A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distant vision on a scale 200 cm away from the objective. Calculate the separation between objective and eyepiece:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 70.8 cm
The objective forms a real image of the scale, and this image acts as the object for the eyepiece, which gives the final image at the least distance of distinct vision, 25 cm. The separation is \( v_o + |u_e| \). Sign convention: distances measured from the lens, positive along the incident light.
Objective (\( u_o = -200\,\mathrm{cm} \), \( f_o = 50\,\mathrm{cm} \))
\( \displaystyle \frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} = \frac{1}{50} - \frac{1}{200} = \frac{3}{200} \)
\( \displaystyle v_o = \frac{200}{3}\,\mathrm{cm} = 66.7\,\mathrm{cm} \)
Eyepiece (\( v_e = -25\,\mathrm{cm} \), \( f_e = 5\,\mathrm{cm} \))
\( \displaystyle \frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \)
\( \displaystyle u_e = -\frac{25}{6}\,\mathrm{cm} = -4.17\,\mathrm{cm} \)
Separation
\( L = 66.7 + 4.17 = 70.8\,\mathrm{cm} \)