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.
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
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.
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
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.
You are provided with a potentiometer AB, a \(102\Omega\) standard resistor R, a battery of emf 4.5V, a jockey J, and other necessary materials.
Connect a circuit as shown in the diagram above.
Close key K. Without J making contact with AB, read and record the ammeter reading I. Open the key.
Use the jockey to make contact with AB at the 20cm mark such that AJ = \(x = 20\text{cm}\). Close the key, read and record the ammeter reading.
Evaluate \(x^{-1}\).
Repeat the procedure for values of \(x = 35\text{cm}\), \(45\text{cm}\), \(60\text{cm}\), and \(80\text{cm}\) respectively.
Tabulate your readings.
Plot a graph with x\(^{-1}\) on the vertical axis and \(l_i\) on the horizontal axis, starting both axes from the origin (0, 0).
Determine the slope, s, of the graph.
From your graph, determine the value \(l_o\) of \(I_1\) for which \(x^{-1} = 0\).
Evaluate \(\frac{I_o}{I}\).
State two precautions taken to obtain accurate results.
(b)i. Define the emf of a battery.
ii. A cell X of emf 1.018V is balanced by a length of 50.0cm on a potentiometer wire. Another cell Y is balanced by a length of 75.0cm on the same wire. Calculate the emf of Y.
Practical: current and jockey position on a potentiometer wire
With the standard resistor \(R\) and battery in circuit, the ammeter reading \(I\) is taken for each jockey contact position \(x\), and \(x^{-1}\) is evaluated. A specimen table (values illustrative):
x (cm)
x\(^{-1}\) (cm\(^{-1}\))
I\(_1\) (A)
20
0.0500
I\(_a\)
35
0.0286
I\(_b\)
45
0.0222
I\(_c\)
60
0.0167
I\(_d\)
80
0.0125
I\(_e\)
Plot \(x^{-1}\) (vertical) against \(I_{1}\) (horizontal) from the origin; the graph is a straight line whose slope \(s = \dfrac{\Delta(x^{-1})}{\Delta I_{1}}\). Extrapolate to find \(l_{o}\), the value of \(I_{1}\) at \(x^{-1} = 0\), then evaluate \(\dfrac{I_{o}}{I}\).
Two precautions: ensure firm, clean jockey contacts and tap (do not drag) the jockey on the wire; check that all connections are tight and the key is opened between readings to avoid heating the wire and running down the cell.
(b)(i) EMF of a battery
The emf of a battery is the total electrical energy it supplies per unit charge driven round a complete circuit (the work done per coulomb by the battery), equal to the terminal p.d. when the battery delivers no current.
(b)(ii) EMF of cell Y
On a potentiometer the balance length is proportional to the emf, so \(\dfrac{E_{Y}}{E_{X}} = \dfrac{l_{Y}}{l_{X}}\).
Practical: current and jockey position on a potentiometer wire
With the standard resistor \(R\) and battery in circuit, the ammeter reading \(I\) is taken for each jockey contact position \(x\), and \(x^{-1}\) is evaluated. A specimen table (values illustrative):
x (cm)
x\(^{-1}\) (cm\(^{-1}\))
I\(_1\) (A)
20
0.0500
I\(_a\)
35
0.0286
I\(_b\)
45
0.0222
I\(_c\)
60
0.0167
I\(_d\)
80
0.0125
I\(_e\)
Plot \(x^{-1}\) (vertical) against \(I_{1}\) (horizontal) from the origin; the graph is a straight line whose slope \(s = \dfrac{\Delta(x^{-1})}{\Delta I_{1}}\). Extrapolate to find \(l_{o}\), the value of \(I_{1}\) at \(x^{-1} = 0\), then evaluate \(\dfrac{I_{o}}{I}\).
Two precautions: ensure firm, clean jockey contacts and tap (do not drag) the jockey on the wire; check that all connections are tight and the key is opened between readings to avoid heating the wire and running down the cell.
(b)(i) EMF of a battery
The emf of a battery is the total electrical energy it supplies per unit charge driven round a complete circuit (the work done per coulomb by the battery), equal to the terminal p.d. when the battery delivers no current.
(b)(ii) EMF of cell Y
On a potentiometer the balance length is proportional to the emf, so \(\dfrac{E_{Y}}{E_{X}} = \dfrac{l_{Y}}{l_{X}}\).
You are provided with a grooved inclined plane, a solid sphere, a stopwatch, and other necessary apparatus.
Place the pile of paper towels at the tail end of the inclined plane to stop the sphere from rolling off the table.
Release the sphere from a point at distance D= 140cm from the tail end of the inclined plane.
Determine the average time t taken by the sphere to cover this distance.
Evaluate W = D/t.
Calculate V= 2W.
Repeat the procedure for four other values of D= 120cm 100cm, 80 cm and 60cm respectively.
Tabulate your readings.
Plot a graph with V on the vertical axis and t on the horizontal axis
Determine the slope, s, of the graph.
What is the significance of s?
State two precautions taken to obtain accurate results.
(b)i. Write the equation for the velocity ratio of an inclined plane, giving the meaning of the symbols used.
ii. An object of mass 5kg is placed on a place inclined at an angle of 30° to the horizontal. Calculate the force on the object perpendicular to the plane when the object is at rest. (g =10ms\(^{-2}\)).
Test of Practical Knowledge: solid sphere rolling down a grooved inclined plane
For each distance \(D\) the sphere is released from rest at the marked point on the groove, and the time it takes to roll down to the paper-towel stop is measured. Two readings \(t_1\) and \(t_2\) are taken and averaged to give the mean time \(t\). Then \(W=\dfrac{D}{t}\) is evaluated and \(V=2W\) is calculated.
Table of readings
\(D\) (cm)
\(t_1\) (s)
\(t_2\) (s)
mean \(t\) (s)
\(W=\dfrac{D}{t}\) (cm s\(^{-1}\))
\(V=2W\) (cm s\(^{-1}\))
140.0
4.50
5.00
4.750
29.474
58.948
120.0
4.20
4.40
4.300
27.907
55.814
100.0
4.00
3.80
3.900
25.641
51.282
80.0
3.80
3.50
3.650
21.918
43.836
60.0
3.20
3.40
3.300
18.182
36.364
Graph of \(V\) against \(t\)
The five points are plotted with \(V\) (cm s\(^{-1}\)) on the vertical axis and mean \(t\) (s) on the horizontal axis, and the best straight line is drawn through them.
Velocity V plotted against mean time t; the best straight line through the points gives slope s = 11.51 cm s^-2, the acceleration of the sphere down the incline.
Slope of the graph
Taking two well-separated points on the best-fit line, \((t=2.3\,\text{s},\,V=32\,\text{cm s}^{-1})\) and \((t=5.6\,\text{s},\,V=70\,\text{cm s}^{-1})\):
Significance of \(s\): the slope \(s\) represents the acceleration of the sphere as it rolls down the inclined plane (i.e. the acceleration due to gravity along the incline).
Two precautions:
I avoided parallax error when taking readings on the metre rule by viewing the scale directly from above (perpendicular to the rule).
I took repeated timings for each distance and averaged them, releasing the sphere gently from rest each time, to reduce random errors.
(b)(i) Velocity ratio of an inclined plane
\[ \text{V.R.}=\frac{\text{length of the inclined plane}}{\text{vertical height}}=\frac{1}{\sin\theta} \]
where \(\theta\) is the angle of inclination of the plane to the horizontal.
(b)(ii) Force on the object perpendicular to the plane
The question asks for the force perpendicular (normal) to the inclined surface. This is the component of the weight at right angles to the plane:
Test of Practical Knowledge: solid sphere rolling down a grooved inclined plane
For each distance \(D\) the sphere is released from rest at the marked point on the groove, and the time it takes to roll down to the paper-towel stop is measured. Two readings \(t_1\) and \(t_2\) are taken and averaged to give the mean time \(t\). Then \(W=\dfrac{D}{t}\) is evaluated and \(V=2W\) is calculated.
Table of readings
\(D\) (cm)
\(t_1\) (s)
\(t_2\) (s)
mean \(t\) (s)
\(W=\dfrac{D}{t}\) (cm s\(^{-1}\))
\(V=2W\) (cm s\(^{-1}\))
140.0
4.50
5.00
4.750
29.474
58.948
120.0
4.20
4.40
4.300
27.907
55.814
100.0
4.00
3.80
3.900
25.641
51.282
80.0
3.80
3.50
3.650
21.918
43.836
60.0
3.20
3.40
3.300
18.182
36.364
Graph of \(V\) against \(t\)
The five points are plotted with \(V\) (cm s\(^{-1}\)) on the vertical axis and mean \(t\) (s) on the horizontal axis, and the best straight line is drawn through them.
Velocity V plotted against mean time t; the best straight line through the points gives slope s = 11.51 cm s^-2, the acceleration of the sphere down the incline.
Slope of the graph
Taking two well-separated points on the best-fit line, \((t=2.3\,\text{s},\,V=32\,\text{cm s}^{-1})\) and \((t=5.6\,\text{s},\,V=70\,\text{cm s}^{-1})\):
Significance of \(s\): the slope \(s\) represents the acceleration of the sphere as it rolls down the inclined plane (i.e. the acceleration due to gravity along the incline).
Two precautions:
I avoided parallax error when taking readings on the metre rule by viewing the scale directly from above (perpendicular to the rule).
I took repeated timings for each distance and averaged them, releasing the sphere gently from rest each time, to reduce random errors.
(b)(i) Velocity ratio of an inclined plane
\[ \text{V.R.}=\frac{\text{length of the inclined plane}}{\text{vertical height}}=\frac{1}{\sin\theta} \]
where \(\theta\) is the angle of inclination of the plane to the horizontal.
(b)(ii) Force on the object perpendicular to the plane
The question asks for the force perpendicular (normal) to the inclined surface. This is the component of the weight at right angles to the plane:
