Question

(a) An astronomical telescope uses an eyepiece of power +80.000 D. If the length of the tube is 1.6125 m, what is the magnification of the telescope? (b) A 15x Galilean telescope adjusted for a relaxed eye is 42.0 cm long. Find the power of each lens
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Answer #1

Here

Power = 1/fe = 80.000

Therefore'

fe = 0.0125 m

Therefore

fo + fe = 1.6125

Therefore

fo = 1.6125 - 0.0125

= 1.6 m

Magnification = fo/fe

= 1.6/0.0125

= 128

(B)

Magnification is given as.

m= fo/fe

fo =m*fe

Here

fe +fo = d

So,

fe = d/(1+m)

Given m = 15

d = 42 cm

fe = 2.625 cm = 0.02624 m

Calculate the power of each lens.

Pe = 1/fe = 38.1 D

fo = 15 * 0.02624 = 0.3936 m

Po = 1/fo = 2.54 D

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