WAEC SSCE - Physics - 2008 (Practical)

Principle (lens displacement method). With the illuminated object and the screen a fixed distance \(d\) apart (and \(d>4f\)), there are two positions of the converging lens, \(l_1\) and \(l_2\), that each throw a sharp image on the screen – one diminished and one magnified. If \(L=(l_1-l_2)\) is the separation of these two positions, the focal length \(f\) is given by

\[ f=\frac{d^{2}-L^{2}}{4d}. \]

Writing \(D=d^{2}-L^{2}\), this rearranges to

\[ D=4f\,d, \]

so a graph of \(D\) (vertical axis) against \(d\) (horizontal axis) is a straight line through the origin whose slope is \(s=4f\).

Table of readings.

d (cm)	d² (cm²)	l₁ (cm)	l₂ (cm)	L = l₁ − l₂ (cm)	L² (cm²)	D = d² − L² (cm²)
100	10000	86.0	13.9	72.1	5198.41	4801.59
85	7225	70.6	14.5	56.1	3147.21	4077.79
75	5625	60.0	15.0	45.0	2025.00	3600.00
65	4225	49.1	15.9	33.2	1102.24	3122.76
55	3025	37.3	17.7	19.6	384.16	2640.84

Graph of D against d.

The points lie on a straight line through the origin, confirming \(D=4f\,d\).

Slope of the graph. Taking two widely separated points on the line of best fit, \((d_1,D_1)=(55,\,2640)\) and \((d_2,D_2)=(100,\,4800)\):

\[ s=\frac{D_2-D_1}{d_2-d_1}=\frac{4800-2640}{100-55}=\frac{2160}{45}=48.0. \]

Evaluating k.

\[ k=\frac{s}{4}=\frac{48.0}{4}=12.0\ \text{cm}. \]

Since \(s=4f\), we have \(k=\dfrac{s}{4}=f\); the value \(k=12.0\ \text{cm}\) is therefore the focal length of the converging lens.

Two precautions.

1. The eye was placed directly in front of the scale when reading the lens position, so that parallax error in reading the metre rule was avoided.
2. The illuminated object, the lens and the screen were kept centred and vertical with their optical centres in one straight horizontal line throughout the experiment.

(b)(i) The statement “the focal length of a converging lens is 15 cm” means that a beam of light travelling parallel to the principal axis is refracted by the lens and brought to a focus at the principal focus, which lies a distance of 15 cm from the optical centre of the lens.

(b)(ii) Distinction between a real image and a virtual image.

Real image	Virtual image
Formed by the actual intersection of refracted (or reflected) light rays.	Formed where the light rays only appear to intersect when produced backwards.
Can be caught and displayed on a screen.	Cannot be caught on a screen.
Usually inverted relative to the object.	Erect (upright) relative to the object.

Metre-bridge determination of R₁ and R₂

Principle. The bridge wire NQ is 100 cm long and uniform. At the balance (null) point P no current flows through the galvanometer, so, with the standard resistor \(R_x = 1\,\Omega\) in the left gap and the wire in the right gap,

\[ \frac{R_x}{R} = \frac{l_x}{l_y} \quad\Rightarrow\quad R = R_x\,\frac{l_y}{l_x}, \qquad l_x = NP,\; l_y = PQ. \]

For wire A this gives \(R_1 = \dfrac{l_y}{l_x}\,R_x\); repeating with wire B at the same lengths gives the corresponding values \(R_2 = \dfrac{l_y'}{l_x'}\,R_x\).

Table of readings (\(R_x = 1\,\Omega\)):

L /cm	lx = NP /cm	ly = PQ /cm	R1 = (ly/lx)Rx /Ω	l′x = NP /cm	l′y = PQ /cm	R2 = (l′y/l′x)Rx /Ω
100	5.2	94.8	18.23	21.5	78.5	3.65
95	6.0	94.0	15.67	22.5	77.5	3.44
85	6.5	93.5	14.38	23.0	77.0	3.35
75	7.7	92.3	11.99	23.6	76.4	3.24
65	8.2	91.8	11.20	24.4	75.6	3.10

Sample evaluation (L = 100 cm row).

\[ R_1 = \frac{l_y}{l_x}R_x = \frac{94.8}{5.2}\times 1 = 18.23\,\Omega, \qquad R_2 = \frac{l_y'}{l_x'}R_x = \frac{78.5}{21.5}\times 1 = 3.65\,\Omega. \]

The remaining rows are evaluated in the same way, as tabulated above.

Graph of R₂ against R₁.

Slope. Reading two widely spaced points on the line of best fit:

\[ (R_1, R_2) = (11.0,\;3.12)\ \text{and}\ (18.5,\;3.66). \] \[ s = \frac{\Delta R_2}{\Delta R_1} = \frac{3.66 - 3.12}{18.5 - 11.0} = \frac{0.54}{7.5} = 0.072. \]

Evaluation of k.

\[ k = \sqrt{s} = \sqrt{0.072} = 0.27. \]

Two precautions.

1. The jockey was tapped gently on the bridge wire (never dragged) so that the wire was not scratched or its cross-section altered.
2. All connecting terminals were kept clean and tight, and readings of the balance point were taken with the eye placed vertically above the wire to avoid parallax error.

(b)(i) Two advantages of a potentiometer over a voltmeter

1. At the balance point the potentiometer draws no current from the cell, so it measures the true e.m.f./potential difference without the loading error that a voltmeter introduces.
2. It gives readings of very high accuracy and its sensitivity (range) can be extended as required, since it depends only on lengths and a standard cell rather than on a moving pointer.

(b)(ii) Internal resistance of a cell

The internal resistance of a cell is the opposition offered by the materials inside the cell (its electrolyte and electrodes) to the flow of current through the cell itself.

Test of Practical Knowledge - Balancing a Metre Rule

Set-up. The uniform metre rule balances horizontally on the knife edge at its centre of gravity, so the point of balance is G = 48.0 cm. The object W is suspended at the 15 cm mark throughout, and the mass M is suspended on the opposite side of G and adjusted to position Y until the rule is again horizontal.

The distance of the load from the pivot is fixed:

\[ D = G - 15 = 48.0 - 15.0 = 33.0\ \text{cm} \]

The distance of the mass from the pivot in each trial is

\[ L = Y - G \]

Table of readings

S/N	M (g)	Y (cm)	L = Y - G (cm)	D (cm)	L-1 (cm-1)
1	20	97.50	49.50	33.00	0.02
2	30	81.00	33.00	33.00	0.03
3	40	72.75	24.75	33.00	0.04
4	50	67.80	19.80	33.00	0.05
5	60	64.50	16.50	33.00	0.06

Theory

Taking moments about G, at balance the anticlockwise moment of W equals the clockwise moment of M:

\[ W \times D = M \times L \]

Making M the subject:

\[ M = (W\,D)\,\frac{1}{L} = (W\,D)\,L^{-1} \]

This has the form \(M = S\,L^{-1}\), a straight line through the origin of gradient \(S = W\,D\).

Graph of M against L^-1

Slope of the graph

Reading two widely separated points on the line of best fit, \((L^{-1}_1, M_1) = (0.02,\ 20)\) and \((L^{-1}_2, M_2) = (0.06,\ 60)\):

\[ S = \frac{M_2 - M_1}{L^{-1}_2 - L^{-1}_1} = \frac{60 - 20}{0.06 - 0.02} = \frac{40}{0.04} = 1000\ \text{g cm} \]

Evaluation of S/D

\[ \frac{S}{D} = \frac{1000}{33.0} = 30.3\ \text{g} \]

Since \(S = W\,D\), the quantity \(S/D\) equals the load W, so the weight of the object is W = 30.3 g-force.

Two precautions

1. The eye was placed vertically above the mark being read so as to avoid error due to parallax.
2. Draught was avoided and the rule was allowed to settle to true horizontal, with the suspended masses hanging freely and not touching the bench, before each reading was taken.

(b)(i) Principle of moments

The principle of moments states that when a body is in equilibrium under the action of parallel forces, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

(b)(ii) Centre of gravity

The centre of gravity of a body is the single point through which the whole weight (the resultant weight) of the body appears to act, whatever the position of the body.