I have read this:
Because the spacetime curvature at the horizon is so great that there is no light-like world line the extends beyond the horizon.
Why does time stop in black holes?
If the curvature is extreme, that would that mean that laser beams (for the sake of argument let's assume they are visible because of some scattering like from the atmosphere) appear to be bent?
Question:
Yes, the curvature is enough to bend lasers. In fact, there is an interesting feature in Schwarzschild spacetime that might answer your question well. If a non-rotating, uncharged black hole (i.e., a Schwarzschild black) has mass $M$, then the event horizon is at $R = 2M$. At $r = 3M$ one has what is called the photon ring. At $r = 3M$, light rays can follow (unstable) circular orbits, meaning it is possible for a light ray to stay indefinitely orbiting the black hole at constant $r = 3M$. This orbit is unstable, so if it is slightly perturbed the light ray will either escape to infinity or fall into the black hole, but in principle you can arrange light to orbit the black hole indefinitely.
I should add that this does not necessarily means the curvature is large. For very big black holes, the curvature at the horizon can be as small as the curvature in your living room (or smaller).
Yes, this is true. Whilst laser light differs from ordinary light in being coherent, it is still light and so the effect of black hole still holds.
Its worth noting that close to black hole we will have light orbiting the black hole. But just away from this orbit the light will be able to escape. This means when you look at a closeby black hole with an accretion disk, then you will see a ring of light. This is an optical illusion, no matter where you stand you will still see this ring. It is this optical illusion that was famously photographed recently.
