You are provided with a metre rule, lens, screen, ray box, and other necessary apparatus. i. Set up the experiment as shown in the diagram above. Measure an...
Question 1 Report
You are provided with a metre rule, lens, screen, ray box, and other necessary apparatus.
i. Set up the experiment as shown in the diagram above. Measure and record the diameter \(a_{0}\), of the illuminated object.
ii. Place the object at a distance \(x = 25\text{cm}\) from the lens. Adjust the screen until a sharp image is obtained on the screen.
iii. Measure and record the diameter, \(a\), of the image.
iv. Measure and record the distance \(v\) between the lens and the screen.
v. Evaluate \(y = P = \frac{1+y^{2}}{y}\) and \(T = x+v\).
vi. Repeat the procedure for \(x = 30\text{cm}\), \(35\text{cm}\), \(40\text{cm}\) and \(45\text{cm}\). In each case, determine the corresponding values of \(a,v,y, P\) and \(T\).
vii. Tabulate your results.
viii. Plot a graph of \(P\) on the vertical axis against \(T\) on the horizontal axis starting both axes from the origin \((0,0)\).
ix. Determine the slope, \(s\), of the graph.
x. Determine the intercept, \(c\), on the horizontal axis.
xi Evaluate \(K = \frac{c}{2}\)
xii. State two precautions taken to ensure accurate results.
(b)i. Explain the statement, the focal length of a converging lens is 20cm.
ii. An object is placed at a distance x from a converging lens of focal length 20cm. If the magnification of the real image is 5, calculate the value of x.
Answer Details
(a) Determination of image magnification with a converging lens
(i) Diameter of the illuminated object, \(a_{0} = 2.00\ \text{cm}\).
(vii) Table of results
| S/N | x (cm) | a (cm) | v (cm) | y = a/a₀ | P = (1+y²)/y | T = x+v (cm) |
|---|---|---|---|---|---|---|
| 1 | 25.0 | 2.60 | 49.0 | 1.30 | 2.07 | 74.00 |
| 2 | 30.0 | 2.40 | 46.0 | 1.20 | 2.03 | 76.00 |
| 3 | 35.0 | 2.10 | 45.0 | 1.05 | 2.00 | 80.00 |
| 4 | 40.0 | 1.80 | 43.0 | 0.90 | 2.01 | 83.00 |
| 5 | 45.0 | 1.60 | 40.0 | 0.80 | 2.05 | 85.00 |
Sample evaluations (S/N 1)
\[ y = \frac{a}{a_{0}} = \frac{2.60}{2.00} = 1.30 \] \[ P = \frac{1+y^{2}}{y} = \frac{1+(1.30)^{2}}{1.30} = \frac{1+1.69}{1.30} = \frac{2.69}{1.30} = 2.07 \] \[ T = x + v = 25.0 + 49.0 = 74.00\ \text{cm} \]Sample evaluations (S/N 3)
\[ y = \frac{2.10}{2.00} = 1.05,\qquad P = \frac{1+(1.05)^{2}}{1.05} = \frac{2.1025}{1.05} = 2.00,\qquad T = 35.0 + 45.0 = 80.00\ \text{cm} \](viii) Graph of P against T
(ix) Slope of the graph
Two points are read on the line of best fit:
\[ (T_{1},\,P_{1}) = (74.00\ \text{cm},\ 2.070) \qquad (T_{2},\,P_{2}) = (80.00\ \text{cm},\ 2.037) \] \[ s = \frac{T_{2}-T_{1}}{P_{2}-P_{1}} = \frac{80.00 - 74.00}{2.070 - 2.037} = \frac{6.00}{0.033} = 181.8\ \text{cm} \](x) Intercept on the horizontal (T) axis
Extending the line of best fit to the horizontal axis gives
\[ c = 81.0\ \text{cm} \](xi) Evaluation of K
\[ K = \frac{c}{2} = \frac{81.0}{2} = 40.5\ \text{cm} \](xii) Two precautions
- The object, the lens and the screen were kept coaxial, with their centres at the same height on a straight line.
- Parallax error was avoided when reading the metre rule, and the screen was adjusted until the sharpest possible image was obtained.
(b)(i) Meaning of “focal length of a converging lens is 20 cm”
It means that the distance between the optical centre of the lens and its principal focus is 20 cm; that is, a beam of light travelling parallel to the principal axis is converged (brought to a focus) at a point 20 cm from the optical centre of the lens.
(b)(ii) Value of x for a real image of magnification 5
For a real image the magnification is
\[ m = \frac{v}{u} = 5 \quad\Rightarrow\quad v = 5u \]Applying the lens formula with \(f = 20\ \text{cm}\):
\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \] \[ \frac{1}{20} = \frac{1}{u} + \frac{1}{5u} = \frac{5+1}{5u} = \frac{6}{5u} \] \[ 5u = 20 \times 6 = 120 \] \[ u = \frac{120}{5} = 24\ \text{cm} \]Therefore \(x = 24\ \text{cm}\).
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