Using the diagram above as a guide, carry out the following instructions: Fix. the drawing paper provided on the drawing board Place the mirror vertically w...
Question 1 Report
Using the diagram above as a guide, carry out the following instructions:
- Fix. the drawing paper provided on the drawing board
- Place the mirror vertically with its longer side resting on the drawing paper. Trace the outline AB of the mirror. Remove the mirror.
- Draw a normal PQ to meet the outline at the middle Q.
- Draw a straight line through A to meet the outline of the mirror at the right angle.
- Trace the incident ray, Q with pins \(P_{1}\) and \(P_{2}\) so that it meets the perpendicular line through A at C such that \(CA=X= 1.0m\).
- Replace the mirror on its outline. Locate the images of \(P_{1}P_{2}\) through the mirror using two other pins \(P_{3}\) and \(P_{4}\) so that \(P_{3}\) and \(P_{4}\) and the images of \(P_{1}\) and \(P_{2}\) are in a straight line.
- Remove the mirror and pins \(P_{3}\) and \(P_{4}\).
Draw a straight line through the pin points to meet AB at Q and CA produced at D. - Measure and record angle ACQ as \(è_{1}\), and angle ADCQ as \(è_{2}\). Also, record the value of x. Evaluate \(è = (è_{1} + è_{2})\), \(x^{-1}\) and tan è.
- Repeat the procedure for four other values of x = 2.0, 3.0, 4.0 and 5.0cm. Tabulate your readings.
- Plot a graph of tan è on the vertical axis against x on the horizontal axis.
- Determine the slope, s, of the graph. Evaluate k = 2s
- State two precautions taken to ensure accurate results. Attach your traces to your answer booklet.
(b)i. Distinguish between regular and diffused reflections.
ii. An object is situated 25cm in front of a plane mirror. Determine the distance of the image from the object. What is the size of the image relative to the object?
Answer Details
(a) Plane-mirror reflection experiment
An incident ray is traced from C to a point on the mirror using pins \(P_1\) and \(P_2\), and its reflected ray is located behind the mirror with pins \(P_3\) and \(P_4\). For each offset \(x = CA\) the two base angles \(\theta_1\) and \(\theta_2\) are measured, the mean angle \(\theta = \tfrac{1}{2}(\theta_1+\theta_2)\) is found, and \(x^{-1}\) and \(\tan\theta\) are evaluated. A complete set of readings is shown below.
| S/N | \(\theta_1\) (°) | \(\theta_2\) (°) | x (cm) | \(x^{-1}\) (cm-1) | \(\theta=\tfrac{1}{2}(\theta_1+\theta_2)\) (°) | \(\tan\theta\) |
|---|---|---|---|---|---|---|
| 1 | 83.0 | 81.0 | 1.0 | 1.00 | 82.0 | 7.115 |
| 2 | 76.0 | 74.0 | 2.0 | 0.50 | 75.0 | 3.732 |
| 3 | 69.0 | 70.0 | 3.0 | 0.33 | 69.5 | 2.675 |
| 4 | 62.0 | 63.0 | 4.0 | 0.25 | 62.5 | 1.921 |
| 5 | 57.0 | 58.0 | 5.0 | 0.20 | 57.5 | 1.570 |
Since \(\tan\theta\) rises steadily as \(x^{-1}\) increases (the product \(x\tan\theta\) stays close to a constant of about 7.6), the straight-line relationship is obtained by plotting \(\tan\theta\) against \(x^{-1}\), and the graph passes close to the origin.
Graph of \(\tan\theta\) against \(x^{-1}\):
Slope and value of k. Reading two well-separated points on the line of best fit, \((x^{-1}=0,\ \tan\theta \approx 0.15)\) and \((x^{-1}=1.0,\ \tan\theta \approx 7.05)\):
\[ s = \frac{\Delta(\tan\theta)}{\Delta(x^{-1})} = \frac{7.05 - 0.15}{1.0 - 0} = 6.9 \] \[ k = 2s = 2 \times 6.9 = 13.8 \approx 14 \]Two precautions:
- Fix the pins truly vertical (not bent) and reasonably far apart so that each ray is well defined.
- Sight the pins and their images with one eye at the level of the paper, and draw thin lines with a sharp pencil, to avoid errors due to parallax.
(b)(i) Regular versus diffused reflection
Regular (specular) reflection occurs at a smooth, polished surface such as a plane mirror: a parallel beam of incident rays is turned back as a parallel beam in one definite direction, so a clear image is formed. Diffused (irregular) reflection occurs at a rough surface such as paper or a wall: parallel incident rays are scattered in many different directions, so no clear image is formed. In both cases the laws of reflection are obeyed at every point; only the surface differs.
(b)(ii) Distance of the image from the object
For a plane mirror the image is formed as far behind the mirror as the object is in front of it. The object is \(25\ \text{cm}\) in front, so the image is \(25\ \text{cm}\) behind the mirror. The distance between the object and its image is therefore:
\[ d = 25 + 25 = 50\ \text{cm} \]The image is the same size as the object (magnification = 1); it is virtual, erect and laterally inverted.
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