A telescope with a Barlow lens
- Round
- Theory
- Points
- 20
- Level
- Silver
- Taught in
- Section 5.3
Adapted from IOAA 2016 (Bhubaneswar), theory problem 7. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.
A refracting telescope of focal ratio f/5 has an objective of focal length 100 cm and an eyepiece of focal length 1 cm.
(1) What is its angular magnification
To double the magnification, a Barlow lens of focal length −1 cm is placed between the objective and the focus of the objective.
(2) How far in front of the objective's focus must the Barlow lens be placed? (6 points)
(3) By how much does the telescope become longer? (4 points)
(4) The eyepiece and the Barlow lens are removed, and a camera with pixels 10 µm wide is placed at the objective's focus. How many pixels apart are the images of two stars 20″ apart? (6 points)
-
By (5.5),
, and the length is cm. -
Let the Barlow lens be a distance
in front of the focus P of the objective. The light is converging towards P, behind the lens, so in the lens equation of Section S4 the object distance is . A Barlow lens that doubles the magnification forms its image of P twice as far from the lens as P, so . With cm,so
cm, and the new image lies cm behind the lens. -
The image has moved from 0.5 cm behind the lens to 1.0 cm behind it, so the eyepiece must move back by
cm, and the telescope becomes 0.5 cm longer. -
By (5.3), each pixel covers
, so the two stars are pixels apart.
AnswerThe telescope magnifies 100 times and is 101 cm long. The Barlow lens goes 0.5 cm in front of the objective's focus, the telescope becomes 0.5 cm longer, and the two stars are about 10 pixels apart on the camera.
This problem also appears among the olympiad problems of Section 5.3, Eyepieces and magnification, which teaches the ideas it uses. Equation and figure numbers refer to the course.