A telescope has an objective of focal length 30 cm and an eyepiece of focal length 3.0 cm. It is focused on a scale distant 2.0 m. For seeing with relaxed eye, the separation between the objective and eyepiece would be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 38.3 cm
The scale is at a finite distance, so the objective forms its image beyond its focus, at a distance found from the lens formula. For a relaxed eye the final image is at infinity, so this image must lie at the focus of the eyepiece.
Sign convention: distances measured from the lens, positive in the direction of the incoming light.
\( u = -2.0\,\mathrm{m} = -200\,\mathrm{cm} \)
\( f_o = +30\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} = \frac{1}{f_o} + \frac{1}{u} = \frac{1}{30} - \frac{1}{200} = \frac{20 - 3}{600} = \frac{17}{600} \)
\( \displaystyle v = \frac{600}{17}\,\mathrm{cm} = 35.3\,\mathrm{cm} \)
The eyepiece must be \( f_e = 3.0\,\mathrm{cm} \) beyond this image.
\( \displaystyle L = v + f_e = 35.3\,\mathrm{cm} + 3.0\,\mathrm{cm} = 38.3\,\mathrm{cm} \)