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)

a)   = 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 /  

The length of the tube is

        L = f₀ +

a) The focal length of the eyepiece

       L = - m   +  

       L = (1-m)

        = L / (1-m)

Let's calculate

        = 81.0 / (1 - (-23.0)

        = 3,375 cm

b) the focal length of the target

       f₀ = -m  

       f₀ = 23 (3.68)

       f₀ = 77.625 cm

the first step is making objective observations.

they have to be based on a certain event and can be confirmed by others if it is right or not.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.