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
In normal adjustment the final image is at infinity, so the lens separation is \( L = f_o + f_e \) and the magnifying power is \( m = f_o/f_e \).
\( f_e = L - f_o = 32\,\mathrm{cm} - 30\,\mathrm{cm} = 2\,\mathrm{cm} \)
\( \displaystyle m = \frac{f_o}{f_e} = \frac{30}{2} = 15 \)