TEST OF PRACTICAL KNOWLEDGE QUESTION
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.
Practical: converging lens, image size and distances
For each object distance \(x\), the screen is adjusted for a sharp image; the image diameter \(a\), the image distance \(v\), then \(y\), \(P\) and \(T = x + v\) are evaluated and tabulated (values depend on the actual measurements):
| x (cm) | a (cm) | v (cm) | y | P | T = x+v (cm) |
|---|
| 25 | a\(_1\) | v\(_1\) | y\(_1\) | P\(_1\) | T\(_1\) |
| 30 | a\(_2\) | v\(_2\) | y\(_2\) | P\(_2\) | T\(_2\) |
| 35 | a\(_3\) | v\(_3\) | y\(_3\) | P\(_3\) | T\(_3\) |
| 40 | a\(_4\) | v\(_4\) | y\(_4\) | P\(_4\) | T\(_4\) |
| 45 | a\(_5\) | v\(_5\) | y\(_5\) | P\(_5\) | T\(_5\) |
Plot \(P\) (vertical) against \(T\) (horizontal) from the origin, read the slope \(s\) and the intercept \(c\) on the \(T\)-axis, then \(K = \dfrac{c}{2}\).
Two precautions: keep object, lens and screen centres at the same height on a straight line; obtain the sharpest possible image and read the metre rule without parallax.
(b)(i) "The focal length of a converging lens is 20 cm" means that a beam of light travelling parallel to the principal axis is brought to a focus (converges to the principal focus) at a point 20 cm from the optical centre of the lens.
(b)(ii) For a real image, magnification \(m = \dfrac{v}{u} = 5\), so \(v = 5u\). Using \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\):
\(\dfrac{1}{20} = \dfrac{1}{u} + \dfrac{1}{5u} = \dfrac{6}{5u}\), so \(5u = 6\times20 = 120\), giving \(u = 24\ \text{cm}\).
Therefore \(x = 24\ \text{cm}\).
Practical: converging lens, image size and distances
For each object distance \(x\), the screen is adjusted for a sharp image; the image diameter \(a\), the image distance \(v\), then \(y\), \(P\) and \(T = x + v\) are evaluated and tabulated (values depend on the actual measurements):
| x (cm) | a (cm) | v (cm) | y | P | T = x+v (cm) |
|---|
| 25 | a\(_1\) | v\(_1\) | y\(_1\) | P\(_1\) | T\(_1\) |
| 30 | a\(_2\) | v\(_2\) | y\(_2\) | P\(_2\) | T\(_2\) |
| 35 | a\(_3\) | v\(_3\) | y\(_3\) | P\(_3\) | T\(_3\) |
| 40 | a\(_4\) | v\(_4\) | y\(_4\) | P\(_4\) | T\(_4\) |
| 45 | a\(_5\) | v\(_5\) | y\(_5\) | P\(_5\) | T\(_5\) |
Plot \(P\) (vertical) against \(T\) (horizontal) from the origin, read the slope \(s\) and the intercept \(c\) on the \(T\)-axis, then \(K = \dfrac{c}{2}\).
Two precautions: keep object, lens and screen centres at the same height on a straight line; obtain the sharpest possible image and read the metre rule without parallax.
(b)(i) "The focal length of a converging lens is 20 cm" means that a beam of light travelling parallel to the principal axis is brought to a focus (converges to the principal focus) at a point 20 cm from the optical centre of the lens.
(b)(ii) For a real image, magnification \(m = \dfrac{v}{u} = 5\), so \(v = 5u\). Using \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\):
\(\dfrac{1}{20} = \dfrac{1}{u} + \dfrac{1}{5u} = \dfrac{6}{5u}\), so \(5u = 6\times20 = 120\), giving \(u = 24\ \text{cm}\).
Therefore \(x = 24\ \text{cm}\).
