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= 25cm 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 = 30cm, 35cm, 40cm and 45cm. 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.
Table of value/observation
a\(_{0}\) = 2.00cm
| S/N | x(cm) | a(cm) | v(cm) | y=\(\frac{a}{a_{0}}\) | p=\(\frac{1+y^{2}}{y}\) | 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 | 2.01 | 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 |
viii.
viii. Slopes(s) = \(\frac{x_{2} - x_{1}}{y_{2} - y_{1}}\) = \(\frac{76 - 70}{2.073 - 2.040}\)
= \(\frac{6}{0.033}\) = 181.8
x. Intercept (c) = 81
xi. Evaluation k = \(\frac{c}{2}\)
= \(\frac{81}{2}\)
= 40.5
xii. Precautions;
(a) Conxial arrangement of optical instruments
(b) Avoided parallax error in reading metre rule
(c) Lens kept upright
(d) Repeated readings shown on the table
(e) Surface of lens cleaned
(f) Avoided zero error on metre rule
(b)i. The distance between the optical centre and the principal focus of the lens is 20cm or if the parallel beam of light is incident on the lens, it is brought to a focus at a point 20cm from the lens.
ii. M = 5cm, F = 20c, x = u
M = \(\frac{v}{u}\)
5 = \(\frac{v}{u}\)
v = 5u
Using the formula
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{5u} + \frac{1}{u} = \frac{1}{20}\)
\(\frac{1+5}{5u} = \frac{1}{20}\)
5u = 20(1 + 5)
5u = 20 x 6
5u = 120
u = \(\frac{120}{5}\) = 24
x = 24cm
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