A highly reliable laser emits a steady beam of light at a fixed ...

[tsand2]

A highly reliable laser emits a steady beam of light at a fixed frequency f that hits a photodetector a light year away in what we can assume is empty space. The laser/detector setup is now placed as a system with a black hole directly behind the laser. What happens to the frequency of the light beam?
  a. The detector measures the frequency of the beam as less than f.
  b. The detector measures the frequency of the beam as greater than f.
  c. The laser emits light at a frequency less than f (as measured at the laser)
  d. The laser emits light at a frequency greater than f (as measured at the laser)
  e. Both a and c

Question 2

Which of the following is not a good conservation law?
 

a. Baryon conservation.
  b. Lepton conservation.
  c. Strangeness in the weak interaction.
  d. Strangeness in the strong interaction.

[Jordin Calloway]

Answer to Question 1

a.
The laser emits the same light at its position because nothing changes with respect to the laser. One can imagine the light being stretched as it climbs out of the potential well of the black hole. A longer wavelength means a lower frequency of light. This is called gravitational redshift, because the light loses energy due to gravity and shifts toward the red end of the spectrum.

Answer to Question 2

c. trangeness is not conserved in weak interactions.

[tsand2]

Thanks for the timely response, appreciate it

[robbielu01]

Great answer, keep it coming