# Q 9.13 A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

P Pankaj Sanodiya

The magnifying power of the telescope is given by

$m=\frac{f_{objective}}{f_{eyepiece}}$

Here, given,

focal length of objective lens =  $f_{objective}=$   144 cm

focal length of eyepiece lens = $f_{eyepiece}=$  6 cm

$m=\frac{f_{objective}}{f_{eyepiece}}=\frac{144}{6}=24$

Hence magnifying power of the telescope is 24.

in the telescope distance between the objective and eyepiece, the lens is given by

$d= f_{objective}+ f_{eyepiece}$

$d = 144+6=150$

Therefore, the distance between the two lenses is 250 cm.

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