Determine and record the approximate focal length f\(_{o}\) of the concave mirror provided. Arrange the ray box, the mirror, and the screen as shown in the ...
Question 1 Report
- Determine and record the approximate focal length f\(_{o}\) of the concave mirror provided.
- Arrange the ray box, the mirror, and the screen as shown in the diagram above.
- Adjust the ray box to a distance b = 20.0cm from the mirror.
- Adjust the position of the screen until a sharp image of the cross wire of the ray box is formed on it.
- Measure and record the distance, a, of the screen from the mirror. Evaluate \(\frac{a}{a}\) =1.
- Repeat the procedure for four other values of b = 25.0, 30.0, 35.0, and 40.0cm. Tabulate your readings.
- Plot a graph of l on the vertical axis against a on the horizontal axis.
- Determine the slope,.s, of the graph. Evaluate S\(^{-1}\).
- State two precautions taken to ensure accurate results.
(b)i. An object is placed at a distance of 10cm in front of a concave mirror of focal length of 15cm. Determine the characteristics of the image formed.
ii. Briefly describe how you obtained f\(_{o}\) in (a)i) above.
Answer Details
The diagram shows the apparatus: a ray box carrying an illuminated cross-wire object, a concave mirror facing it, and a small screen between them. The distance from the ray box (object) to the mirror is b and the distance from the screen (image) to the mirror is a.
(a) The practical
For each object distance \(b\) (20.0, 25.0, 30.0, 35.0, 40.0 cm) the screen is moved until the cross-wire image is sharp, and \(a\) is read off. The mirror equation is
\[\frac{1}{a}+\frac{1}{b}=\frac{1}{f_o}\]Writing \(l=\dfrac{1}{a}\) and rearranging,
\[l=\frac{1}{a}=\frac{1}{f_o}-\frac{1}{b}.\]A graph of \(l=\dfrac{1}{a}\) (vertical) against \(\dfrac{1}{b}\) (horizontal) is a straight line of slope \(S=-1\) whose intercept on the vertical axis equals \(\dfrac{1}{f_o}\). Hence the focal length is obtained from that intercept, \(f_o=\dfrac{1}{\text{intercept}}\), and \(S^{-1}=-1\).
Sample table (illustrative)
| b/cm | a/cm | l = 1/a (cm-1) |
|---|---|---|
| 20.0 | ~30.0 | 0.033 |
| 25.0 | ~26.0 | 0.038 |
| 30.0 | ~24.0 | 0.042 |
| 35.0 | ~22.5 | 0.044 |
| 40.0 | ~21.5 | 0.047 |
Two precautions:
- The image on the screen was made as sharp as possible before each reading to avoid parallax and focusing error.
- The distances \(a\) and \(b\) were measured horizontally from the pole of the mirror, with the ray box, screen and mirror kept on the same straight line (aligned axis).
(b)(i) Object 10 cm in front of a concave mirror, f = 15 cm
Here \(u=10\text{ cm}\), \(f=15\text{ cm}\) (object inside the focal point).
\[\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{10}=\frac{2-3}{30}=-\frac{1}{30}\]\[v=-30\text{ cm}.\]Magnification:
\[m=\left|\frac{v}{u}\right|=\frac{30}{10}=3.\]The negative \(v\) means the image is behind the mirror. Characteristics of the image: it is virtual, erect (upright), magnified (three times the object size), and formed 30 cm behind the mirror.
(b)(ii) How \(f_o\) was obtained in (a): A graph of \(l=\tfrac{1}{a}\) against \(\tfrac{1}{b}\) was plotted; the intercept on the vertical axis is \(\tfrac{1}{f_o}\), so \(f_o\) is the reciprocal of that intercept.
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