Using the above diagram as a guide, carry out the following instructions. Fix a metre rule on the bench with its graduated side facing up. Place the illumin...
Question 1 Report
Using the above diagram as a guide, carry out the following instructions.
- Fix a metre rule on the bench with its graduated side facing up.
- Place the illuminated object at the 0cm end and the screen at the 100cm end of the rule such that the distance d between the illuminated object and screen is 100cm.
- Record the distance d. Also, evaluate and record d\(^{2}\).
- Place and move the Converging lens between the illuminated object and the Screen until a sharp diminished image of the object is formed on the screen.
- Read and record the position Ih of the lens.
- Now move the lens towards the object until another sharp image of the object is formed on the Screen.
- Read and record the new position l\(_{2}\) of the lens.
- Evaluate and record L= (l\(_{1}\) - I\(_{2}\)), also evaluate L\(^{2}\) and D = d\(^{2}\) - L\(^{2}\).
- Repeat the procedure for four values of d= 85, 75, 65 and 55cm. Tabulate your readings.
- Plot a graph or Don the vertical axis against d on the horizontal axis.
- Determine the slope, s, of the graph.
- Evaluate k = \(\frac{s}{4}\).
- State two precautions taken to obtain accurate results.
(b)i. Explain the statement the focal length of a converging lens is 15cm.
ii. Distinguish between a real image and a virtual image.
Answer Details
Principle (lens displacement method). With the illuminated object and the screen a fixed distance \(d\) apart (and \(d>4f\)), there are two positions of the converging lens, \(l_1\) and \(l_2\), that each throw a sharp image on the screen – one diminished and one magnified. If \(L=(l_1-l_2)\) is the separation of these two positions, the focal length \(f\) is given by
\[ f=\frac{d^{2}-L^{2}}{4d}. \]Writing \(D=d^{2}-L^{2}\), this rearranges to
\[ D=4f\,d, \]so a graph of \(D\) (vertical axis) against \(d\) (horizontal axis) is a straight line through the origin whose slope is \(s=4f\).
Table of readings.
| d (cm) | d² (cm²) | l₁ (cm) | l₂ (cm) | L = l₁ − l₂ (cm) | L² (cm²) | D = d² − L² (cm²) |
|---|---|---|---|---|---|---|
| 100 | 10000 | 86.0 | 13.9 | 72.1 | 5198.41 | 4801.59 |
| 85 | 7225 | 70.6 | 14.5 | 56.1 | 3147.21 | 4077.79 |
| 75 | 5625 | 60.0 | 15.0 | 45.0 | 2025.00 | 3600.00 |
| 65 | 4225 | 49.1 | 15.9 | 33.2 | 1102.24 | 3122.76 |
| 55 | 3025 | 37.3 | 17.7 | 19.6 | 384.16 | 2640.84 |
Graph of D against d.
The points lie on a straight line through the origin, confirming \(D=4f\,d\).
Slope of the graph. Taking two widely separated points on the line of best fit, \((d_1,D_1)=(55,\,2640)\) and \((d_2,D_2)=(100,\,4800)\):
\[ s=\frac{D_2-D_1}{d_2-d_1}=\frac{4800-2640}{100-55}=\frac{2160}{45}=48.0. \]Evaluating k.
\[ k=\frac{s}{4}=\frac{48.0}{4}=12.0\ \text{cm}. \]Since \(s=4f\), we have \(k=\dfrac{s}{4}=f\); the value \(k=12.0\ \text{cm}\) is therefore the focal length of the converging lens.
Two precautions.
- The eye was placed directly in front of the scale when reading the lens position, so that parallax error in reading the metre rule was avoided.
- The illuminated object, the lens and the screen were kept centred and vertical with their optical centres in one straight horizontal line throughout the experiment.
(b)(i) The statement “the focal length of a converging lens is 15 cm” means that a beam of light travelling parallel to the principal axis is refracted by the lens and brought to a focus at the principal focus, which lies a distance of 15 cm from the optical centre of the lens.
(b)(ii) Distinction between a real image and a virtual image.
| Real image | Virtual image |
|---|---|
| Formed by the actual intersection of refracted (or reflected) light rays. | Formed where the light rays only appear to intersect when produced backwards. |
| Can be caught and displayed on a screen. | Cannot be caught on a screen. |
| Usually inverted relative to the object. | Erect (upright) relative to the object. |
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