You are provided with an illuminated object, converging lens, screen, metre rule, and other necessary materials. Measure and record the size a\(_{o}\) of th...
Question 1 Report
You are provided with an illuminated object, converging lens, screen, metre rule, and other necessary materials.
- Measure and record the size a\(_{o}\) of the illuminated object.
- Place the object O and the screen S on Opposite sides of the converging lens L.
- Set the distance between the object and the lens U = 30cm.
- Adjust the screen until a sharp image of the illuminated object is obtained on the screen.
- Measure and record the size a of the image.
- Evaluate m = \(\frac{a}{a_{0}}\), and m\(^{-1}\)
- Repeat the procedure for four other values of U=35cm, 40cm, 45cm and 50cm respectively.
- Tabulate your readings.
- Plot a graph with m\(^{-1}\) on the vertical axis and U on the horizontal axis.
- Determine the slope, s, of the graph and intercept, C, on the vertical axis.
- Determine the value of U for which m\(^{-1}\)=0.
- State two precautions taken to obtain accurate results.
(b)i. Using your graph, determine the value of m for which U= 37cm.
ii. Sketch a diagram to illustrate how a converging lens may be used to produce a real diminished image of an object.
Answer Details
Practical: linear magnification of a converging lens
For each object distance \(U\) the object and screen are placed on opposite sides of the lens and the screen adjusted until a sharp image forms; the image size \(a\) is measured, then \(m = \dfrac{a}{a_{0}}\) and \(m^{-1}\) are evaluated. A specimen table (values illustrative):
| U (cm) | a (cm) | m = a/a\(_0\) | m\(^{-1}\) |
|---|---|---|---|
| 30 | a\(_1\) | m\(_1\) | 1/m\(_1\) |
| 35 | a\(_2\) | m\(_2\) | 1/m\(_2\) |
| 40 | a\(_3\) | m\(_3\) | 1/m\(_3\) |
| 45 | a\(_4\) | m\(_4\) | 1/m\(_4\) |
| 50 | a\(_5\) | m\(_5\) | 1/m\(_5\) |
Theory of the graph. For a real image, \(m = \dfrac{v}{u}\) and \(\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\). Combining, \(m = \dfrac{f}{u - f}\), so
\[ m^{-1} = \frac{u - f}{f} = \frac{1}{f}\,U - 1. \]
Hence a graph of \(m^{-1}\) (vertical) against \(U\) (horizontal) is a straight line of slope \(s = \dfrac{1}{f}\) and vertical intercept \(C = -1\). The value of \(U\) for which \(m^{-1} = 0\) is \(U = f\) (image formed at infinity).
Two precautions: ensure the object, lens and screen centres are at the same height and lie on a straight line; focus for the sharpest image and avoid parallax when reading the sizes.
(b)(i) Read from the straight-line graph the value of \(m^{-1}\) at \(U = 37\text{ cm}\), then \(m\) is its reciprocal (this reading comes from the candidate's own graph).
(b)(ii) To produce a real, diminished image, the object is placed beyond twice the focal length (\(u > 2f\)); the image then forms between \(F\) and \(2F\) on the other side, real, inverted and smaller than the object.
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