I\'ve worked on these and cannot figure them out! Please help and show steps! 1.

Question

I've worked on these and cannot figure them out! Please help and show steps!

1.             A 198x astronomical telescope is adjusted for a relaxed eye when the two lenses are 1.06m apart. What is the focal length of the eyepiece?

2.             What is the focal length of the objective?

3.             What is the magnifying power of a +3.40 D lens used as a magnifier? Assume a relaxed normal eye.

Bonus Question: What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.77m and an eyepiece whose focal length is 3.20cm?

Explanation / Answer

for a telelscope the length of the telelscope = sum of the focus of the objective and the eye piece

magnification = fo/fe

L = 1.06 m

Power = 1/f in m

thus

fo + fe = 1.06.....................1

fo / fe = 198...........................2

fo = 198 fe

putting in 1

199 fe = 1.06 m

fe = 0.0053266 m

fo = 1.0546 m

magnifiying power of a magnifying glass = 25/f for a relaxed eye

here f is in cm

1/f = 3.4 here f in m

f = 29.4 cm

magnifying power = 25/29.4 = 0.85

for astromomical telescope magnifying power = fo / fe

R = 5.77

R = 2f

fo = 2.885 m

fe = 3.2cm

M = fo/fe = 2.885 x 100 / 3.2 = 90.156

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