Date | May 2011 | Marks available | 2 | Reference code | 11M.3.SL.TZ2.20 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 2 |
Command term | Define | Question number | 20 | Adapted from | N/A |
Question
This question is about a converging lens.
Define angular magnification.
A thin converging lens of focal length 4.5 cm is to be used as a magnifying glass. The observer places the lens close to her eye. The least distance of distinct vision is 24 cm.
(i) Show that the distance of the object from the lens is 3.8 cm.
(ii) Determine the angular magnification produced by the lens.
Suggest two reasons why, for high magnifications, a combination of lenses is used rather than a single lens.
Markscheme
ratio of angle subtended by image and angle subtended by object;
angles measured at eye;
(i) \(\frac{1}{u} - \frac{1}{{24}} = \frac{1}{{4.5}}\);
distance=3.8 cm
(ii) angular magnification \( = \frac{{{h_{\rm{I}}}}}{D} \div \frac{{{h_{\rm{O}}}}}{D}\);
= linear magnification;
\( = \frac{v}{u}\);
\( = \frac{{24}}{{3.8}}\)
=6.3;
Award [2 max] for use of linear magnification alone.
less spherical/chromatic aberration (than single lens);
greater aperture can be used/greater light-collecting ability (than single lens);