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  4. An astronomical telescope is set for normal adjustment and the distance between its objective and eye piece is 1.05 cm. The magnifying power of the telescope is 20. What is the focal length of the objective ?

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Q. An astronomical telescope is set for normal adjustment and the distance between its objective and eye piece is 1.05 cm. The magnifying power of the telescope is 20. What is the focal length of the objective ?

Ray Optics and Optical Instruments

Solution:

Here M = $f_0 / f_e \,and\, L= f_ 0 + f_e= 1 .05.$
$Hence \, f_e = 105 - f_ o.\, And 20 =f_o /(105- f _0). $
$This \,gives\, f_0 -- 1m \,and\, f_e = 0 .05 m.$