Use the diagram above as a guide to carry out the following experiment.
Trace the outline ABCD of the rectangular glass prism on the drawing paper provided.
Remove the prism. Select a point N on AB such that AN is about one-quarter of AB.
Draw the normal LNM. Also, draw a line RN to make an angle \(\theta = 75^\circ\) with AB at N.
Fix two pins at P\(_1\) and P\(_2\) on line RN. Replace the prism on its outline.
Fix two other pins at P\(_3\) and P\(_4\) such that they appear to be in a straight line with the images of the pins at P\(_1\) and P\(_2\) when viewed through the prism from DC.
(vi) Remove the prism and the pins at P\(_3\) and P\(_4\). Draw a line to join P\(_3\) and P\(_4\)
Produce line P\(_4\) P\(_3\) to meet the line DC at O. Draw a line to join NO.
Measure and record the values of MO and NO.
Evaluate \(\theta = \frac{MO}{NO}\) and \(\cos \theta\).
Repeat the procedure for four other values of \(\theta = 65^\circ\) \(55^\circ\), \(45^\circ\), and \(35^\circ\). In each case, evaluate \(\theta\) and \(\cos \theta\).
Tabulate your readings.
Plot a graph with \(\cos \theta\) on the vertical axis and \(\theta\) on the horizontal axis.
Determine the slope, s, of the graph.
State two precautions taken to ensure accurate results
(b)i. State Snell's law of refraction.
ii. Calculate the critical angle for the glass prism used in the experiment above if its refractive index is 1.5.
(a) Refraction of light through a rectangular glass block
The ray RN strikes face AB at N, refracts into the glass and travels to O on face DC. LNM is the normal at N and M is the foot of the normal on DC, so NM is perpendicular to DC and triangle NMO is right-angled at M.
Ray tracing through rectangular glass block ABCD: incident ray RN (glancing angle theta at N), normal LNM meeting DC at M, refracted ray NO reaching O on DC, and emergent ray OP3P4. NM is the block thickness, MO the base and NO the hypotenuse of right-angled triangle NMO.
In right-angled triangle NMO the refracted ray NO makes the angle of refraction \( r \) with the normal NM, so \( \sin r = \dfrac{MO}{NO} \). The glancing angle at AB is \( \theta \), hence the angle of incidence is \( i = 90^\circ - \theta \) and \( \sin i = \cos\theta \). Applying Snell's law \( \sin i = n\sin r \):
\[ \cos\theta = n\,\frac{MO}{NO} \]
A graph of \( \cos\theta \) against \( \varphi = \dfrac{MO}{NO} \) is therefore a straight line through the origin whose slope is the refractive index \( n \).
Observation / table of readings
S/N
θ (°)
MO (cm)
NO (cm)
φ = MO/NO
cos θ
1
75.0
1.1
6.1
0.18
0.26
2
65.0
1.8
6.3
0.29
0.42
3
55.0
2.5
6.5
0.38
0.57
4
45.0
3.2
6.8
0.47
0.71
5
35.0
3.9
7.2
0.54
0.82
Worked check for row 1: \( \varphi = \dfrac{MO}{NO} = \dfrac{1.1}{6.1} = 0.18 \) and \( \cos 75^\circ = 0.26 \). The remaining rows are evaluated in the same way.
Graph of cos θ against φ = MO/NO
Straight line through the origin. Slope s = (0.80 - 0.30)/(0.52 - 0.20) = 1.56, which equals the refractive index n of the glass.
Slope of the graph
Taking two well-separated points on the line of best fit, \( (0.20,\ 0.30) \) and \( (0.52,\ 0.80) \):
Since \( \cos\theta = n\varphi \), the slope of the line equals the refractive index of the glass, so \( n = 1.56 \).
Two precautions
The pins were fixed vertically (upright) and their images viewed at eye level to avoid parallax error.
The pins were kept reasonably far apart (about 4 cm) and a sharp pencil was used so that the traced lines were thin and neat.
(b)(i) Snell's law of refraction
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media; and the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
\[ \frac{\sin i}{\sin r} = n \]
where \( i \) is the angle of incidence and \( r \) is the angle of refraction.
(b)(ii) Critical angle
For the glass of refractive index \( n = 1.5 \), at the critical angle \( C \) the ray is refracted along the surface \( (r = 90^\circ) \), so:
\[ \sin C = \frac{1}{n} = \frac{1}{1.5} = 0.6667 \]\[ C = \sin^{-1}(0.6667) = 41.8^\circ \]
(a) Refraction of light through a rectangular glass block
The ray RN strikes face AB at N, refracts into the glass and travels to O on face DC. LNM is the normal at N and M is the foot of the normal on DC, so NM is perpendicular to DC and triangle NMO is right-angled at M.
Ray tracing through rectangular glass block ABCD: incident ray RN (glancing angle theta at N), normal LNM meeting DC at M, refracted ray NO reaching O on DC, and emergent ray OP3P4. NM is the block thickness, MO the base and NO the hypotenuse of right-angled triangle NMO.
In right-angled triangle NMO the refracted ray NO makes the angle of refraction \( r \) with the normal NM, so \( \sin r = \dfrac{MO}{NO} \). The glancing angle at AB is \( \theta \), hence the angle of incidence is \( i = 90^\circ - \theta \) and \( \sin i = \cos\theta \). Applying Snell's law \( \sin i = n\sin r \):
\[ \cos\theta = n\,\frac{MO}{NO} \]
A graph of \( \cos\theta \) against \( \varphi = \dfrac{MO}{NO} \) is therefore a straight line through the origin whose slope is the refractive index \( n \).
Observation / table of readings
S/N
θ (°)
MO (cm)
NO (cm)
φ = MO/NO
cos θ
1
75.0
1.1
6.1
0.18
0.26
2
65.0
1.8
6.3
0.29
0.42
3
55.0
2.5
6.5
0.38
0.57
4
45.0
3.2
6.8
0.47
0.71
5
35.0
3.9
7.2
0.54
0.82
Worked check for row 1: \( \varphi = \dfrac{MO}{NO} = \dfrac{1.1}{6.1} = 0.18 \) and \( \cos 75^\circ = 0.26 \). The remaining rows are evaluated in the same way.
Graph of cos θ against φ = MO/NO
Straight line through the origin. Slope s = (0.80 - 0.30)/(0.52 - 0.20) = 1.56, which equals the refractive index n of the glass.
Slope of the graph
Taking two well-separated points on the line of best fit, \( (0.20,\ 0.30) \) and \( (0.52,\ 0.80) \):
Since \( \cos\theta = n\varphi \), the slope of the line equals the refractive index of the glass, so \( n = 1.56 \).
Two precautions
The pins were fixed vertically (upright) and their images viewed at eye level to avoid parallax error.
The pins were kept reasonably far apart (about 4 cm) and a sharp pencil was used so that the traced lines were thin and neat.
(b)(i) Snell's law of refraction
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media; and the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
\[ \frac{\sin i}{\sin r} = n \]
where \( i \) is the angle of incidence and \( r \) is the angle of refraction.
(b)(ii) Critical angle
For the glass of refractive index \( n = 1.5 \), at the critical angle \( C \) the ray is refracted along the surface \( (r = 90^\circ) \), so:
\[ \sin C = \frac{1}{n} = \frac{1}{1.5} = 0.6667 \]\[ C = \sin^{-1}(0.6667) = 41.8^\circ \]
You are provided with cells, a potentiometer, an ammeter, a voltmeter, a bulb, a key, a jockey, and other necessary materials.
Measure and record the emf E of the battery.
Set up a circuit as shown in the diagram above.
Close the key K and use the jockey to make a firm contact at J on the potentiometer wire such that PJ = x = 10cm.
Take the record of the voltmeter reading V and the corresponding ammeter reading I.
Evaluate log V and log I.
Repeat the procedure for other values of x =20, 30. 40, 50, and 60cm.
Tabulate your readings.
Plot a graph with the log I on the vertical axis and log V on the horizontal axis.
Determine the slope, s, of the graph.
Determine the intercept, c, on the vertical axis.
State two precautions taken to ensure accurate results.
(b)i. How is the brightness of the bulb affected as x increases? Give a reason for your answer.
ii. List two electrical devices whose actions do not obey ohm's law
(a) Potentiometer measurement of V and I
The emf of the battery is first measured directly with the voltmeter across the battery terminals on open circuit, giving \(E = 3.0\ \text{V}\).
The circuit is connected as shown below. The key K is closed and the jockey is pressed firmly at J so that \(PJ = x\). For each length x the voltmeter reading V (across the bulb) and the ammeter reading I are recorded, and \(\log V\) and \(\log I\) are evaluated.
Circuit: driver battery E and key K send current through the potentiometer wire PQ; the jockey taps at J so that PJ = x. The ammeter A reads the bulb current I and the voltmeter V reads the pd across the bulb.
Observation / table of readings
x (cm)
V (V)
I (A)
log V
log I
10
0.30
0.06
-0.523
-1.222
20
0.40
0.09
-0.398
-1.046
30
0.50
0.13
-0.301
-0.886
40
0.60
0.18
-0.222
-0.745
50
0.70
0.24
-0.155
-0.620
60
0.79
0.30
-0.102
-0.523
Graph of log I against log V
A straight line of best fit; slope s = 1.66, intercept on the vertical axis c = -0.35.
A straight line of best fit is drawn through the plotted points.
Slope, s
Taking two widely separated points on the line of best fit, \((\log V_1, \log I_1) = (-0.523,\ -1.222)\) and \((\log V_2, \log I_2) = (-0.102,\ -0.523)\):
Using the line equation \(\log I = s\,\log V + c\) with the point \((-0.102,\ -0.523)\):
\[ c = \log I - s\,\log V = -0.523 - (1.66)(-0.102) = -0.523 + 0.169 \]\[ c = -0.35 \]
Two precautions
The jockey was pressed on the potentiometer wire only momentarily (and the key opened between readings) to prevent heating of the wire and the cells.
The eye was placed directly in line with the pointer when reading the ammeter and voltmeter to avoid error of parallax.
(b)(i) Effect on the brightness of the bulb as x increases
The brightness of the bulb increases as x increases. This is because a longer length PJ of the potentiometer wire delivers a larger potential difference to the bulb, so both the voltage across the bulb and the current through it increase, raising the electrical power \(P = VI\) dissipated in the filament.
(b)(ii) Two electrical devices that do not obey Ohm's law
The emf of the battery is first measured directly with the voltmeter across the battery terminals on open circuit, giving \(E = 3.0\ \text{V}\).
The circuit is connected as shown below. The key K is closed and the jockey is pressed firmly at J so that \(PJ = x\). For each length x the voltmeter reading V (across the bulb) and the ammeter reading I are recorded, and \(\log V\) and \(\log I\) are evaluated.
Circuit: driver battery E and key K send current through the potentiometer wire PQ; the jockey taps at J so that PJ = x. The ammeter A reads the bulb current I and the voltmeter V reads the pd across the bulb.
Observation / table of readings
x (cm)
V (V)
I (A)
log V
log I
10
0.30
0.06
-0.523
-1.222
20
0.40
0.09
-0.398
-1.046
30
0.50
0.13
-0.301
-0.886
40
0.60
0.18
-0.222
-0.745
50
0.70
0.24
-0.155
-0.620
60
0.79
0.30
-0.102
-0.523
Graph of log I against log V
A straight line of best fit; slope s = 1.66, intercept on the vertical axis c = -0.35.
A straight line of best fit is drawn through the plotted points.
Slope, s
Taking two widely separated points on the line of best fit, \((\log V_1, \log I_1) = (-0.523,\ -1.222)\) and \((\log V_2, \log I_2) = (-0.102,\ -0.523)\):
Using the line equation \(\log I = s\,\log V + c\) with the point \((-0.102,\ -0.523)\):
\[ c = \log I - s\,\log V = -0.523 - (1.66)(-0.102) = -0.523 + 0.169 \]\[ c = -0.35 \]
Two precautions
The jockey was pressed on the potentiometer wire only momentarily (and the key opened between readings) to prevent heating of the wire and the cells.
The eye was placed directly in line with the pointer when reading the ammeter and voltmeter to avoid error of parallax.
(b)(i) Effect on the brightness of the bulb as x increases
The brightness of the bulb increases as x increases. This is because a longer length PJ of the potentiometer wire delivers a larger potential difference to the bulb, so both the voltage across the bulb and the current through it increase, raising the electrical power \(P = VI\) dissipated in the filament.
(b)(ii) Two electrical devices that do not obey Ohm's law
You are provided with a wooden block to which a hook is fixed, a set of masses, spring balance, and other necessary materials. Using the diagram above as a guide, carry out the following instructions.
Record the mass m\(_{0}\), indicated on the wooden block.
Place the block on the table.
Attach the spring balance to the hook.
Pull the spring balance horizontally with a gradual increase in force until the block just starts to move Record the spring balance reading F.
Repeat the procedure by placing in turn mass m=200, 400, 600, and 800g on top of the block. In each case, read and record the corresponding value of F.
Evaluate M = m\(_{0}\) + m and R = \(\frac{m}{100}\) in each case
Tabulate your readings.
Plot a graph with F on the vertical axis and R on the horizontal axis
Determine the slope, s, of the graph.
State two precautions taken to ensure accurate results.
(b)i. Define coefficient of static friction.
ii. A block of wood of mass 0.5 kg is pulled horizontally on a table by a force of 2.5 N. Calculate the coefficient of static friction between the two surfaces.(g = 10ms\(^{-2}\))
(a) Static friction experiment
The block just begins to move when the horizontal pull \(F\) equals the limiting (static) friction. The limiting friction is proportional to the normal reaction, and the normal reaction equals the total weight \(M = m_0 + m\). The mass marked on the wooden block is \(m_0 = 400\,\text{g}\).
Observation / table of values (\(m_0 = 400.0\,\text{g}\)):
S/N
\(m_0\) (g)
\(m\) (g)
\(M = m_0+m\) (g)
\(R = \dfrac{M}{100}\)
\(F\) (N)
1
400.0
0.0
400.0
4.00
1.70
2
400.0
200.0
600.0
6.00
2.40
3
400.0
400.0
800.0
8.00
3.60
4
400.0
600.0
1000.0
10.00
5.20
5
400.0
800.0
1200.0
12.00
6.30
Graph of \(F\) against \(R\)
Pulling force F (N) plotted against R with a line of best fit; slope s = 0.68.
Slope of the graph
Taking two well-separated points on the line of best fit, \((R_1, F_1) = (6.0,\ 2.3)\) and \((R_2, F_2) = (12.0,\ 6.4)\):
I avoided parallax error by reading the spring balance with my eye directly in line with the pointer.
I ensured the spring balance was pulled strictly horizontally so that the full weight acted as the normal reaction, and I noted the reading at the exact instant the block just started to move.
(b)(i) Definition
The coefficient of static friction is the ratio of the limiting (maximum) frictional force \(F\), acting just as the body is about to move, to the normal reaction \(R\) between the two surfaces in contact:
\[ \mu_s = \frac{F}{R} \]
(b)(ii) Calculation
Normal reaction:
\[ R = mg = 0.5 \times 10 = 5\,\text{N} \]
At the point of moving, the limiting friction equals the applied force, \(F = 2.5\,\text{N}\).
\[ \mu_s = \frac{F}{R} = \frac{2.5}{5} = 0.5 \]
The coefficient of static friction between the two surfaces is 0.5.
The block just begins to move when the horizontal pull \(F\) equals the limiting (static) friction. The limiting friction is proportional to the normal reaction, and the normal reaction equals the total weight \(M = m_0 + m\). The mass marked on the wooden block is \(m_0 = 400\,\text{g}\).
Observation / table of values (\(m_0 = 400.0\,\text{g}\)):
S/N
\(m_0\) (g)
\(m\) (g)
\(M = m_0+m\) (g)
\(R = \dfrac{M}{100}\)
\(F\) (N)
1
400.0
0.0
400.0
4.00
1.70
2
400.0
200.0
600.0
6.00
2.40
3
400.0
400.0
800.0
8.00
3.60
4
400.0
600.0
1000.0
10.00
5.20
5
400.0
800.0
1200.0
12.00
6.30
Graph of \(F\) against \(R\)
Pulling force F (N) plotted against R with a line of best fit; slope s = 0.68.
Slope of the graph
Taking two well-separated points on the line of best fit, \((R_1, F_1) = (6.0,\ 2.3)\) and \((R_2, F_2) = (12.0,\ 6.4)\):
I avoided parallax error by reading the spring balance with my eye directly in line with the pointer.
I ensured the spring balance was pulled strictly horizontally so that the full weight acted as the normal reaction, and I noted the reading at the exact instant the block just started to move.
(b)(i) Definition
The coefficient of static friction is the ratio of the limiting (maximum) frictional force \(F\), acting just as the body is about to move, to the normal reaction \(R\) between the two surfaces in contact:
\[ \mu_s = \frac{F}{R} \]
(b)(ii) Calculation
Normal reaction:
\[ R = mg = 0.5 \times 10 = 5\,\text{N} \]
At the point of moving, the limiting friction equals the applied force, \(F = 2.5\,\text{N}\).
\[ \mu_s = \frac{F}{R} = \frac{2.5}{5} = 0.5 \]
The coefficient of static friction between the two surfaces is 0.5.
