assume that arcsin always results in values between $-\pi/2$ and $\pi/2$. You'll note that for $R=r_s$, you find that $\theta=0$, which means only photons directed radially outwards escape. For $R=\frac{3}{2}r_s$, $\theta=\frac{\pi}{2}$ as my first paragraph stated. And that a radially inward photon ($\theta=\pi$) is always absorbed (this angle is asymptotically approached as $R\to\infty$).
