Date | May 2012 | Marks available | 4 | Reference code | 12M.3.SL.TZ2.18 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 2 |
Command term | Calculate and Determine | Question number | 18 | Adapted from | N/A |
Question
This question is about an astronomical telescope.
A particular astronomical telescope is being used to observe the Moon. The ray diagram shows the position P of the intermediate image of the Moon formed by the objective lens.
The telescope is in normal adjustment.
On the diagram above,
(i) label with the letter F the two focal points of the eyepiece lens.
(ii) draw rays to determine the location of the final image of the Moon.
The diameter of the Moon subtends an angle of 8.7×10–3 rad at the unaided eye.
(i) Determine the diameter of the image of the Moon formed by the objective lens.
(ii) The focal length of the eyepiece is 30 cm. Calculate the angle that the final image of the Moon subtends at the eyepiece.
Markscheme
(i) F at P and second F at the same distance to the right of the eyepiece; (judge by eye)
(ii)
first construction line or ray; (judge by eye)
second construction line or ray;
extension of these to the left as parallel lines;
(i) d=ƒ0 tanθ≈ƒ0θ ;
d=90×tan(8.7×10-3)=0.78cm;
(ii) angular magnification is \(M = \left( { - \frac{{{f_0}}}{{{f_{\rm{e}}}}} = } \right)\left( - \right)3\);
hence \(\theta ' = - 3\theta = \left( - \right)0.026{\rm{rad}}\);
or
\(\theta ' = \frac{{0.78}}{{30}}\);
\(\theta ' = \left( - \right)0.026{\rm{rad}}\);
Examiners report