An astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance
- 50.0 cm
- 54.0 cm
- 37.3 cm
- 46.0 cm
Answer
(2) 54.0 cm
Solution
For the objective: u = -200 cm, fo = 40 cm
1v - 1u = 1fo
1v = 140 - 1200 = 5 - 1200 = 150
v = 50 cm
For the final image at infinity (normal adjustment), this image must lie at the focus of the eyepiece.
Separation = v + fe = 50 + 4 = 54.0 cm
Answer: (2)