An astronomical telescope uses an objective lens of focal length 30 cm. In normal adjustment, distance between the two lenses is 32 cm. The magnifying power of telescope is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 15
Distance between the objective and eyepiece lens, $d=32cm$
focal length of the objective lens, $f_0=30cm$
$d=f_0+f_e$
$f_e=d-f_0=32-30=2cm$
$⇒f_e=2cm$
Magnifying Power, $M=\frac{f_0}{f_e}=\frac{30}{2}=15$