COLLEGE OF ENGINEERING

INSTRUCTOR’S MANUALPHYF121

Compiled ByDr Zolman Hari

Mr Liaw Hock Sang

Table of Contents

Page

INTRODUCTION 3

EXPERIMENTS

1 Electric Fields and Potentials 4

2 Ohm's Law 6

3 RC Circuit 10

4 Wheatstone Bridge 12

5 Electromagnetism 14

6 Interactive Physics: Electrodynamics 17

7 LR Circuit 21

8 LRC Circuit 23

9 Thermal Expansion 26

10 Geometric Optics: Reflection and Refraction 28

2

INTRODUCTION

a) Instruction To Demonstrators1. Try the related experiment before the lab session begins.2. The lab session (experiment and report) must be finished within 3 hours.3. Make sure you can come 10 minutes before each lab session starts.4. Report your attendance and any problem of every experiment that you conducted to the lab

assistant.5. Report to the coordinator the name of the student, who does not attend the lab more than 3

times without MC or valid reason.6. Take the student attendance for each lab session.7. Return the report to students on each lab session so that they know their mistakes. 8. Collect the graded report at the end of every lab session, but return all the reports to students

so that they can get prepared for the common written test at the end of semester.

b) Instruction To Students1. Attendance: Compulsory

a. The instructor / lab assistant will take the attendance for each lab session.b. Attendance less than 80% - consider failed (Grade E) c. 1 group (2-3 students) submitted 1 report only.

2. The students must study the manual before coming to the lab.3. The students must finish the lab session (experiment and report) within 3 hours.4. The students must follow the ‘report writing format’.5. The instructor must explain Report writing format to the students.6. Submit the lab report during lab session.7. Explain marking scheme and rotating schedule.8. Student’s mark will be deducted if they enter the lab late, for instance 10 marks for each 10

minutes.

c) Marking Scheme (Lab report)

1) Cover Sheet 5%2) Purpose 5%3) Equipment 5%4) Procedure 5%5) Data 10%6) Analysis 15%7) Discussion 10%8) Conclusion 5%

Total: 60%

d) Laboratory Assessment Methods

Lab report (10 experiments) 70%

Common written test 30%

Total 100%

3

Experiment 1: Electric fields and potentials

Cover sheet:

Purpose:To investigate the electric field lines and the equipotential surfaces around charged conductors, and the relationship between them.

Equipment:A sheet of carbon-impregnated paper with a configuration of conductors, power supply, multimeter, corkboard, patch cord wire, alligator clips.

Procedure:

Data: (drawing)

Analysis and Discussion:Answers to Questions1. Look at the drawing you made. What can you say about the directions of the field lines

relative to the equipotentials? The direction of the field lines is perpendicular to the equipotential lines.

Do your result approximately with theory? Yes, the result is approximately with theory.

Think about how the magnitude of the electric field is related to the change of potential with distance.The change of potential between any two adjacent equipotential curves drawn in the experimental result are constant, that is . At places where the magnitude of electric field is greater, the separation of equipotential is less distance apart. In other words, the magnitude of electric field is directly proportional to the change of

potential with distance: .

2. Observe on the separation of the equipotentials. Are they same distance apart everywhere? Why?No, This is because the electric field is not uniform.

If is not uniform, . Then separation d is constant.

4

3. Use the technique you have learnt to draw equipotential and electric field for the following figures:a)

b)

c)

Other Discussions

Conclusion:

5

+ +

+

-----

+++++

-----

Experiment 2: Ohm’s Law

Cover sheet:

Purpose :Section A: To verify the expressions for the equivalent resistance for resistors connected in parallel and in series.

Section B and C: To investigate the relationship between current and voltage for Ohmic and non-Ohmic resistive elements.

Equipment:3 resistors (220 , 330 , and 560 )1 multimeter1 breadboardScience Workshop 750 Interface SystemVoltage Sensor (CI-6503)Data Studio Program (installed)RLC Network (CI-6512) – 10 Resistor, 100 Resistor, 0.22 A BulbPatch Cords

Procedure:

Data:Section A

Table 1Resistance Theoretical value

()Experimental

value ()% Difference

R1 220 217.7R2 330 327.4R3 560 555

Three resistors in series

1104

Three resistors in parallel

106

Two largest resistors in parallel and the smallest one in series

425

Section B: Activity P48 – Resistor

Table 2

10 ohmNo. X (Current) Y (Voltage) R = V/I1 0.291 2.9732 0.093 0.9803 -0.198 -2.027

Average Resistance

6

Table 3

Table 5

AnalysisTable 1Resistance Theoretical value

()Experimental value

()% Difference

R1 220 217.7 1.05R2 330 327.4 0.79R3 560 555 0.89Three resistors in series

1110 1104 0.54

Three resistors in parallel

106.8 106 0.75

Two largest resistors in parallel and the smallest one in series

427.6 425 0.61

Table 2

Table 3

100 ohmNo. X (Current) Y (Voltage) R = V/I1 0.030 2.9732 9.884E-03 0.9803 -0.020 -2.027

Average Resistance

Light bulbNo. X (Current) Y (Voltage) R = V/I1 0.109 1.4862 -0.070 -1.1823 -0.108 -1.385

10 ohmNo. X (Current) Y (Voltage) R = V/I1 0.291 2.973 10.22 0.093 0.980 10.53 -0.198 -2.027 10.2

Average Resistance 10.3

100 ohmNo. X (Current) Y (Voltage) R = V/I1 0.030 2.973 99.12 9.884E-03 0.980 99.23 -0.020 -2.027 101.35

Average Resistance 99.9

7

Table 4

Table 5

Average resistance (ten-ohm resistor) = 10.3 volt/amp

Average resistance (100 ) = 99.9 volt/amp

Sample Calculation:

Three resistors in series :R = R1 + R2 + R3 = 220 + 330 + 560 = 1110

Three resistors in parallel :

Twp largest resistors in parallel and the smallest one in series :

% Difference =

Discussion:

Ohmic ResistorTheory Experiment % Difference10 10.3 3.0100 99.9 0.1

Light bulbNo. X (Current) Y (Voltage) R = V/I1 0.109 1.486 13.632 -0.070 -1.182 16.893 -0.108 -1.385 12.82

8

1. What is the relationship between current and voltage in a simple resistor? What is the relationship between current and voltage in the filament of an incandescent light bulb?In a simple resistor, the current is directly proportional to voltage. In the filament of an incandescent light bulb, the relationship between current and voltage is not linear.

2. Compare the ratio of voltage and current from the Scope display to the resistance of the resistor(s) used.The ratio of voltage and current from the scope display is almost equivalent to the resistance of the resistor(s) used.

3. Does each resistor you used have a constant resistance?Yes, the 10 resistor and the 100 resistor have a constant resistance.

4. Does the light bulb filament have a constant resistance (constant ratio of voltage to current? Why or why not?)No, the light bulb filament does not have a constant resistance. This is because the resistance of the filament will change as it heats up and cools down.

5. The slope of the graph for the light bulb is not symmetric. Why is the trace on the Scope different when the filament is heating up compared to the trace when the filament is cooling down?For light bulb, the resistance of the filament will change as it heats up and cools down. As a result, the voltage across the filament also changes. The graph of voltage versus current might show a curve with a changing slope.

Conclusion:1. The expressions for the equivalent resistance for resistors connected in parallel and in

series are verified, as stated in Section A (Data and Analysis).For ohmic resistive elements, the current is proportional to voltage, but for non-ohmic resistive elements, the relationship is not linear.In this experiment, the simple 10 ohm and 100 ohm resistors are an ohmic resistive elements. The bulb is non-ohmic resistive element.

9

Exp 3: RC circuit

Cover sheet:

Purpose: 1. Explain the charging characteristics of a capacitor with ac voltage.2. Appreciate how the oscilloscope can be used to monitor electrical characteristics and to

make electrical measurements.3. Describe how an RC time constant may be measured from an oscilloscope trace.

Equipment:Function generator (square wave), oscilloscope, three resistors (5 k, 10 k, and 20 k, or resistance box), three capacitors (0.05 F, 0.1 F, and 0.2 F), Connecting wires, 2 sheets of Cartesian graph paper, (Optional) unknown resistor wrapped in masking tape to conceal value.

Procedure:

Data:Table 1 (Purpose: To determine the effect of R on the time constant.)

R( )

C( F )

Divisions for 0.63 rise

Sweep time/div

Exp. time constant

Computed RC

Percent error

Case 1R1C1

560 0.1 5.4 (from Oscilloscope)

10s (from Oscilloscope)

5.4 X 10-5 5.6 X 10-5

Case 2R2C1

220 0.1

Case 3R3C1

1000 0.1

Slope of the versus R plot: [C1]Percent difference between slope and C1: _________________

Table 2 (Purpose: To determine the effect of C on the time constant)R

( )C

( F )Divisions

for 0.63 riseSweep

time/divExp. time constant

Computed RC

Percent error

Case 1R1C2

560 0.056

Case 2R1C3

560 0.5

Slope of the versus C plot: [R1]Percent difference between slope and R1: _____________________

10

Analysis and Discussion:Answers to Question:1. Based on experimental results, under what conditions are the charging times of

different RC circuits the same?The answer based on your data.

2. In the form , the = RC in the exponential must have units of time. (Why?) Show that this is the case.Rearrange the equation,

V – Vo = Vo e-t/

V - Vo = e-t/

Vo

- ln V - Vo = t Vo

= t Or students can use dimensional analysis to prove out (time constant) has the same unit as t (time). Therefore, (time constant) has the same unit as t (time).

3. How could the value of an unknown capacitance be determined using the experimental procedures? Show explicitly by assuming a value for an experimentally determined time constant. Refer to table 1; explain how you get the time constant from the procedure, after plotting versus R graph, the slope of the graph is the unknown capacitance. It is because = RC.

Conclusion: 1. Conclude the percent difference between unknown C1 and given C1 2. Conclude the percent difference between unknown R1 and given R1

11

EXP 4: Resistivity Measurement

Cover sheet:

Purpose:To determine the resistivity of a piece of wire using the Wheatstone bridge

Equipment: 1 set Wheatstone bridge with sliding jockey1 switch1 piece 5 resistorsPower supply 50-80 cm Constantan SWG 26 wire1 Micrometer screw gaugeGalvanometer

Procedure:

Data: [Measured values in blue; Calculated values in red]

Resistance for P: 5

a) Average measurement for l1 and l2

l1 measurement (cm)a1 a2 b3 b4 Average l1

30.6 30.5 30.6 30.7 30.6

l2 measurement (cm)b1 b2 a3 a4 Average l2

69.4 69.5 69.4 69.3 69.4

b) Wire length measurement Length of wire X (x 0.05) cm 77.00

c) Diameter measurementWire Diameter (D 0.005) mm

1 2 3 Average Diameter, D0.450 0.450 0.450 0.450

Analysis and Discussion:

12

Resistivity of wire X

Other Discussions

Conclusion:

13

EXP 5: Electromagnetism

Cover Sheet:

Purpose:To get a clearer picture of what actually electromagnetism is all about and to relate what have been taught in their lectures with this experiment.

Equipment:Compass, solenoids, DC power supply, AC power supply (0-6 V), galvanometer, digital multimeter, 2 bar magnets, spring, iron core, blank A4 papers, 2 retort stands, connecting wires.

Procedure:

Data:

Part I: Constructing magnetic field lines produced by a permanent magnet and an electromagnet.Give marks to the sketch (i) using the bar magnet, and (ii) using the 400-turn solenoid

Part III:

Table 1Number of Turns Input V Output V Core Voltage Ratio

Primary Coil

Secondary Coil

400 400 6.0 V 0.41 V NoCore

0.07

400 400 6.0 V 0.52 V Straight-shaped

0.09

400 400 6.0 V 3.64 V U-shaped 0.61400 400 6.0 V 8.86 V -shaped 1.48

Core configuration: -shaped

Table 2Number of Turns Input V Output V Voltage Ratio

Primary Coil Secondary Coil400 400 6.0 V 8.86 1.48200 400 6.0 V 18.01 3.00200 800 6.0 V 36.35 6.06400 800 6.0 V 17.80 2.97400 200 6.0 V 4.40 0.73800 200 6.0 V 20.9 3.48800 400 6.0 V 3.61 0.60

Analysis and Discussion

14

Answers to Questions in Part IWhat can you deduce from these constructions?From the field lines constructed, we can deduce that a current carrying solenoid acts as a permanent magnet.

Answers to Questions in Part II1. What can you deduce about the strength of the magnetic field with and without the core?

The field strength produced by a solenoid with core is higher compared to the one without core.

2. Which solenoid produces higher field strength?The solenoid with 800 turns produces higher field strength.

3. What will happen to the iron core when the power supply is turned on?The iron core will be attracted into the solenoid.

4. What happens to the indicator of the galvanometer? Explain your observation using Faraday’s and Lenz’s law.The indicator of the galvanometer deflects in one direction when the bar magnet is moved upward, and deflects in the opposite direction when the bar magnet is moved downward. These observations indicate that a current is induced in the solenoid. Changing the direction of the magnet’s motion changes the direction of the current induced by that motion.

5. What is the effect of the induced emf in the coil on the oscillation? Hint: compare its oscillation without the solenoid.The spring oscillates faster compare to the one without the solenoid.

6. Explain what you observe.After a moment, the stationary magnet starts to oscillate at the same rate as the magnet in the first solenoid. This is due to the induced current from the first solenoid, which has been transferred to the second solenoid with the stationary magnet.

Answers to Questions in Part III

1. Which core configuration gives the maximum transfer of electromagnetic effect to the secondary coil? Develop a theory to explain the differences between configurations.The -shaped core configuration gives the maximum transfer of electromagnetic effect to the secondary coil.

2. From your data in Table 2, for a primary having a constant number of turns, graph the resulting output voltage versus the number of turns in the secondary. What type of mathematical relationship exists between numbers of turns of wire and the resulting output voltage? Is the data ideal? Why or why not?

15

The output voltage increased when the number of turns of wire increased. The data is ideal. The relationship can be represented by the expression below:

3. Consider further improvements to your transformer. What additional changes might you make to increase the transfer from one coil to the other?Use the appropriate shape of the core, for example when using the " " and " I " cores together will accommodate a larger coil former and can produce a larger transformer. The air gap in the middle of the coil can reduce electromagnetic interference.

Conclusion:

16

EXP 6: Interactive Physics: Electrodynamics

Cover Sheet:

Purpose:To investigate the behavior of a charged particle in electric and magnetic fields

Equipment:Computer, CD of Explorer 3.04 Electrodynamics, LOGAL Physics Software, Inc

Procedure:The procedure of the experiment is no needed since the software has been installed, as student version, in the computer prior to the laboratory session. The students only need to insert the CD and click on appropriate shortcut to run the program.

Data, Analysis and Discussion:

Electric Field 1:IntroductionQ: What happens to the particle?

The particle moves (accelerates) downwards.

Change DirectionQ: What do you think will happen if you change the field direction to UP?

When the field is directed upwards, the particle moves upwards.

Change ChargeQ: In which direction do you think the charged particle will move?

The charged particle moves upwards. (A negative charge moves in the opposite direction to a positive charge in the same field.)

Q: What will happen if you now change the field direction to UP?The charged particle moves downward.

Velocity and EnergyQ: What does the trace tell you about the velocity of a charged particle in a uniform electric field?

The velocity of the charged particle increases with time (particle accelerating).

Q: How does the particle's kinetic energy change during the motion?The KE increases as the particle moves in the field.

MotionQ: How does an electric field affect the motion of a charged particle?

An electric field accelerates a charged particle.

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Electric Field 2:

-1 N/CQ: On the diagram, sketch your prediction of the path of the particle before running the simulation.

Kinetic EnergyQ: How do you expect the kinetic energy of a moving charged particle to change: (in the presence of an electric field?)

KE eventually increases in the presence of an electric field. If qv (conventional current) is opposite in direction to the electric field, it will lose kinetic energy initially, slowing down and stopping. Then it will begin to move in the direction of the electric field, picking up kinetic energy.

(in the absence of an electric field?)KE remains constant in the absence of an electric field.

Magnetic Field:

1.0 TeslaQ: If the particle is positively charged and is initially at rest, predict how it behaves.

Students may expect the particle to move under the influence of the field. They may be surprised to see that the particle remains stationary.

The PathQ: Describe the path of the particle.

The particle moves in a counterclockwise circle.

Q: What happens to the kinetic energy?The KE remains constant (unlike the situation with an electric field, where it increases). (The program actually shows a small change over time, due to algorithm approximations.)

Electric and Magnetic Fields:

IntroductionQ: The particle is positively charged. If a small electric field acting vertically downwards is applied while the simulation runs, predict what will happen to:

i) the path of the particle? Draw a sketch.Student prediction.

ii) the kinetic energy of the particle?Student prediction.

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Electric Field TrialsQ: Your predictions:i) the path of the particle? Draw a sketch.

ii) the kinetic energy of the particle?The KE oscillates, but over time increases.

Q: Does the position of the particle when switching the electric field ON affect the general shape of the final trace?

The general shape is not affected.

Magnetic Field TrialsQ: If you switch the magnetic field ON after a short time, predict what will happen to:i) the path of the particle. Draw a sketch.

(The arrow shows the point where the magnetic field was switched on.)

ii) the kinetic energy of the particle?When the electric field only is on, the KE increases monotonically. When the magnetic field is switched on as well, the KE oscillates, but over time increases.

Q: Does the position of the particle when switching the magnetic field ON affect the general shape of the final trace?

The general shape is not affected.

Q: From your results, summarize the effect on a charged object when there is a magnetic and an electric field acting at right angles.

When both fields are present, a charged particle spirals at right angles to the direction of the electric field, and the particle gains KE.

19

Additional Questions

The Mass Spectrometera) If FB = FE , show the trace of electron in Figure 1

Figure 1

a) If FB > FE , show the trace of electron in Figure 2

Figure 2

Conclusion:(As refer back to the purpose of the experiment: The conclusion might be written as: The behavior of charged particle in electric field can be understood with expression F = qE = ma, and of that in magnetic fields with expression F = q v B = ma.)

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

Bin

E

e

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

x x x x x x x x

Bin

E

e

20

EXP 7: LR circuit

Cover Sheet:

Purpose: To study the behavior of an inductor as a time-dependent voltage is applied.

Equipment: Science Workshop 750 Interface SystemVoltage Sensor (CI-6503)Data Studio Program (installed)RLC Network (CI-6512) – Inductor Coil, Iron Core, 10 ResistorPatch CordsMultimeterLCR meter

Procedure:

Data: [Measured values in blue; Calculated values in red]

The measured inductance = 19.2 mH

Table 1Item ValueInductor resistance, RL 5.65 Resistor resistance, RR 10.20 Total resistance, R = RL + RR 15.85 Experimental time constant, 1.1610-3 s

Theoretical time constant, 1.2110-3 s

Percentage error 4.13%

Analysis:

Sample Calculation

Time to reach half-maximum = 0.0 739 s 0.0725 s = 0.0014 s

21

Discussion:Answers to Questions

1. What is the relationship between the voltage across the inductor and the voltage across the resistor in an inductor-resistor circuit?As the voltage across the inductor increases, the voltage across the resistor decreases, or vice versa. The sum of the voltages across the inductor and the resistor equals the supplied voltage.

2. What is the relationship between the current through the inductor and the behavior of an inductor in a DC circuit?The current through the inductor does not increase instantly as the voltage changes instantly from zero to maximum, but it increases exponentially according to the equation

.

The current in the circuit takes certain time to reach the maximum (saturation). This is because the inductor creates a back-emf in response to the rise in current.

3. How does the inductive time constant found in this experiment compare to the theoretical value given by = L/R?

4. Does Kirchhoff’s Loop Rule hold at all times? Use the graphs to check it for at least three different times: Does the sum of the voltages across the resistor and the inductor equal the source voltage at any given time?Yes. Yes, the sum of the voltages across the resistor and the inductor always equals the source voltage at any given time.

Conclusion:

22

EXP 8: LRC circuit

Cover Sheet:

Purpose: To study the electromagnetic resonance in an inductor-resistor-capacitor circuit

Equipment: Science Workshop 750 Interface SystemVoltage Sensor (CI-6503)Data Studio Program (installed)RLC Network (CI-6512) – Inductor Coil, Iron Core, 10 Resistor, 100 F CapacitorPatch CordsGraph paper (optional)LCR meter (optional) [5 marks]

Procedure:

Data: [Measured values in blue; Calculated values in red]

Table 1Freq (Hz) VR Current

(VR / R)Freq (Hz) VR Current

(VR / R)10 0.170 V 0.017 A 90 1.599 V 0.157 A20 0.374 V 0.037 A 100 1.531 V 0.150 A30 0.612 V 0.060 A 110 1.429 V 0.140 A40 0.850 V 0.083 A 120 1.327 V 0.130 A50 1.088 V 0.107 A 130 1.224 V 0.120 A60 1.327 V 0.130 A 140 1.122 V 0.110 A70 1.497 V 0.147 A 150 1.054 V 0.104 A80 1.599 V 0.157 A

Table 2Item Measured valueInductance 19.10 mHResistance 10.18 Capacitance 99.9 F

Table 3Resonant frequency (linear) 86.6 HzResonant angular frequency 544.12 rad/sTheoretical resonant angular frequency 723.94 rad/sPercentage difference of angular frequency 24.8%

23

Analysis and Discussion:Calculation:

Resonant angular frequency = res = 2vres = 2(86.6Hz) = 544.12 rad/sTheoretical resonant angular frequency res

723.94 rad/s

Answers to Questions1. The purpose of this activity is to study resonance in an inductor-resistor-capacitor circuit

(LRC circuit) by examining the current through the circuit as a function of the frequency of the applied voltage. What will happen to the amplitude of the current in the LRC circuit when the frequency of the applied voltage is at or near the resonant frequency of the circuit?The amplitude of the current in the LRC circuit increases when the frequency of the applied voltage is near the resonant frequency of the circuit, and reaches maximum at resonant frequency.

2. How does your measured value for resonant angular frequency compare to the theoretical value for resonant angular frequency? Remember,

The measured value of resonant angular frequency is 25% less than the theoretical value of resonant angular frequency.

3. Is the plot of current versus frequency symmetrical about the resonant frequency? Explain.From Table 1, the plot of current versus frequency is as shown in Figure 1.

24

0 20 40 60 80 100 120 140 1600.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0.18C

urre

nt

Frequency

From Figure 1, the plot of current versus frequency is symmetrical about the resonant frequency.

4. At resonance, the reactance of the inductor and the capacitor cancel each other so that the impedance (Z) is equal to just the resistance (R). Calculate the resistance of the circuit by using the amplitude of the current at resonance in the equation R = V/I (where V is the amplitude of the applied voltage). Is this resistance equal to 10 ohms? Why not?

In the LRC circuit, the impedance (Z) of the circuit is given by the expression below:

At resonance, XC = XL, the reactances of the inductor and capacitor cancel each other. Therefore, the impedance Z is equal to the total resistance R of the circuit:

With the amplitudes of the applied voltage V and current I, we get the total resistance R equals 19.1 :

The resistance R is not equal to 10 ohms because 10 ohms is just the resistance of the resistor. However, R is the total resistance of the circuit, which includes the resistances of the resistor and inductor.

Figure 1

25

Conclusion:

26

EXP 9: Thermal Expansion

Cover Sheet:

Purpose:To measure the coefficient of linear expansion for Copper and Steel.

Equipment:Copper and steel tubes (with bracket)Expansion base (consist of Dial gauge spring arm, Slotted bracket, Thumbscrew, Stainless steel pin, Thermistor Lug, and Banana connectors)Foam InsulatorSteam containerLatex tubesSteam generatorMultimeter

Procedure:

Data:

Table 1Data Calculations

Lo (mm) Rrm () L (mm) Rhot () Trm (C) Thot (C) T (C)Copper

Steel

Analysis and Discussions:Note: Accepted values for the linear expansion coefficient:

Cu = 17.6 10-6 C 1 steel = 11.4 10-6 C 1

Using the equation , calculate linear expansion coefficient, , for copper and steel.

After that,Calculate the percentage different of Cu and steel by using the following equation, experimental value – theoretical value X 100% = theoretical value

Answers to Questions1. Look up the accepted values for the linear expansion coefficient for copper and steel.

Compare these values with your experimental values. What is the percentage difference in each case? Is your experimental error consistently high or low?Answer this question based on the percentage different that you calculate out.If the percentage different is small, that is approximately 10%, therefore the experimental error is consistently low.

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2. On the basis of your answers in question 1, speculate on the possible sources of error in your experiment. How might you improve the accuracy of the experiment?The possible sources or error in this experiment are:The uncertainty of the meter tape and the uncertainty of the dial gauge made the length readings limited to certain significant figures.Parallax error occurred during reading the valuesLost of heat from the tube to the surrounding, which cause the change temperature inaccurate.The temperature calculated using resistance-temperature conversion is only7 an approximation because we are using linear regression method, that is the scale is assumed to be divided equallyThe unstable room temperature in the physic lab due to the air-condition could change the initial temperature of the tube rapidlyThe steel consists of mixture of carbon and iron. The percentage composition of steel may be varying from sample to sample. Thus, the theoretical value of steel, which is 11 X 10-6 oC-1 might be an average value only and not a definite (universal) value. Therefore, the percentage difference calculated was not accurate.

In order to improve the accuracy of the experiment, the following steps were suggested:The experiment is to be conducted in a room with a stable (constant) room temperature.Make sure the eye is perpendicular to the dial gauge or meter tape when taking the values.Use a digital meter that can read the expansion of the length of the tube.The whole tube should be insulated to prevent heat lost to surrounding.Use a more accurate measuring device to measure the length of the tube.

3. From your result, can you calculate the coefficients of volume expansion for copper and steel? (ie. )

From the result, the coefficient of volume expansion () for copper and steel can be calculated (approximate) by assuming that the tube expand in all dimensions.

Conclusion:

28

EXP 10: Geometrical Optics: Reflection and Refraction

Purpose:To investigate the law of reflection and the law of refraction, and to observe total internal reflection and color dispersion

Equipment:PASCO Introductory Optics System

Procedure:

Data:Table 1: Reflection

Angle of incidence, 1 ()

Angle of reflection, 1

’ ()10.0 10.020.0 20.030.0 30.040.0 40.050.0 50.060.0 60.070.0 70.080.0 80.0

Table 2: RefractionAngle of incidence,

1 ()Angle of refraction,

2()0.0 0.010.0 6.020.0 13.030.0 20.040.0 25.050.0 30.060.0 35.070.0 39.080.0 41.0

Table 3: Total Internal Reflection and DispersionAngle of incidence,

1 ()Angle of refraction,

2 ()Angle of reflection,

1’ ()

0.0 0.0 0.010.0 15.0 10.020.0 30.0 20.030.0 48.0 30.040.0 73.0 40.042.0 90.0 42.050.0 50.060.0 60.070.0 70.080.0 80.0

29

The largest angle in which all the components of the refracted rays can still be seen is at the angle of incidence of 42o.

The angle of refraction for red component is 80o , and blue component is 84o.

Analysis and Discussion:

Answers to Questions in ProcedureIs there a reflected ray?Yes

Note any dispersion your observed, and at which angle you can first notice it. Does it increase with increasing angle?The first dispersion observed is at the incident angle of 20o. The dispersion increases with increasing angle of incidence.

Observe what happen to the reflected ray. What happens as the refracted ray approaches 90 degrees?The reflected ray is brighter as the angle of incidence increases. When the refracted ray approaches 90o, the angle of incidence is known as critical angle.

What happens if you increase the angle of incidence past the point where the angle of refraction is 90 degrees?The ray is totally reflected and no refracted ray is observed. This phenomenon is called the total internal reflection.

Analysis and Discussion:1. Do the data of part I verify the law of reflection as stated in equation (1)?

Yes. The angle of reflection equals the angle of incidence. So the law of reflection is verified.

2. Using the data of part II, plot a graph of sin 1 versus sin 2. From the graph, obtain the index of refraction of the acrylic material of which the lens is made. Is Snell’s law verified? What criterion do you use to determine this?The plot of sin 1 versus sin 2 is shown in Figure 1 below.

Table 2Angle of incidence,

1 ()Angle of refraction,

2()sin 1 sin 2

0.0 0.0 0.0000 0.00010.0 6.0 0.174 0.10520.0 13.0 0.342 0.22530.0 20.0 0.500 0.34240.0 25.0 0.643 0.42350.0 30.0 0.766 0.50060.0 35.0 0.866 0.57470.0 39.0 0.940 0.62980.0 41.0 0.985 0.656

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0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.2

0.4

0.6

0.8

Sin

The

ta 1

Sin Theta 2

.

Therefore, The index of refraction of acrylic material,

n2 = slope of the graph

= [2 marks]

The percent error as compared to the literature value, 1.4917, is

Thus, the Snell’s law is verified. The criterion used to determine the index of refraction of the material is n2 = slope of the graph.

3. From the data of part III, calculate the index of refraction for red light and for blue light. How do these two values compare with the value you obtained from the data in part II? Does the critical angle you observed experimentally agree with the theoretical value? Given nair = 1.00From Part III,

Figure 1

Red Blue

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From Part II,

The index of refraction = slope of the graph =

Percent difference between the refractive indexes obtained in Part II and Part III is about 2 % for the red light, and 1 % for the blue light.

Yes, the critical angle observed experimentally agrees with the literature value:

Critical angle,

Conclusion:

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