Respuesta :
Answer:
a) [tex]f_{e}[/tex] = 3,375 cm , b) f₀ = 77.625 cm
Explanation:
The magnification of a telescope is, to see at the far point of vision (infinity image)
m = - f₀ / [tex]f_{e}[/tex]
The length of the tube is
L = f₀ + [tex]f_{e}[/tex]
a) The focal length of the eyepiece
L = - m [tex]f_{e}[/tex] + [tex]f_{e}[/tex]
L = [tex]f_{e}[/tex] (1-m)
[tex]f_{e}[/tex] = L / (1-m)
Let's calculate
[tex]f_{e}[/tex] = 81.0 / (1 - (-23.0)
[tex]f_{e}[/tex] = 3,375 cm
b) the focal length of the target
f₀ = -m [tex]f_{e}[/tex]
f₀ = 23 (3.68)
f₀ = 77.625 cm
Assuming that this telescope would allow an observer to view a lunar crater in focus with a completely relaxed eye, the focal length fe of the eyepiece is:
- a) = 3,375 cm
The focal length of the objective lens (fo = ?cm) is:
- b) f₀ = 77.625 cm
What is Focal Length?
This refers to the distance from a lens centre and the focus.
Therefore, to calculate
a) The focal length of the eyepiece
- L = - m Fe+ Fe
- L = Fe (1-m)
- Fe = L / (1-m)
Let's calculate
Fe= 81.0 / (1 - (-23.0)
Fe = 3,375 cm
b) the focal length of the target
- f₀ = -m Fe
- f₀ = 23 (3.68)
- f₀ = 77.625 cm
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