- Fix a metre rule on the bench with the graduated face up.
- Place the illuminated object at the zero end of the rule and the screen at the other end as illustrated in the diagram above.
- Measure and record D, the distance between the object and the screen. Evaluate D\(^{2}\).
- Place and move the converging lens between the illuminated object and the screen until a diminished sharp image of the object is formed on the screen. Read and record the position, X\(_{1}\), of the lens. From this position, move the lens towards the object until another sharp image of the object is formed on the screen. Read and record the new position x\(_{2}\), of the lens.
- Evaluate and record L (x\(_{1}\) – x\(_{2}\)), L\(^{2}\)) and (D\(^{2}\) – L\(^{2}\))
- Repeat the procedure for D = 90, 80, 70 and 60 cm. In each case, evaluate, L L\(^{2}\) and (D\(^{2}\) – L\(^{2}\)). Tabulate your readings.
- Plot a graph of D\(^{2}\) – L\(^{2}\) on the vertical axis against D on the horizontal axis.
- Determine the slope, S, of the graph and evaluate K = \(\frac{s}{4}\). State two precautions taken to ensure accurate results.
(b)i. Distinguish between a real image and a virtual image.
Draw a ray diagram to show how a converging lens may be used to form a real diminished image of an object.
Explanation
Table os values/observation
| D(cm) | D\(^{2}\)cm | X\(_{1}\)(cm) | X\(_{2}\)(cm) | L=(x\(_{1}\)-x\(_{2}\))cm | L\(^{2}\)cm | D\(^{2}\)-L\(^{2}\)cm |
| 100 | 10000 | 80.50 | 18.30 | 62.20 | 3868.8 | 6131.16 |
| 90 | 8100 | 70.20 | 19.00 | 51.20 | 2621.4 | 5478.56 |
| 80 | 6400 | 59.10 | 20.00 | 39.10 | 1528.8 | 4871.19 |
| 70 | 4900 | 46.70 | 22.00 | 24.70 | 610.09 | 4289.91 |
| 60 | 3600 | 29.00 | 7.60 | 21.40 | 457.96 | 3142.04 |
Slope (s) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{6000-3000}{100-50} = \frac{300}{50}\) = 60
Evaluate K = \(\frac{8}{4} = \frac{60}{4}\) = 15
Precaution:
- I avoided parallax when reading the metre rule.
- I ensured that a sharp image is formed on the Screen.
(b) Real image is formed by the actual intersection of rays whereas a virtual image is formed by the apparent intersection of rays when their directions have been produced backwards, or a Real image is formed at the front of the mirror while a virtual image is formed at the back of the mirror.
ii
An object beyond 2F, the image is real, inverted, smaller than an object, between F and 2F, ray parallel to the principal axis and ray passes through the optical centre.