A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
(magnifying_power) The magnifying power of the telescope is 24.
(separation) The separation between the objective and the eyepiece is 150 cm.
Explanation
This question tests the understanding of telescope optics. For a telescope viewing distant objects in normal adjustment, the magnifying power is the ratio of the focal length of the objective to that of the eyepiece. The tube length (separation between lenses) equals the sum of both focal lengths since the image formed by the objective must be at the focal point of the eyepiece for the final image to be at infinity.
Solution Steps
Step 1: Calculate magnifying power using m = f_o/f_e = 144 cm / 6.0 cm = 24
Step 2: Calculate separation using L = f_o + f_e = 144 cm + 6.0 cm = 150 cm