00:01
To produce the same light energy per unit area on the film, what is the required exposure time to photograph the same nebula with a smaller telescope, which has an objective lens with an aperture diameter of 60 millimeters and a focal length of 900 millimeters.
00:16
So i not is the intensity of the light from the nebula.
00:20
Theta not is its angular diameter.
00:24
So with the first telescope, we got theta knot is equal to negative h prime.
00:31
Over f -not where h prime is the image diameter so the light the power of the light captured by the telescope so the power is equal to the intensity at times the area since the intensity in the area is pi 200 millimeters squared over four so the energy is equal to the power times the change in time, which is i .0 .2 .200 millimeters squared over 4, and then the time is 1 .5 minutes.
01:31
So for the second telescope, p2 is equal to i .0 .a2, which is i .0.
01:48
Pi 60 millimeters squared over 4.
01:54
Similarly, the energy is equal to the power...