The largest refracting telescope in the world is at yerkes observatory in wisconsin. the objective lens is 1.02 in diameter and has a focal length of 19.4 . suppose you want to magnify jupiter, which is 1.38×105 in diameter, so that its image subtends an angle of (about the same as the moon) when it is 6.28×108 from earth.
a) what focal-length eyepiece do you need?

fe = 2.1 cm

Explanation:

Given that,

focal length of objective = fo = 19.4 ;

diameter of Jupiter = d = 1.38 x 10^5 km

θo = 12°

distance from earth = o = 6.28 x 10^8 km

we need to determine the focal length of eye piece fe.

The given diameter of Jupiter will act as the perpendicular of right angled triangle, and the base will be the distance. Using trigonometry we can find the angle subtended by objective.

tanθe = 1.38 x 10^5 km / 6.28 x 10^8 km = 0.0002197

θe = 0.013°

we know that, magnification is given by:

m = θe/θo = 12 / 0.013 = 923

Also, m = fo / fe

=> fe = fo / m = 19.4/923 = 0.021 m = 2.1 cm

Hence, fe = 2.1 cm

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6