A telescope has an eye piece of focal length 5.0 cm. The objective lens has focal length 12 times that of the eye-piece. In normal adjustment, what is the separation between the two lenses?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 65 cm
In a telescope in normal adjustment,
$d=f_{objective}+f_{eyepiece}$
$=5+12×f_{eyepiece}$
$=5+12×5$
$=65cm$