1. On one of the shelves in your physics lab is displayed an antique telescope. A sign underneath the instrument says that the telescope has a magnification of 23.0 and consists of two converging lenses, the objective and the eyepiece, fixed at either end of a tube 81.0cm long. Assuming that this telescope would allow an observer to view a lunar crater in focus with a completely relaxed eye, what is the focal length fe of the eyepiece?
Note that to view the crater with a completely relaxed eye, the eyepiece must form its image at infinity and that images formed by telescopes of this nature are always inverted.
fe = ?cm
2. What is the focal length of the objective lens? (fo = ?cm)

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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