Focal length of the objective of an astronomical telescope is 1 m. Under normal adjustment, length of the telescope is 1.05 m. Calculate focal length of the eyepiece and magnifying power under normal adjustment.

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1 Answers

Given: f0 = 1 m, L = 1.05 m

To find:

i. Focal length of eyepiece (fe)

Magnifying power under normal adjustment (M)

Formula:

i. L = f0 + fe

ii. M = \(\frac{f_0}{f_e}\)

Calculation: From formula (i),

fe = L – f0 = 1.05 – 1 = 0.05 m

From formula (ii),

M = \(\frac{1}{0.05}\) = 20

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