| Date | November 2016 | Marks available | 1 | Reference code | 16N.3.SL.TZ0.12 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Identify | Question number | 12 | Adapted from | N/A |
Question
A lamp is located 6.0 m from a screen.
Somewhere between the lamp and the screen, a lens is placed so that it produces a real inverted image on the screen. The image produced is 4.0 times larger than the lamp.
Identify the nature of the lens.
Determine the distance between the lamp and the lens.
Calculate the focal length of the lens.
The lens is moved to a second position where the image on the screen is again focused. The lamp–screen distance does not change. Compare the characteristics of this new image with the original image.
Markscheme
converging/positive/biconvex/plane convex
Do not accept convex.
\(\frac{v}{u} = 4\)
Award [3] for a bald correct answer.
v + u = 6
Allow [1] if the answer is 4.8 «m».
so lens is 1.2 «m» from object or u = 1.2 «m»
«\(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\), so \(\frac{1}{{1.2}} + \frac{1}{{4.8}} = \frac{1}{f}\), so» f = 0.96 «m» or 1 «m»
Watch for ECF from (b)
real AND inverted
smaller OR diminished
