Question

A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

1. the telescope is in normal adjustment (i.e., when the final image is at infinity)?
2. the final image is formed at the least distance of distinct vision (25 cm)?

Answer 1

Focal length of the objective lens, f_o = 140 cm

Focal length of the eyepiece, f_e = 5 cm

Least distance of distinct vision, d = 25 cm

(a) When the telescope is in normal adjustment, its magnifying power is given as:

${\displaystyle \text{m}=\frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}}}$

= ${\displaystyle \frac{140}{5}}$

= 28

(b) When the final image is formed at d, the magnifying power of the telescope is given as:

${\displaystyle \frac{{\text{f}}_{\text{o}}}{{\text{f}}_{\text{e}}} [1+\frac{{\text{f}}_{\text{e}}}{\text{d}}]}$

= ${\displaystyle \frac{140}{5}[1+\frac{5}{25}]}$

= 28[1 + 0.2]

= 28 × 1.2

= 33.6