PhysicsDepartment - Brock University · c BrockUniversity,2015-2016. Contents ... units and physical quantities ... Your graphs will be sent there for later inclusion in your online

Brock University Physics DepartmentSt. Catharines, Ontario, Canada L2S 3A1

PHYS 1P92 Laboratory Manual

Physics Department

Copyright c© Brock University, 2015-2016

Contents

1 Capacitance 1

2 Faraday rotation 12

3 Resistance 20

4 Electron charge-to-mass ratio 30

5 Diffraction of light by a grating 40

A Review of math basics 48

B Error propagation rules 51

C Graphing techniques 53

i

ii

(ta initials)

first name (print) last name (print) brock id (ab17cd) (lab date)

Experiment 1

Capacitance

In this Experiment you will learn

• the relationship between the voltage and charge stored on a capacitor;

• how to compensate for the effect of a measuring instrument on the system being tested;

• to visualize data in different ways in order to improve the undwerstanding of a physical system;

• to extend your data analysis capabilities with a computer-based fitting program;

• to apply different methods of error analysis to experimental results.

Prelab preparation

Print a copy of this Experiment to bring to your scheduled lab session. The data, observations andnotes entered on these pages will be needed when you write your lab report and as reference materialduring your final exam. Compile these printouts to create a lab book for the course.

To perform this Experiment and the Webwork Prelab Test successfully you need to be familiar with thecontent of this document and that of the following FLAP modules (www.physics.brocku.ca/PPLATO).Begin by trying the fast-track quiz to gauge your understanding of the topic and if necessary review themodule in depth, then try the exit test. Check off the box when a module is completed.

FLAP PHYS 1-1: Introducing measurement

FLAP PHYS 1-2: Errors and uncertainty

FLAP PHYS 1-3: Graphs and measurements

FLAP MATH 1-1: Arithmetic and algebra

FLAP MATH 1-2: Numbers, units and physical quantities

WEBWORK: the Prelab Capacitance Test must be completed before the lab session

! Important! Bring a printout of your Webwork test results and your lab schedule for review by theTAs before the lab session begins. You will not be allowed to perform this Experiment unless therequired Webwork module has been completed and you are scheduled to perform the lab on that day.

! Important! Be sure to have every page of this printout signed by a TA before you leave at the endof the lab session. All your work needs to be kept for review by the instructor, if so requested.

CONGRATULATIONS! YOU ARE NOW READY TO PROCEED WITH THE EXPERIMENT!

1

2 EXPERIMENT 1. CAPACITANCE

The capacitor

A capacitor is a device that stores electric charge (electrons). It consists of two electrically conductiveparallel metal plates separated by an insulating layer of air or other dielectric material, a material thatcan support an electric field. The total amount of charge q stored is proportional to the electric potentialdifference, or voltage, VC between the plates, so that

q = CVC . (1.1)

The capacitance C of a parallel plate capacitor is proportional to the plate area and the dielectric constantof the medium between the plates, and inversely proportional to the plate separation. Capacitance ismeasured in units of Farads (F), microFarads (µF = 10−6 F) and picoFarads (pF = 10−12 F).

Figure 1.1: Basic capacitor circuit

A series circuit provides only one path for movement of charge. Figure 1.1 shows a series circuitconsisting of a source of electric potential energy V of voltage V , a switch S, a current limiting resistor Rand a capacitor C.

Kirchoff’s Voltage Law (KVL) states that the algebraic sum of the voltages in any closed circuit loopis zero,

∑

V = 0. With voltage sources considered positive and voltage drops considered negative, weestablish that the source voltage V will be equal to the voltage drops VR across R and VC across C, so that

V = VR + VC . (1.2)

The capacitor plate material consists of a conductive lattice of atoms. The positive nucleus is fixed inplace while some of the negative electrons surrounding an atom are free to migrate from the atom whensubjected to an external force.

Let us assume that initially there is no charge stored on the capacitor plates so that the plates areelectrically neutral and the voltage across the capacitor VC = 0.

When the switch S is closed, the positive terminal of the voltage source attracts electrons away fromthe upper plate of the capacitor, leaving the upper plate with a net positive electric charge. The positivecharge now present on the upper plate attracts electrons from the voltage source negative terminal to thelower capacitor plate, giving it a net negative charge.

This flow of charge dq through the circuit during a time interval dt defines the electric current i = dq/dt.The current i is directly proportional to the voltage VR across the resistor R and inversely proportionalto the circuit resistance R. This relationship between current, voltage and resistance is known as Ohm’sLaw:

i = VR/R. (1.3)

3

The charge separation q at the two capacitor plates establishes a voltage, or potential difference,VC = q/C across the capacitor.

As VC increases, the difference VR = V − VC across R decreases, as does the current i = (V − VC)/Rflowing through R. This process continues until the voltage VC across C is equal to the voltage V of thesource, at which time charge no longer flows and the current i = 0.

From Kirchoff’s Voltage Law, we know that the sum of all the voltage sources minus all the voltage dropsin a circuit equals to zero. To examine the capacitor charging process, we traverse the circuit of Figure 1.1clockwise from the negative (-) terminal of the battery, adding each voltage source and subtracting thevoltage drop across each component:

V − VR − VC = 0

V − iR− q

C= 0 (1.4)

A current i that varies as a function of time t is symbolized by i = i(t). Substituting this relationshipin Equation 1.4 and rearranging yields the charging equation for the circuit:

i(t) =dq

dt=

V

R−(

1

RC

)

q (1.5)

The solution of this differential equation in terms of q is given by

dq

dt=

(

V

R

)

e−t/RC (1.6)

Here, e = 2.718 . . . is the base of the natural logarithms (ln), not the elementary charge. We note inEquation 1.6 that at time t = 0 the exponential term is e0 = 1 and i = V/R does not have any dependenceon time. Let this initial constant current be I0. Then the current i(t) flowing in the circuit at any time tis given by

i(t) = I0e−t/RC (1.7)

Capacitors in parallel

The charge stored on a capacitor is directly proportional to the surface area of the capacitor plates.Referring to Figure 1.2 we note that putting two capacitors in parallel results in an equivalent plate surfacearea that is the sum of the individual plate areas. This qualitative result can be expressed mathematically.The voltage across each capacitor is V . Applying Equation 1.1 to the charge stored in each capacitor:

Figure 1.2: Capacitors in parallel

q1 = C1V, q2 = C2V. (1.8)

The total charge q stored in the parallel arrange-ment of capacitors is the sum of the charges storedin each capacitor,

q = q1 + q2 = (C1 + C2)V (1.9)

The equivalent capacitance Cp with the same totalcharge q and voltage V is then

Cp =q

V= C1 + C2 (1.10)


The relationship can be extended to any number of capacitors in parallel by simply adding the contributionsfrom the charge stored in each of the capacitors:

Cp = C1 + C2 + . . . + CN =N∑

i=1

Ci (1.11)

Capacitors in series

Figure 1.3: Capacitors in series

When a voltage V is applied across several ca-pacitors connected in series, a charge separationq = C1V1 = C2V2 = . . . will be induced across

each capacitor. Using KVL, the sum of the volt-

ages across each capacitor is equal to the appliedvoltage: V = VC1

+ VC2+ . . ., then the equivalent

capacitance Cs of two or more capacitors in seriesis given by

1

Cs=

1

C1+

1

C2+ . . . =

N∑

i=1

1

Ci(1.12)

Procedure

Figure 1.4: Schematic diagram of experimental setup, connected to measure a single capacitor

Manufacturer’s values of components used in the experimental circuit:

Rd ±∆(Rd) = (100± 5)Ω, Rc±∆(Rc) = (1.00± 0.05)× 105 Ω, C ±∆(C) = (2.2± 0.2)× 10−6 F.

Figure 1.4 shows the schematic diagram of the electrical circuit used in this experiment. The circuituses one or two removable jumper wires to electrically arrange the capacitors in series, parallel, or to onlyinclude a single capacitor as in Figure 1.4. The capacitors C1 = C2 = C have the same nominal value.

5

A computer USB port provides the voltage source V=5V used to power the circuit. This connection isalready made across terminals A and P. With the USB ground, or zero voltage reference point, connectedto the input terminal A, there are 5V present at the terminal P.

To measure voltage properly, the LabPro voltage probe Rp should be connected across Rc so that it’szero reference point, the black clip, is also connected to the A terminal. The red clip is connected to theB side of Rc as shown in Figure 1.4.

The LabPro unit acts as a resistance of Rp = 107Ω in parallel with resistance Rc. The effective circuitresistance of these two resistors in parallel is given by 1

1

R=

1

Rc+

1

Rp(1.13)

• Calculate R and ∆R using the given values of Rc, ∆Rc, and Rp. Assume ∆Rp = ±0.005 ×Rp.

R = .................... = .................... = ....................

∆R = R2

√

(

∆Rc

R2c

)2

+

(

∆Rp

R2p

)2

= .................... = ....................

R = .................± ................. Ω

When the normally-open switch S is pressed, Rd connects across C and any charge present on the capac-itor plates discharges very quickly through Rd so that VC → 0. Since R and Rd are in series, V = VR+VRd

.With R ≫ Rd, the voltage drop across Rd is nearly zero and can be ignored. Now, VR = VAB = V and byOhm’s Law, a steady current I0 = VAB/R flows through R.

When the switch S is released, the time-dependent current i(t) = VAB/R decreases exponentiallywith time as a voltage VC develops across the test capacitor(s), and hence VAB decays exponentially toapproximately zero. Replacing i(t) in Equation 1.7 we get an expression for the voltage VAB across R:

VAB = I0R e−t/RC . (1.14)

Part 1: Single capacitor

• Assemble the circuit board as shown in Figure 1.4, with a jumper wire connecting the terminal P toterminal P2. Verify the connection of the board to a USB port and the LabPro voltage probe.

• Close any open Physicalab programs, then start a new PhysicaLab session and enter your emailaddress in the email entry box. Your graphs will be sent there for later inclusion in your online labreport. Email yourself a copy of all graphs.

• Check the Ch1 box and choose to collect 50 points at 0.05 s/point. Select scatter plot. Press and

hold the switch S to discharge the capacitor. Click Get data . As soon as data appears, release theswitch.

1for a derivation of the parallel-resistor equation, refer to the Resistance experiment


Your graph should display a horizontal line at V ≈ 5 V followed by an exponential decay to V ≈ 0 V fromthe time that the switch was released.

The LabPro converts the continuously varying analog input voltage into a digital representation con-sisting of discrete and equally spaced increments in V , so that the input voltage is linearly quantized. Thevoltage difference between two adjacent voltage levels defines the internal scale and hence the resolutionof the LabPro.

• Zoom in on the near-zero region of data at the end of the decay curve by unchecking Autoscale and

adjusting the X-axis scale values, then click Draw .

Your graph should display these points placed on one of several horizontal equally spaced lines that repre-sent the the discrete voltage steps Vs of the converter output. From the graph and the corresponding datatable values, select two adjacent voltage steps V1 and V2, then determine the voltage resolution dV = V2−V1

of the Labpro converter scale and the error ∆VAB of a reading:

V1 = ............., V2 = ............., dV = ............., ∆VAB = .............

• Display the complete data set by checking Autoscale, then check the X grid and Y grid boxes to

display a grid on your graph. Select scatter plot, then click Draw .

You will be simultaneously fitting two separate equations to your data. The first equation is given byY = A and will fit a straight line at Y = VAB to the initial portion of your data, from time t = 0 to therelease of the switch at time t0 = C, where A,C represent parameters of the fitting equations.

The second equation will attempt to fit the exponential portion of the data, from the time t0 = Cand amplitude VAB = A to a final value of VAB ≈ 0 at some later time. This equation is given byY = A exp(−B∗(x−C)). The fitting parameter B determines the decay rate of the exponential curve, andcomparison with equation 1.14 shows that B = 1/RC and x= t. To summarize, we can express the timedependence of the voltage VAB across R by the expression

VAB(t) =

I0R, t < t0

I0R exp(−B (t− t0)), t ≥ t0

• To fit your data, check Fit to: y= and select from the drop-down list or enter the following string,without spaces, in the fitting equation box: A*(x<C)+A*(exp(-B*(x-C)))*(x>=C).

The fit parameters A, B and C are initially set to one. These values may be too distant from the actualfit values to allow the fitting algorithm to converge and provide a valid result. If you attempt to performa fit and get an error message, review the scatter plot of your data to estimate some more appropriatevalues for A and C.

To estimate the parameter B, you can use the fact that a capacitor discharge curve decays to 1/e =1/2.718 . . . of the original voltage level after a time ∆t = RC, defined as the time constant of the circuit.Choose a time t1 along the exponential portion of the curve and a time t2 at the point where the curvehas decreased to approximately 1/3 of the level at t1. Since ∆t = t2 − t1 ≈ RC then B = 1/RC ≈ 1/∆t.

• Label the axes and title the graph with your name and circuit arrangement used. Click the Send to:button to email yourself a copy of your graph for the exponential decay of a single capacitor.

• Check the Y log box to display the voltage in logarithmic units, then uncheck autoscale and setY= -3 to 1 in 4 steps to range y from 10−3 to 101. Redraw and email your graph.

7

? What does the exponential decay look like and why is this so?

? What feature of the graph does the fit parameter B represent? How would you prove it? Recall thatyou have taken the log of the voltage.

? Why are the points at the bottom right corner of the graph scattering? Use values from your dataset to support your conclusions.

? Do any negative voltage points from your data set appear on the graph? Can a logarithmic plotdisplay values of y ≤ 0 ?

• Record below the initial voltage A and decay parameter B

B = 1/RC = .................± .................1/s

A = I0R = .................± .................V

• Calculate the experimental value C and ∆C for the capacitor:

C = ................... = .................... = ....................

∆C = C

√

(

∆B

B

)2

+

(

∆R

R

)2

= .................... = ....................

C = .................± .................F

• Calculate the initial current I0 and its error ∆I0:

I0 = ................... = .................... = ....................

∆I0 = I0

√

(

∆A

A

)2

+

(

∆R

R

)2

= .................... = ....................

I0 = .................± .................

• The manufacturer’s value of the capacitance C used in this part of the experiment is

C = .................± .................F

Part 2: Capacitors in parallel

• Remove all jumper wires. Connect a jumper wire from P1 to P3 and another from P2 to P.

• Acquire a data set, then fit and send the exponential decay graph for two capacitors in parallel.Calculate the following parameters, then enter the results below.


B = 1/RCp = .................± .................1/s

A = I0R = .................± .................V

Cp = .................± .................F

I0 = .................± .................A

• Calculate the effective capacitance Cp and the error, or tolerance ∆Cp of the two capacitors in parallelusing the manufacturer’s values of C1 and C2.

Cp = ................... = .................... = ....................

∆Cp = ................... = .................... = ....................

Cp = .................± .................F

Part 3: Capacitors in series

• Remove all jumper wires. Connect a jumper wire from P3 to P.

• Acqiure a data set, then fit and print the exponential decay graph for two capacitors in series. Usea separate sheet to calculate the following parameters, then enter the results below.

B = 1/RCs = .................± .................1/s

A = I0R = .................± .................V

Cs = .................± .................F

I0 = .................± .................A

• Calculate the effective capacitance Cs and the error, or tolerance ∆Cs of the two capacitors in seriesusing the manufacturer’s values of C1 and C2.

Cs = ................... = .................... = ....................

∆Cs = ................... = .................... = ....................

Cs = .................± .................F

9

Part 4: Circuit time constants

• For the three circuits, use your experimental C value to calculate the tc time constant of each chargingcircuit. Also calculate the td, the time constant of the discharging circuit, when the switch S is closedand the charge stored in the capacitor discharges through Rd.

? How do these time constants vary with C, R and Rd?

tc(1) = ................... = .................... = ....................

td(1) = ................... = .................... = ....................

tc(2) = ................... = .................... = ....................

td(2) = ................... = .................... = ....................

tc(3) = ................... = .................... = ....................

td(3) = ................... = .................... = ....................

IMPORTANT: BEFORE LEAVING THE LAB, HAVE A T.A. INITIAL YOUR WORKBOOK!

Lab report

Go to your course homepage on Sakai (Resources, Lab templates) to access the online lab report worksheetfor this experiment. The worksheet has to be completed as instructed and sent to Turnitin before the labreport submission deadline, at 11:00pm six days following your scheduled lab session. Turnitin will notaccept submissions after the due date. Unsubmitted lab reports are assigned a grade of zero.

................................................................................

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................................................................................

................................................................................

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Check your schedule!

This is a reminder that Experiment 2 does not necessarily follow Experiment 1.

You need to login to your course homepage on Sakai and check your lab schedule to determine theexperiment rotation that you are to follow.

My Lab dates: Exp.2:......... Exp.3:......... Exp.4:......... Exp.5.........

Note: The Lab Instructor will verify that you are attending on the correct date and have prepared for thescheduled Experiment; if the lab date or Experiment number do not match your schedule, or the reviewquestions are not completed, you will be required to leave the lab and you will miss the opportunity toperform the experiment. This could result in a grade of Zero for the missed Experiment.

To summarize:

• There are five Experiments to be performed during this course, Experiment 1, 2, 3, 4, 5.

• Everyone does the first experiment on their first scheduled lab session.

• The next four experiments are scheduled concurrently on any given lab date.

• To distribute the students evenly among the scheduled experiments, each student is assigned to oneof four groups, by the Physics Department. Your schedule is shown in your course homepage onSakai.

11

(ta initials)


Experiment 2

Faraday rotation


• that a beam of laser light can be affected by a magnetic field;

• how a polarizer interacts with a plane polarized incident beam of light;

• how to determine the limits of validity for an experimental result;



Prelab preparation








WEBWORK: the Prelab Faraday Test must be completed before the lab session




12

13

The Faraday effect

The Faraday effect, discovered by Michael Faraday in 1845, was the first experimental evidence that lightand electromagnetism are related. This effect occurs in most optically transparent dielectric materials(including liquids) when they are subjected to strong magnetic fields.

Light, and in general, electromagnetic radiation (EMR) takes the form of self-propagating waves invacuum or in matter. These waves consist of alternating magnetic and electric field components thatoscillate perpendicular to one another and to the direction of motion of the wave.

By convention, the electric field vector ~E defines the polarization angle of the wave at any instantof time. A beam is said to be unpolarized when the ~E orientation of the component waves is a randommixture of all possible angles.

Figure 2.1: Polarization of light

Figure 2.1 depicts some electric field ~E oscilla-tions striking a polarizer grid with a vertical polar-ization axis. A polarizer selectively transmits onlythe component of ~E that is parallel to the polariza-tion axis of the polarizer, in this case ~Ey.

Recalling that ~E = ~Ex + ~Ey, then the vertical

wave is transmitted fully ( ~Ey = ~E), the horizontal

wave is attenuated fully ( ~Ex = 0), and the diagonal

waves transmit only their ~Ey component, althoughthis is not shown in the diagram.

The transmitted beam is said to be plane-polarized because all the ~Ey point in the same direc-tion, as shown by the arrow on the viewing screen.

The Faraday effect or Faraday rotation is a magneto-optical phenomenon, or an interaction betweenlight and the magnetic field in a dielectric, or non-conducting, medium. A magnetic field induces a rotationof the atomic magnetic dipoles in the dielectric, making it dielectrically polarized.

Figure 2.2: Faraday rotation of plane-polarized waveby angle β.

This causes a beam of EMR entering the mate-rial to split into two beams by the effect of doublerefraction, or circular birefringence.

These beams propagate throught the materialat different speeds so that upon emerging from thematerial, they recombine with a phase shift that isexpressed as a rotation in the polarization angle ofthe beam, as shown in Figure 2.2.

The rotation angle β of the plane of polarizationis proportional to the intensity of the component ofthe magnetic field ~B in the direction of the beamof light, as well as the length l of the sample:

β = νBl (2.1)

The Verdet constant ν is an optical parameterthat describes the strength of the Faraday effect fora particular material; it varies with the temperatureof the sample and the wavelength of the incidentlight.

14 EXPERIMENT 2. FARADAY ROTATION

Malus’ Law of polarization

In 1809, Etienne-Louis Malus (1775-1812) observed that when a polarizer is placed in front of a beam ofplane polarized incident light of intensity I0, the intensity I of the plane polarized transmitted beam isgiven by

I = I0 cos2 β, (2.2)

where β is the angle between the light’s initial polarization ~E and the polarization axis of the polarizer.From Equation 2.2 it is apparent that when β = 0, I = I0 and the light is fully transmitted, whenβ = 90, I = 0 and the light is fully blocked, and when β = 45, I = I0/2. Equation 2.2 can be easily

derived from the previous discussion of ~E components and by recalling that the intensity of a wave ofamplitude A is I = A2.

Procedure

The Faraday rotation apparatus consists of four basic components: the light source, the solenoid and powersupply, the analyzer polariod and the optical detector.

1: The light source

The rectangular enclosure on the right side of the Faraday apparatus contains the light source, a redlaser pointer of 650 nm wavelength. The laser light exits the enclosure through an integral polarizingfilter so that the output of the light source is a 95% plane polarized wave. By manipulating the fournylon thumb screws on the laser mount, the laser beam can be adjusted to traverse the apparatusalong its central axis and properly arrive at the optical detector. The beam should be adjusted toyield a maximum meter reading.

2: The Solenoid

The solenoid is a multilayer coil of wire 150 mm long that surrounds a sample of dielectric material, al = 100 mm long rod of SF-59 high-index glass. When a current i flows throught the coil, a magneticfield develops around the coil. Inside the coil, this field vector ~B points along the axis of the coil, inthe direction of the analyzer polaroid, as shown in Figure 2.2.

While the magnetic field does vary along the coil axis, this variation is not significant for samplesshorter than and properly centered in the solenoid. The calibration constant for the solenoid is:

B = 0.0111i (2.3)

where i is in Amperes (A) and B in Tesla (T).

To generate a magnetic field, set a voltage on the external power supply, then press the pushbuttonon the Faraday apparatus to energize the solenoid briefly. The current flowing throught the solenoidis displayed on the power supply current meter. The coil resistance is R ≈ 2.6 Ω so that accordingto Ohm’s Law, the coil current is i = V/R.

Avoid prolonged current flow through the solenoid; the coil will heat up and increase the temperatureof the sample, altering your results.

3: The analyzer polaroid

This component is a polaroid film that can be rotated 360 in a calibrated mount graduated at 5

intervals. A set screw is used to lock the protractor at a specific angle. Hold the protractor flatagainst the mount to prevent the angle, or meter reading, from changing as you gently tighten theset screw. Do not overtighten the assembly.

15

4: The Detector

The intensity of the transmitted beam is measured with a photodiode detector. The detector issensitive to the visible as well as some of the infrared spectrum. The output of the detector is acurrent id directly proportional to the input intensity I. The current flows through a resistor R,generating a voltage V = idR that is displayed in units of 0.1 mV on the digital readout. A gainswitch on the detector and gain adjustment knob on the front panel are used to set R and scale thisoutput to the 0 − 1999 range of the four-digit 7-segment display, the region for which the detectoroutput is linear. Since only relative intensity (I/I0) measurements will be made, calibration of thebeam intensity is not required.

Part 1: Verification of Malus’ Law

The task is to verify that Equation 2.2 is valid for this apparatus.

• Turn on the bench-top power supply, then switch on the laser.

The red laser beam should be visible at the polarizer. If there is no visible beam or the glass rodis protruding from either end of the coil, then the apparatus needs to be re-aligned. See the labinstructor.

Allow 5 minutes for the laser to warm up.

• Turn off the laser. The meter should display a zero reading, but may not.

? Why might the detector show a non-zero value with the laser is off? Explain how you would testyour hypothesis.

• Turn on the laser, then for each of the three detector switch settings slowly rotate the analyzerpolaroid over 360 and note how the intensity readout varies. Select the switch position that gives amaximum reading that does not exceed the range of the display, 1999. This will yield the best displayresolution for the measurement of I. This maximum reading is the unattenuated beam intensity I0.

? How many maxima do you note as you rotate the protractor over 360?

• Record below an I0 value and the corresponding protractor angle:

I0 = ...................., β0 =....................

? What is the periodicity, in degrees, of the cos2(θ) function? At which angles of θ is the cos2(θ)function a maximum? What is the conversion from degrees to radians?

• In 5 increments over a 180 range that includes the above I0 angle, record the beam intensity I asa function of the polarizer angle setting. Set the angle carefully; you should not need to tighten thepolarizer set screw as you take these measurements.

• Close any open Physicalab programs, then start a new PhysicaLab session and enter your emailaddress and that of your partner, if any, in the email entry box. Your graphs will be sent there forlater inclusion in your online lab report.

• Enter the data pairs (β, I) in the data window. Select scatter plot. Click Draw to plot a graphof your data. Select fit to: y= and enter A*(cos(B*x-C))**2+D in the fitting equation box.

Note that the cosine function in the fitting equation expects an argument in radians while your datais in degrees.


β ()

I (mV)

β ()

I (mV)

β ()

I (mV)

β ()

I (mV)

Table 2.1: Intensity as a function of polarization angle

? What are the dimensions of the four fit parameters A, B, C, D ?

? What is the physical meaning of the four fit parameters A, B, C, D ?

• From your graph, estimate and enter values for the fitting parameters A, B, C, D. Click Draw toperform a fit on your data. If the fit fails, you may need to reconsider some of your guesses.

• Label the axes and enter your name and a description of the data as part of the graph title. Click

Send to to email the group a copy of the graph. Record the fit results below:

A = ..........± ............... B = ..........± ...............

C = ..........± ............... D = ..........± ...............

? For which value of x = β0 does the cos2(Bx+ C) function yield a maximum I?

• Determine β0 from the fit results, then compare this value to that obtained from a visual estimate ofβ0 from the graph (it may help to check the X grid box) and to the value obtained previously whenrotating the polarizer to I0:

β0 (fit) = ............... = ............... = ...............

β0 (graph) = ............... β0 (I0) = ...............

The angle of interest is not actually β0, since I does not depend much on β near I0. The greatestchange, hence the best resolution, in intensity I with β occurs when β = βm = β0 ± 45 andI = Im = I0/2.

? Set β = βm. How did the intensity change? Is this as you expected?

? Does it matter which of the two angles is used? What differences would you note as β is increased?

17

Part 2: Determination of Verdet constant

As shown in Equation 2.1, the change in polarization angle β is proportional to the magnetic field B andhence to the solenoid current i. To optimize the measurement of this rotation in the plane of polarization:

1. set the analyzer polaroid to βm, at 45 relative to the incident beam, as follows:

(a) rotate the analyzer to get a maximum reading on the display;

(b) record the maximum intensity I0 = ..............., when β = β0.

(c) rotate the analyzer until Im = I0/2 = ................ Now, β = βm.

2. Record the intensity at βm without a magnetic field, when i = 0 and B = 0;

3. Apply a current i to the solenoid to generate a magnetic field B. The intensity reading will change.

4. Rotate the analyzer polaroid until the intensity reading matches the previous B = 0 value, thenestimate the new angle β.

The coil current is only readable during the time that the button is pressed and the coil has current flowingthrough the windings. This makes it difficult to set a specific current value, quickly, so that the coil doesnot heat up and hence increase the temperature of the glass sample. The Verdet constant varies withtemperature as well as with the frequency of light transmitted.

As already mentioned, the coil has a resistance R ≈ 2.6 Ω and obeys Ohm’s Law, V = iR. Hence theequation

i = V/2.6

can be used to predict the value of voltage V to set on the power supply so that the coil will be energizedwith a current i when the button is pressed.

• Record below β and i for set voltages V corresponding to nominal currents of i ≈ 0, 1, 2, 3 A.

V (V)

i (A)

β ()

Table 2.2: Rotation data, measured with protractor

• Enter the four points (i, β), then fit a straight line to your data by entering A+B*x in the fittingequation box. Label the graph, then email a copy.

Combining Equations 2.1 and 2.3 yields β = (0.0111νl)i and the slope from your fit can then beused to get a value for the Verdet constant:

A = ..........± .......... B = ..........± ..........

ν =B

0.0111 ∗ l = .....................= ......................

∆(ν) = ν

√

(

∆B

B

)2

+

(

∆l

l

)2

= .....................= ......................

ν = ...............± ...............


Part 3: Determination of Verdet constant, a better method

You may have noticed that the rotation angles measured in the previous section are very small for therange of currents available. With the coarse 5 resolution of the analyzer scale, the measured angles exhibitan error of 2.5, large enough to make these results, as well as the estimate for the Verdet constant ν,meaningless.

This being said, you can still guess the angle values by interpolating between the scale increments toget a qualitative feel for the relationship between the solenoid current i and the resulting rotation β.

You will now apply an indirect method that does not require angle measurements to determine therotation angle and hence the Verdet constant. Using the Malus Equation 2.2 and solving for β yields:

I = I0 cos2 β → β = arccos

√

I

I0(2.4)

The rotation angle can thus be calculated using only the values of intensity I0 and I. This approach leadsto a substantial improvement in resolution since both I and I0 are measured precisely with the digitalmeter. To determine a value for the Verdet constant:

• Rotate the analyzer to get a maximum I0 = ............... reading on the display, as before;

• rotate the analyzer until I = I0/2, then secure it with the thumb screw;

• Fill Table 2.3 with a series of voltage values 0 < V ≤ Vmax, where as before Vmax yields a currenti ≈ 3 A. For each V entry in the table, adjust the power supply to set this output voltage. Press thebutton to energize the solenoid and note the solenoid current i and the resulting beam intensity I,then release the button and record the data in Table 2.3;

• Use Equation 2.4 to calculate the rotation angle β. Enter your data in Table 2.3;

V (V)

i (A)

I

β ()

Table 2.3: Rotation data calculated from intensity ratio

• Plot your data as (i, β). The graph should approximate a straight line. Select fit to: y= and enterA*x+B in the fitting equation box and perform a linear fit on your data. Email a copy of yourgraph, then from the slope, determine a value and error estimate for the Verdet constant:

A = ..........± .......... B = ..........± ..........

ν = = .....................= .....................= ......................

∆(ν) = = .....................= .....................= ......................

ν = ...............± ...............

19

Part 4: Verdet constant of SF-59 glass

From the Internet or another source, determine the value of Verdet constant ν for SF-59 glass at roomtemperature and incident light of 650 nm. Include the appropriate units and an estimate of the error inthe value, if available:

ν(SF − 59) = ...............± ...............


Lab report


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Experiment 3

Resistance


• the concept of resistance, a characteristic of materials that resists the flow of electrons;

• about the behaviour of a semiconducting device, the diode;

• that a physical system is dependent on many factors such as temperature;



Prelab preparation








WEBWORK: the Prelab Resistors Test must be completed before the lab session




20

21

Resistance

When electrons, or other electric charge carriers (e.g. ions in a solution), are forced to move through amedium by an applied electric field (voltage, V ), their motion is in most cases retarded by scattering offimperfections (impurities) and vibrating atoms in the medium. This resistance to the movement of chargeis defined as

Figure 3.1: IV relationship for Ohmic resistor

Figure 3.2: Basic resistor circuit

R =V

I

where V is the voltage, or potential difference, appliedacross the material and I is the current, or rate of the

movement of electric charge (electrons) in the material.The resistance R of a medium (resistor) is dependenton its chemical properties, geometry, external magneticfield, temperature (the magnitude of atomic vibrationsincreases with temperature), etc.

The value of resistance may also show a dependenceto the magnitude and polarity of the voltage V appliedacross its terminals, as is observed with a device made ofsemiconducting material. Semiconductor resistance de-creases with temperature, an effect known as thermalrunaway.

A resistor that is independent of the voltage appliedacross it is called an Ohmic resistor after George Si-mon Ohm (1787-1854) who described mathematicallythe electrical characteristics of such a device. Ohm’s Lawstates that the electric current I that flows in a conduc-tor is proportional to the potential difference V betweenthe ends of the conductor, and is inversely proportionalto its resistance R.

I =V

R(3.1)

The unit for resistance is the ohm (Ω), and is derived from the units of voltage and current:

1 ohm =1 volt

1 ampere.

Equation 3.1 for an Ohmic resistor is the equation of a straight line, with the slope equal to theresistance R, as shown in Figure 3.1. By varying the voltage across a resistor and recording the current ineach case, a IV graph can be plotted, and from that graph, the resistance of an unknown resistor can beestablished. A schematic representation of the simplest electric circuit is given in Figure 3.2.

Ohmic resistors are used primarily to limit the current flow in an electric circuit. Several methods areused in their construction. For example, some resistors consist of a fine wire wound on an insulating core.The ones that you will use are formed from various carbon compounds.

Kirchoff’s Laws

The behaviour of any electric circuit can be examined with the aid of two rules developed by Gustav Kirchoff(1824-1887). These rules arise from the application of fundamental physical laws to electric circuits.

Kirchoff’s Voltage Law, or loop rule, states that the total work done on an electron by the voltagesources in a circuit equals the total work extracted from the electron while traversing the circuit. In

22 EXPERIMENT 3. RESISTANCE

following any such closed circuit loop, the gains in potential energy will be equal to the losses, so that∑

V = 0. This is the principle of conservation of energy.

A junction is a point in a circuit where a number of wires are connected together. Kirchoff’s current

law, or junction rule, states that the total electron current entering a junction, or node, equals the totalelectron current leaving the junction,

∑

I = 0. In effect, it states that no electrons are created or destroyed.This is the principle of conservation of electric charge.

Effective resistance of resistors in series

Figure 3.3: Resistors in series

The effective resistance for R1 and R2 connectedin series is RS . Applying Kirchoff’s Voltage Law(starting at O, traversing the loop clockwise) to theclosed circuit loop in Figure 3.3 yields:

V − IR1 − IR2 = 0

V = IR1 + IR2

V

I= RS = R1 +R2.

Therefore, for two resistors connected in series,

RS = R1 +R2. (3.2)

and for any number N of resistors in series

RS = R1 +R2 + · · ·+RN =N∑

i=1

Ri (3.3)

Effective resistance of resistors in parallel

Figure 3.4: Resistors in parallel

The effective resistance RP of resistors R1 and R2

in Figure 3.4 can be determined by noting that thevoltage V is the same across both resistors and ap-plying Kirchoff’s Current Law at Junction O:

I − I1 − I2 = 0

Then, for two resistors connected in parallel:

1

RP=

1

R1+

1

R2. (3.4)

and for any number N of resistors in parallel

1

RP=

1

R1+

1

R2+ · · · + 1

RN=

N∑

i=1

1

Ri. (3.5)

23

Procedure

The output terminals of the power supply constitute an open circuit. The test circuit, or load, of effectiveresistance R is connected across the power supply to establish a closed circuit through which a current Ican flow. This load may be one or more resistors in parallel or series, a diode, light bulb, motor or anyother electrical component or system.

To monitor the amount of current flowing in this closed circuit at a set voltage V , the power supplyincludes a current meter (ammeter) connected in series between the voltage source and the resistive load.Refer to Figure 3.5 for the circuit schematic.

The digital meters have a measurement error of ± 0.5% plus one least significant digit (LSD) on thedisplay.

Figure 3.5: Experimental setup

Part 1: Single resistors

In this exercise, you will determine R for a resistor from the slope of a line of best fit through a series of(I, V ) data points. You will then compare the result with the nominal resistance of the component.

• With the power supply off, check that the Range button is pressed. Rotate the voltage adjust knob

on the power supply counterclockwise until it stops turning. The output voltage is now set to 0 V.

• Connect the circuit shown in Figure 3.5. Connect the resistor with the 1 Ω nominal resistance as theload resistor. We will call this resistor R1. The error, or tolerance, of the resistors is ± 5%.

• Turn on the power supply. Press and hold the CC set button and use the curent adjust knob toset a maximum current to 2.0 A. This will limit the power supply current in case of a short circuit.

• With the voltage adjust knob, set the output voltage from 0.1 V to 1.0 V in increments of 0.1 V andat each step record in Table 3.1 the magnitude of the current flowing through the resistor.

? With V = 1.0 V and a current of around 1A flowing through the resistor, occasionally touch theresistor. What do you note?

? Does the current value change if this voltage is applied to the resistor for an extended period of time?Why might be the cause of this change? How should the current change?


V (V) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

∆(V ) (V)

I (A)

∆(I) (A)

Table 3.1: Experimental results for resistor R1

• Perform a sample calculation of ∆(V ) and ∆(I), the errors in V and I, at V = 0.5 V and enter theseresults for all the values of V in Table 3.1.

∆(V ) = .................... = .................... = ....................

∆(I) = .................... = .................... = ....................

• Close any open Physicalab programs, then start a new PhysicaLab session and enter the emails ofthe group members. Enter the data pairs and their associated errors (I, V,∆(V ),∆(I)) in the data

window. Select scatter plot. Click Draw to generate a graph of your data. The graph shouldapproximate a straight line. Select fit to: y= and enter A*x+B in the fitting equation box. Click

Draw to perform a linear fit on your data. Label the axes and enter your name and a description

of the data as part of the graph title. Click Send to to email all members a copy of your graph forlater inclusion in your lab report.

• Enter below the experimental values for R1 and ∆(R) from the slope of the graph. Also include forcomparison the nominal value R1 and tolerance ∆(R) of resistor R1.

(slope)R1 = .................± .................Ω

(nominal)R1 = .................± .................Ω

• Repeat the above steps by connecting the circuit shown in Figure 3.5 using the resistor R2 withnominal resistance of 0.75 Ω. as the load resistance. Enter your findings in Table 3.2.

V (V) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

∆(V ) (V)

I (A)

∆(I) (A)

Table 3.2: Experimental results for resistor R2

25

• Summarize below the values of R2 from the slope of the graph and the the value displayed on theresistor R2.

(slope)R2 = .................± .................Ω

(nominal)R2 = .................± .................Ω

Part 2: Resistors in series

• Replace the single resistor used in Part 1 by the two resistors connected in series as shown in Fig-ure 3.3. Set V=0.5 V and measure the corresponding value of I.

V = ..........± .......... V I = ..........± .......... A

Use Equation 3.1 to determine the experimental effective resistance Rs. Apply the proper errorpropagation rule to evaluate the error ∆(Rs) of the two resistors in series.

Rs = .................... = .................... = ....................

∆(Rs) = .................... = .................... = ....................

Rs(Ohm′s Law) = ...............± ...............Ω

• Use the nominal component values for R1 and R2 and Equation 3.2 to calculate the theoreticaleffective resistance Rs and error ∆(Rs) of the two resistors in series:

Rs = .................... = .................... = ....................

∆(Rs) = .................... = .................... = ....................

Rs(Series Law) = ...............± ...............Ω

Part 3: Resistors in parallel

• Connect the two resistors in parallel. Set V = 0.5 V and measure the corresponding value of I.Calculate the experimental effective resistance Rp and error ∆(Rp) of the two resistors in parallel.

Rp = .................... = .................... = ....................

∆(Rp) = .................... = .................... = ....................

Rp(Ohm′s Law) = ...............± ...............Ω


• Use the nominal component values and Equation 3.4 to calculate the theoretical resistance Rp andthe error ∆(Rp) of the two resistors in parallel:

Rp = .................... = .................... = ....................

∆(Rp) = .................... = .................... = ....................

Rp(Parallel Law) = ...............± ...............Ω

Part 4: IV characteristics of a diode

Figure 3.6: Current flow in a diode

As mentioned in the Introduction, many electrical devices do notobey Ohm’s law. A diode is a semiconducting device whose re-sistance not only depends on the voltage V applied across its ter-minals, but also on the polarity, or direction that the voltage isapplied.

The polarity of the diode is identified by a band at one end ofthe diode body. A diode is forward biased when the band end isconnected to the negative (-) terminal of the power supply. It isreverse biased when the band end is connected to the positive (+)terminal of the power supply.

• Set V = 0 V. Replace the resistors with the diode so that itis forward biased. Measure the current in the circuit over arange of voltages from 0.1 V to 1.0 V in steps of 0.1 V. Donot exceed 1 V as the diode may begin to conduct excessively, overheat and burn out. Present yourresults in Table 3.3.

V (V) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

∆(V ) (V)

I (A)

∆(I) (A)

Table 3.3: Experimental results for forward biased diode

? Is it difficult to read the current as the voltage is set to 1.0 V? Why? What would you think ishappening? Does this effect eventually disappear? Why?

• Rearrange the diode so that it is connected in the reverse biased direction. Repeat the series ofmeasurements and enter your data in Table 3.4.

• On the same graph, plot I as a function of −1.0 ≤ V ≤ 1.0 using the two data sets from Tables 3.3and 3.4. Check Line between points to display the trend of your data points. Properly label andtitle the graph, then email a copy.

27

V (V) -0.1 -0.2 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1.0

∆(V ) (V)

I (A)

∆(I) (A)

Table 3.4: Experimental results for reverse biased diode

Part 5: Resistance characteristics of a heated filament

In this exercise you are going to explore the temperature dependence of a resistor. The resistor in this caseis the Tungsten filament of a low voltage light bulb. As the voltage applied across the filament is increased,the current causes the filament to increase in temperature and eventually begin to glow.

V (V) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

I1 (A)

I2 (A)

I3 (A)

〈I〉 (A)

Table 3.5: Experimental results for Tungsten filament

• Replace the diode with the light bulb. Measure I and enter the results in Table 3.5. Estimate thevoltage when the filament begins to glow. Repeat three times, then calculate 〈I〉 and 〈V 〉.

Von1 = ........V, Von2 = ........V, Von3 = ........V, 〈Von〉 = ........V

• Plot the (V, 〈I〉) results for the tungsten filament. Also include a point at (0,0). Connect the pointswith line segments as before, then save a copy of the graph.

? As you analyse the graph, what do you note in the behaviour of the V, I curve in the region wherethe light bulb begins to glow? How would you explain this feature of the graph?


Lab report



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Experiment 4

Electron charge-to-mass ratio


• the relationship between electric and magnetic fields as applied to an electron;

• how to determine the strength of the Earth’s magnetic field;

• to compensate for the effect of undesirable contributions in an experiment;



Prelab preparation








WEBWORK: the Prelab Electron Test must be completed before the lab session




30

31

The electron charge-to-mass ratio

This experiment, performed in 1897 by J.J. Thomson, demonstrated that atoms are not fundamental unitsof matter but are composed of aggregates of charged particles, protons and electrons. It was used todetermine a value for e/m, the ratio of electron charge to electron mass. Commonly called the BainbridgeExperiment in honour of the scientist who developed the method, this experiment exhibits an ingeniouscombination of electromagnetic theory and mechanics.

Figure 4.1: The Bainbridge Tube, with front at bottom and base at left

As shown in Figure 4.1, the central component of the apparatus is a large vacuum tube (Bainbridgetube). Inside the tube, a horizontal metal cylinder with a narrow opening at the conical end, called theanode, encloses an axially mounted wire, the cathode.

The cathode heater power supply provides a voltage across the cathode wire and causes a significantelectron current flow through the wire. Collisions between these energetic electrons and the atoms in thewire ejects some electrons from the material, forming an electron cloud inside the cyclinder, around thecathode wire.

The anode power supply is then used to set the anode at a potential V more positive than that ofthe cathode, to create a radial electric field inside the cylinder that will accelerate the negatively chargedelectrons toward the outer anode. Some of these electrons shoot through the slot in the anode. Afterescaping, they will continue to move in a straight line with a constant speed v since there is no electricfield outside the anode.

Work-energy theorem demands that the work done by the electric field in accelerating an electron isequal to the change in the electron’s potential energy,

∆(K.E.) = W = eV. (4.1)

Assuming that the initial kinetic energy of the electrons near the surface of the cathode is zero, Equation 4.1yields the value for the kinetic energy an electron has after it escapes through the slot in the anode.

32 EXPERIMENT 4. ELECTRON CHARGE-TO-MASS RATIO

Changing the anode voltage V controls this kinetic energy and, therefore, the speed v of the electron,

K.E. =1

2mv2 ⇒ v =

√

2V( e

m

)

. (4.2)

The second stage of this experiment involves magnetic forces. Two large Helmholtz coils, positionedabove and below the Bainbridge tube, are used to create a uniform magnetic field in the tube. With thisarrangement, the direction of the magnetic field intensity ( ~B) is vertical, perpendicular to the velocity ~v

of the electrons. The magnetic force (~F ) experienced by an electron is given by the vector cross-product

~F = e~v × ~B and F =∣

∣

∣

~F∣

∣

∣= evB. (4.3)

Since ~F is always perpendicular to ~v, it does no work on the electron, and cannot change its kinetic energyor its speed. The magnetic force acts as a centripetal force, Fc = mac, making the electrons move in acircle of radius r. The plane of the circle is perpendicular to ~B. From mechanics, the centripetal force is

F = evB = Fc = mac =mv2

r. (4.4)

Rearranging Equation 4.4 so that only e/m remains on the left-hand side yields

e

m=

2V

r2B2. (4.5)

Thus the charge-to-mass ratio for an electron, e/m, can be determined from V , r and B.

The magnetic field strength B depends of the radius R and the number of turns N of wire of theHelmholtz coils and the current I flowing in them:

B =8µ0

5√5

N

RI (4.6)

where B is in units of Tesla (T), I in units of Ampere (A) and µ0 = 4πx10−7 (T-m/A2) is the magneticconstant of a vacuum. For the Helmholtz coils of this apparatus, N = 150 turns and R = 0.140 m, then

B = (9.634 × 10−4) I

The Banibridge tube is filled with mercury vapour. Electrons colliding with mercury atoms inducethese atoms to emit a faint blue-green light, making the electron path visible in a dark room.

By varying the magnetic field strength B at constant anode voltage V , or by varying V at constant B,the electron beam is made to follow a circular path. A horizontal scale with a sliding pointer is used tomeasure the left and right extremes of this circular beam. This yields tha beam diameter from which thebeam radius can be determined.

This is essentially the strategy we will pursue for this experiment: set a current I and voltage V ,determine the size of the circular orbit r, determine the magnetic field strength B from Equation 4.6, thensubstitute these values into Equation (4.5) to calculate the ratio e/m.

The electron beam emerging from the electron gun may spread out slightly as it goes around the orbit.There are two main reasons for this spreading. First, not all the electrons leave the slot with exactly thespeed v given by Equation (4.2), nor are all of the velocities ~v exactly perpendicular to ~B. Also, collisionswith the mercury may change the velocity of the electrons and therefore, the radius of their orbit.

33

Figure 4.2: Schematic Diagram of Electrical Apparatus

Procedure

The experimental electronics are shown in Figure 4.2. Ask the instructor to explain the operation of anyof the electrical components which you do not understand.

1. The Accelerating Voltage dial sets the accelerating potential V applied to the electrons inside thecylinder and hence determines the velocity v of the electrons emitted from the tip of the electrongun. Once outside the gun, the speed of the electrons will not change. The Accelerating Voltage,measured in Volts, is displayed on the analog meter labelled V.

To see a beam, V must be ≈ 80 V or greater, otherwise the electrons will not have enough energy toionize the Mercury gas in the Bainbridge Tube and make it glow.

The analog meters used to measure current and voltage also operate on a magnetic principle: here acoil is mounted on concentric pivots between the poles of a permanent magnet, with a pointer mountedin the plane of rotation. When a current flows through the coil, a magnetic field is produced thatinteracts with the magnetic field of the permanent magnet, causing the coil and attached pointerto rotate. A circular scale can then display the current flowing through the coil. To take a readingfrom an analog meter, view the needle at a right angle to the face of the meter so that it’s profilewill be minimized, then the point on the scale behind the needle represents the best value for themeasurement.

2. The Magnetizing Current dial adjusts the amount of current I flowing through the Helmholtz coilsand hence their magnetic field intensity B. This current, measured in Amperes, is displayed on theanalog meter labelled A.

3. The power supply that sends a current through the cathode filament in order to heat it and make itand release electrons, sets the beam intensity and is not adjustable in this experiment.

The measurement error of an analog meter is equal to one-half of the smallest increment in the in themeter scale.

Note: There are no workstations in the darkroom. As you proceed to do Parts 1-3 of this Experiment,begin by performing all the the data gathering steps in the darkroom with the Bainbridge Tube, thenmigrate to a workstation to complete the calculations, graphing and analysis part of the Experiment.


Part 1: Initial setup and precautions

With the power supply set to ”OFF”, perform the following steps:

• set the Deflecting Voltage polarity to ”OFF” and voltage control fully counterclockwise to minimum.This voltage, when applied to a pair of plates inside the electron gun, will cause the beam to bend.This control is not part of the experiment;

• set the Accelerating Voltage control fully counterclockwise to minimum;

• set the Magnetizing Current direction to ”OFF” and current control fully counterclockwise to mini-mum.

• Turn the power supply to ”ON”. Verify that the filament inside the electron gun illuminates. Allowthe cathode to heat up for a few minutes before applying Accelerating Voltage.

During the operation of the tube, observe the following precautions:

• bring the Accelerating Voltage up slowly;

• do not allow the electron beam to strike the glass envelope of the tube for an extended period of time.This happens when there is an applied Acelerating Voltage and insufficient Magnetizing Current tobend the beam in a circular path inside the tube;

• limit the operating session to one hour to prevent the overheating of the electron gun;

• turn all controls to minimum and all switches to ”OFF” before turning the power ON or OFF!

Part 2: Constant anode voltage

In this section you will measure the diameter of the electron beam as you vary the magnetic field whilekeeping the accelerating voltage constant. You will then determine the average 〈e/m〉 and the standarddeviation σ(e/m) of the distribution of e/m values calculated from this experimental data. Similarly, youwill determine 〈v〉 and σ(v) from the set of v results.

• With the tube warmed, slowly increase the Accelerating Voltage to 150 V. A straight beam shouldemerge from the electron gun.

• Set the current direction to clockwise, then increase the Magnetizing Current I to 1.0 A; the electronbeam should now trace a circular orbit.

• Position your eye directly in front of the left side of the beam’s arc. Slide the vertical pointer alongthe horizontal scale until it is directly in line with the left side of the arc. When your eye, pointer andbeam edge are correctly aligned, the pointer will appear to have a minimum width and will overlapthe edge of the beam.

• Record the index position d1 in Table 4.1, then repeat the above step for the right side of the beamand record this index position d2 in the data table.

• With the radius of the beam r given by r = (d2 − d1)/2, calculate r, then using Equation calculatee/m. Record Im and V in the appropriate row of Table 4.1. Calculate below the values of B, r, e/mand the speed v of the electrons, then enter the results in Table 4.1.

r =d2 − d1

2= ............................ = ..................

35

B = (9.634 × 10−4) I = ............................ = ..................

v =2V

rB= ............................ = ..................

e

m=

2V

r2B2= ............................ = ..................

• Repeat the above measurements and calculations for Magnetizing Currents of 1.2, 1.4, 1.6, 1.8, 2.0and 2.2 A, keeping V constant.

i V (V) I (A) d1 (m) d2 (m) r (m) B (T) v (m/s) e/m (C/kg)

1 150 1.0

2 150 1.2

3 150 1.4

4 150 1.6

5 150 1.8

6 150 2.0

7 150 2.2

Table 4.1: Calculation of e/m and electron velocity v

Since there is a sample of several values for v and e/m, the best error estimate σ is obtained froma statistical calculation that yields the averages 〈v〉, 〈e/m〉 and standard deviations σ(v), σ(e/m). Thestandard deviation of a sample σ provides a measure of how closely the sample values are clustered aboutthe sample average.

i em

em −

⟨ em⟩ ( e

m −⟨ em⟩)2

i v v − 〈v〉 (v − 〈v〉)2

1 1

2 2

3 3

4 4

5 5

6 6

7 7⟨ em⟩

= variance = 〈v〉 = variance =

σ( em)

= σ(v) =

Table 4.2: Calculation of 〈x〉 and σ(x) for e/m and v. Here, variance(x) = σ2(x) = 1N−1

∑Ni=1(xi − 〈x〉)2.


• Use Table 4.2 as your worksheet to calculate the average and the standard deviation of e/m and v.

• In Physicalab, click File, New to clear the data window, then enter in a column the seven e/mvalues. Select Options, Insert X index to insert a column of index values.

• Check bellcurve y data to display a distribution of your e/m values and the Gaussian bellcurvedetermined from the average and standard deviation of the sample. The value of 〈e/m〉 ± σ(e/m)should be identical with the results that you obtained in Table 4.2. Email yourself a copy of thegraph.

• Repeat the above step using the seven values for v.

You may now complete this part of the experiment by reporting the final results, correctly rounded, fore/m and v.

e/m = ..................± .................. C/kg

v = ..................± .................. m/s

Part 3: Constant coil current

In this section you will determine e/m and σ(e/m) by performing a linear least squares fit on a set ofgraphed data points. Equation (4.5) can be rewritten to indicate that V varies with r2.

V =B2

2

( e

m

)

r2 =

(

B2e

2m

)

r2 (4.7)

and is the equation of a straight line through the origin, with B2e/(2m) as the slope.To test this relation, I and hence B will be fixed and V varied to to change the beam diameter.

• Set the Magnetizing Current to 2.0 A and V to 100 V to produce a visible beam.

? As the voltage is decreased below 100 V, how does the beam behave? Why is this so? At whatvoltage is the beam no longer visible?

• Determine the beam radius for V = 100 − 220 V and enter the results in Table 4.3.

• Enter in the Physicalab data window the coordinate pairs (r2, V ). Select scatter plot. Click Drawto generate a graph of your data. Your data should approximate a straight line. If this is not thecase, check your data and then redraw the graph.

• Select fit to: y= and enter A*x+B in the fitting equation box. Here B is the fitting parametercorresponding to the Y-intercept of the straight line, not the magnetic field vector B. Label the axes

and title the graph. Click Draw , then Send to: to email the group a copy of the graph.

• From the fit, record below the value of the slope A and error ∆(A):

A =B2e

2m= ..................± .................. V/m2 (4.8)

• Determine ∆(I) by using the instrument scale error, then B, ∆(B), e/m and ∆(e/m). Finally,summarize these results in the proper format:

37

i V (V) I (A) d1 (m) d2 (m) r (m) r2 (m)

1 100 2.0

2 120 2.0

3 140 2.0

4 160 2.0

5 180 2.0

6 200 2.0

7 220 2.0

Table 4.3: Data for determining e/m and σ(e/m) using a linear least squares fit

∆(I) = ........................ = ...................... = ......................

B = ........................ = ...................... = ......................

∆(B) = ........................ = ...................... = ......................

e/m = 2 ∗A ∗B−2 = ...................... = ......................

∆(e/m) =( e

m

)

√

(

∆A

A

)2

+

(

2∆(B)

B

)2

= ...................... = ......................

I = ..................± .................. A

B = ..................± .................. T

e/m = ..................± .................. C/Kg

Part 4: electron e/m

From the Internet or another source, or using the known values of e and m, determine the value of e/m.

e/m = ..................



Lab report


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(ta initials)


Experiment 5

Diffraction of light by a grating


• the geometical analysis of a diffraction grating;

• to use a diffraction grating to measure the Balmer Spectrum of Hydrogen;

• to visualize data in different ways in order to improve the understanding of a physical system;



Prelab preparation








WEBWORK: the Prelab Diffraction Test must be completed before the lab session




40

41

The diffraction grating

The optical diffraction grating is a glass or plastic plate with many fine, parallel grooves spaced the samedistance d from each other on its surface. According to Huygens’ Principle, when monochromatic lightfrom a distant source or laser hits the grating, each groove behaves as a source of spherical wavelets thatre-radiate from the grating in phase with the incident wave.

The superposition of these secondary waves will contribute to either enhance or diminish the brightnessof ths diffracted beam at a given point in space, resulting in a pattern of bright and dark regions in space.This pattern can be viewed and analysed by placing a screen parallel to and far away from the gratingsomewhere in the path of the diffracted beam. Then,

1. if the incident light beam is a parallel beam and is incident at a right angle to the grating, a diffractedbeam will be offset from its parallel neighbour by a distance d sinα. as shown in Figure 5.1.

2. When this distance is equal to an integer number m of wavelengths λ of the incident light, the twobeams are in phase and will exhibit constructive interference by displaying a series of bright regionson the screen. These interference maxima are given by:

d sinα = mλ, m = 0,±1,±2, . . . (5.1)

where λ is the wavelength, d is the grating spacing, and m is an integer called the order number.The zero-order beam m = 0 is a continuation of the incident beam (i.e. α = 0).

3. When the path difference between adjacent beams is (m + 1/2)λ, then destructive interference willresult in dark regions, or interference minima, on the screen.

4. There are two first order beams, m = ±1 at angles given by sinα = ±λ/d, two second order beamsm = ±2 at sinα = ±2λ/d, et cetera. Hence the measurement of the angle α, together with the ordernumber m, gives the ratio λ/d, and if either λ or d is known, the other can be calculated.

Figure 5.1: Diffraction by a grating

To diffract X-rays, electrons, or neutron matterwaves, one needs a diffraction grating whose d iscomparable with the wavelength of the waves. Itturns out that crystal materials have interatomicspacings comparable with the λ of X-rays. X-raydiffraction is now a standard way of determiningthe atomic arrangements in a crystal.

Note that if λ > d one doesn’t get diffractionmaxima of order m ≥ 1 and only the zero-orderbeam (m = 0) will be visible at the centre of thediffraction pattern. On the other hand, if λ is muchless than d, then the corresponding angles becomemuch too small and the diffraction pattern will notbe resolved, appearing again as a bright spot on thescreen.

Thus, the diffraction effect becomes importantwhen λ is not too small a fraction of d.

To experimentally determine a value for thegrating distance d of a diffraction grating a monochromatic light source of a known wavelength λ is usedas the incident beam. The beam is diffracted from the grating and generates an interference pattern on ascreen a distance D from the grating, as shown in Figure 5.4.

On the screen the distance L between pairs of bright spots from the diffraction of order ±m can bemeasured and Equation 5.1 can be used to calculate the grating distance d. Note that this equation assumesthat the beams from adjacent grating slits are parallel, that is, that D ≫ L.

42 EXPERIMENT 5. DIFFRACTION OF LIGHT BY A GRATING

Energy levels of the Hydrogen atom

A Hydrogen atom consists of a positively charged proton making up the atomic nucleus and a negativelycharged electron orbiting this nucleus. Quantum Theory predicts that the electron may only find itself inone of n = 1, 2, 3, . . . discrete orbits, or energy levels, around the nucleus.

Figure 5.2: Electronic energy transitions of H2

When a Hydrogen atom is subjected to an elec-tric discharge, its one electron may absorb some ofthis energy. When this happens, the electron makesa transition from an orbit n1 to an orbit n2 wheren2 > n1. The electron eventually decays back to alower orbit, releasing this surplus energy in the formof a photon of one of several specific wavelengths λ.Some of these transitions radiate photons that havethe wavelength of visible light and can thus be mea-sured with a grating spectrometer.

Johann J. Balmer (1825-1898) discovered thatfor the Hydrogen atom the series of energy transi-tions wavelengths from an initial energy level ni > 2to the energy level nf = 2, the Balmer series, areapproximately given by:

λ =

[

R

(

1

n2f

− 1

n2i

)]−1

→ λ =

[

R

(

1

22− 1

n2i

)]−1

(5.2)

This equation is a special case of the more general Rydberg formula that accounts for all transitions fromni to nf . For example, the Lyman series outlines the transitions from an initial energy level ni > 1 to a finalenergy level nf = 1 that are not in the visible region of the spectrum. The value of the Rydberg constant Ris determined by fitting this empirical equation to experimental data and is equal to R = 1.097× 107 m−1.

Figure 5.3: Diffraction of the H2 spectrum

The electron energy level transitions that pro-duce wavelengths in the visible region are from aninitial orbit n = 3, 4 or 5 to a final orbit n = 2,as shown in Figure 5.2. These transitions will bevisible as red (n = 3 → 2), blue (n = 4 → 2), andviolet (n = 5 → 2) lines in the spectrum of molecu-lar Hydrogen (H2) from an H2 discharge tube. Thecombination of these three colors, as in the unscat-tered beam, is observed as a pink line.

The spectrometer, shown in Figure 5.5, is an op-tical instrument that uses a prism or a diffractiongrating to scatter an incident light beam of interestinto component wavelengths. The grating used inthis spectrometer is chosen to optimally view thefirst order (m = ±1) diffraction pattern of the inci-dent beam.

When the light from the H2 discharge lamp isviewed through a spectrometer, the three colorslines scatter at different angles symmetrically aboutthe direct path of the incident light, as in Figure 5.3.

43

Procedure

Part 1: Determining the spacing of a diffraction grating

Figure 5.4: Experimental setup for Part I.

In this part of the experiment you will determinethe spacing d of a glass grating by passing a laserbeam through it and examining the diffraction pat-tern projected on a screen. The laser beam is par-allel and monochromatic, with wavelength

λ±∆λ = 632.8 ± 0.5 nm.

• Check that the screen and grating surface areperpendicular to the incident beam, and mea-sure the distance D between the grating andthe screen, as shown in Figure 5.4.

D = ...........± ...........m

? What do you note in the diffraction pattern as you slowly rotate the grating? Based on your obser-vation, if the grating was set to some angle θ from being perpendicular to the beam, how would youadjust your L values to account for this offset?

• Mount a sheet of graph paper on the screen and carefully mark the series of interference maxima.Identify the straight path m = 0 maximum. Each student must make and analyse their owninterference pattern. Measure the distance L between the centres of pairs of spots of order m, (+mto −m) for m = ±1 to m = ±10. Record your results in Table 5.1.

m L (m) d (m) (d− 〈d〉) (m) (d− 〈d〉)2 (m)

±1

±2

±3

±4

±5

±6

±7

±8

±9

±10

〈d〉 = variance =

σ(d) =

Table 5.1: Calculation of 〈x〉 and σ(x) for d. Here, variance(x) = σ2(x) = 1N−1

∑Ni=1(xi − 〈x〉)2.


• With L and D measured, the relationship between these variables and the angle α in Figure 5.4 isgiven by

tan(α) = L/2D. (5.3)

If the angle α is small, then sinα ≈ tanα and equating these two terms in Equations 5.3 and 5.1yields

d = 2mλD/L. (5.4)

• Calculate d for the ten measurements of L, then calculate an average value 〈d〉 and standard deviationσ(d) of d and enter the results in Table 5.1.

• Close any running Physicalab programs, then start a new Physicalab session and enter the emailaddress of the group members. Enter the d values as coordinates (m,d). Select bellcurve and

bargraph, then click draw to display a distribution of your n values with the average 〈d〉 andstandard deviation σ(d) of the sample. Click Send to: to email yourself a copy of the graph.

• Verify that the results from the distribution are identical to those from Table 5.1. If they are not,you need to review your calculations. Report below the grating spacing for this grating:

d = ...........± ...........m

Part 2: Determining the wavelengths of the Balmer spectrum of H2

Figure 5.5: Experimental setup for part 2

The light source is a hydrogen discharge tube. Thissource illuminates a slit at one end of the collimatortube, and a lens at the other end makes a parallelbeam of the light passing through the slit. Thebeam is diffracted by the grating and collected bythe front lens of the telescope, which focuses thelight on a set of cross-hairs.

The telescope can rotate around the grating, itsangular position with respect to an arbitrary zerogiven by an angular scale on the base of the in-strument. Proceed as follows, remembering neverto touch the grating as it is easily damaged and isexpensive to replace.

Note: to extend the life of the H2 discharge tube, keep it from overheating: turn it on for 30 secondsor less ito adjust the crosshairs then turn it off for at least 30 seconds while you read the Vernier scale.

• Rotate the telescope to a position opposite the collimator. Looking through the telescope you shouldsee a sharp image of the slit, its colour the same as that of the light emanating from the dischargetube. Gently lock the telescope in place Looking through the telescope you should see a sharp imageof the slit, its colour the same as that of the light emanating from the discharge tube. Gently lockthe telescope in place with the knob located at the centre of the telescope base.

• Turn the fine-adjust knob located on the right side of the telescope base to centre the crosshairs

diagonally on an edge of the slit image. A screw on the collimator near the light source allowsadjustment of the slit width. Be sure to have the slit and crosshairs in focus.

? Does it matter which edge of the image is used for reference? How does this choice impact theremainder of your experiment?

45

• Read the position of the telescope from the angular scale on the base. This can be done to a precision

of 1′ ( ± one minute, 60′ = 1

) . To read the scale:

1. Locate the “0” line on the vernier scale, and note which main scale division it is immediatelyafter, e.g. 21230′ on the main scale in Figure 5.6. Note that the numbers on the main andvernier scales increase from right to left, and not from left to right as you are used to reading.

2. Scan along the line where the main and vernier scales meet, and note which one vernier scaledivision is directly in line with a main scale division, e.g. 17′ on the vernier scale in Figure 5.6.

3. Add the main and vernier scale readings to obtain the angular scale reading, e.g. 21230′+17′ =21247′ in Figure 5.6. Enter your measurement in Table 5.2.

Figure 5.6: Example of Angular Vernier Scale Reading – 21247′

• Unlock the telescope and slowly rotate it to the right until the first violet slit image is in the field ofview. Lock the telescope, and use the fine-adjust knob until the cross hairs are again situated on the

same edge of the slit image as was used before. Read the position indicated on the angular scale andlet its value be θ+1. This value corresponds to the direction of the diffracted beam with m = +1 forthe violet spectral line.

• Unlock the telescope, rotate it to the left of center until the first violet image is seen again. Determineits angular position θ−1, corresponding to the diffracted beam with m = −1. Repeat the abovemeasurements for the blue and violet spectral lines and enter your data in the first row of Table 5.2.

• Convert your data values from degrees and minutes to decimal degrees, recalling that 1

= 60′, andenter these in the second row of Table 5.2.

Pink (θ0) Violet (θ+1) Violet (θ−1) Blue (θ+1) Blue (θ−1) Red (θ+1) Red (θ−1)

, ′

Table 5.2: Measurements for the spectral lines of H2 in degrees/minutes and decimal degrees


The diffraction angle α±1 for a particular colour is the measured angle of deviation θ±1 of that colourfrom the light’s direct path reference angle θ0. Calculating the difference between the angular positions ofthe pink and coloured lines will give you α±1, i.e.

α±1 = |θ0 − θ±1| (5.5)

• Calculate from the data in Table 5.2 the values of the diffraction angle α±1 for the three lines of theH2 spectrum. There will be two results for each colour, α±1, one for each side of the reference angleθ0. Calculate the average 〈α〉 of these two angles then estimate the error with ∆(α) = 1

2 |α+1 −α−1|.

line α+1 α−1 〈α〉 ∆(α)

Violet (αV )

Blue (αB)

Red (αR)

Table 5.3: Calculated diffraction angles for the spectral lines of H2

The diffraction grating used in the spectrometer is made with a line density of

N ±∆(N) = 600 ± 1 lines/mm.

This is not the same value as the grating spacing in Part 1. The line density and grating spacing are relatedvia d = 1/N . The distance d between the lines and error ∆(d) for the grating used in this spectrometer is

d = ................. = .................= ................. mm

∆(d) = ................. = .................= ................. mm

• Calculate a wavelength λ(α) and the associated error ∆(λ(α)) for the violet, blue and red spectralline of H2. Angle errors must be expressed in radians. Record this data in Table 5.4.

λV =d

msin(αV ) = ...................... = ......................

λB = = ........................... = ...................... = ......................

λR = = ........................... = ...................... = ......................

∆(λV ) = λV

√

(

∆(d)

d

)2

+

(

cos(αV )∆(αV )

sin(αV )

)2

= ...................... = ......................

∆(λB) = = ........................... = ...................... = ......................

∆(λR) = = ........................... = ...................... = ......................

47

• Use the Balmer Equation 5.2 to calculate the theoretical wavelengths λ(Balmer) of the violet, blueand red lines of the H2 spectrum. Append this data to Table 5.4.

λV (Balmer) = ....................= ....................= ....................

λB(Balmer) = ....................= ....................= ....................

λR(Balmer) = ....................= ....................= ....................

line transition λ(α) ∆(λ(α)) λ(Balmer)

violet

blue

red

Table 5.4: Calculated λ(α) from angles 〈α〉 and λ(Balmer) from Equation 5.2


Lab report


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Appendix A

Review of math basics

Fractions

a

c+

b

d=

ad+ bc

cd; If

a

c=

b

d, then ad = cb and

ad

bc= 1.

Quadratic equations

Squaring a binomial: (a+ b)2 = a2 + 2ab+ b2

Difference of squares: a2 − b2 = (a+ b)(a− b)

The two roots of a quadratic equation ax2 + bx+ c = 0 are given by x =−b±

√b2 − 4ac

2a.

Exponentiation

(ax)(ay) = a(x+y) ,ax

ay= ax−y , a1/x = x

√a , a−x =

1

ax, (ax)y = a(xy)

Logarithms

Given that ax = N , then the logarithm to the base a of a number N is given by logaN = x.For the decimal number system where the base of 10 applies, log10 N ≡ logN and

log 1 = 0 (100 = 1)

log 10 = 1 (101 = 10)

log 1000 = 3 (103 = 1000)

Addition and subtraction of logarithms

Given a and b where a, b > 0: The log of the product of two numbers is equal to the sum of the individuallogarithms, and the log of the quotient of two numbers is equal to the difference between the individuallogarithms .

log(ab) = log a+ log b

log(a

b

)

= log a− log b

The following relation holds true for all logarithms:

log an = n log a

48

49

Natural logarithms

It is not necessary to use a whole number for the logarithmic base. A system based on “e” is often used.Logarithms using this base loge are written as “ln”, pronounced “lawn”, and are referred to as natural

logarithms. This particular base is used because many natural processes are readily expressed as functionsof natural logarithms, i.e. as powers of e. The number e is the sum of the infinite series (with 0! ≡ 1):

e =

∞∑

n=0

1

n!=

1

0!+

1

1!+

1

2!+

1

3!+ · · · = 2.71828 . . .

Trigonometry

Pythagoras’ Theorem states that for a right-angled triangle c2 = a2 + b2.Defining a trigonometric identity as the ratio of two sides of the triangle,there will be six possible combinations:

sin θ =b

ccos θ =

a

ctan θ =

b

a=

sin θ

cos θ

csc θ =c

bsec θ =

c

acot θ =

a

b=

cos θ

sin θ

sin(θ ± φ) = sin θ cosφ± cos θ sinφ sin 2θ = 2 sin θ cos θ 180 = π radians = 3.15159 . . .

cos(θ ± φ) = cos θ cosφ∓ sin θ sinφ cos 2θ = 1− 2 sin2 θ 1 radian = 57.296 . . .

tan(θ ± φ) =tan θ ± tanφ

1∓ tan θ tan φtan 2θ =

2 tan θ

1− tan2 θsin2 θ + cos2 θ = 1

To determine what angle a ratio of sides represents, calculate the inverse of the trig identity:

if sin θ =b

c, then θ = arcsin

(

b

c

)

For any triangle with angles A,B,C respectively opposite the sides a, b, c:

a

sinA=

b

sinB=

c

sinC, (sine law) c2 = a2 + b2 − 2ac cosC. (cosine law)

The sinusoidal waveform

Consider the radius vector that describes the circumference of a circle, as shown in Figure A.1 If we increaseθ at a constant rate from 0 to 2π radians and plot the magnitude of the line segment b = c sin θ as afunction of θ, a sine wave of amplitude c and period of 2π radians is generated.

Relative to some arbitrary coordinate system, in this case the X-Y axis shown, the origin of this sinewave is located at a offset distance y0 from the horizontal axis and at a phase angle of θ0 from the verticalaxis.

The sine wave referenced from this (θ, y) coordinate system is given by the equation

y = y0 + c sin(θ + θ0)

50 APPENDIX A. REVIEW OF MATH BASICS

Figure A.1: Projection of a circular motion to an X-Y plane to generate a sine wave

Appendix B

Error propagation rules

• The Absolute Error of a quantity Z is given by ∆Z, always ≥ 0.

• The Relative Error of a quantity Z is given by ∆ZZ , always ≥ 0.

• If a constant k has no error associated with it: constant factors out of relative error

Z = kA ∆Z = k∆A and∆Z

Z=

∆A

A

• Addition and subtraction: note that error terms always add

Z = kA±B ± . . . ∆(Z) =√

(k∆A)2 + (∆B)2 + . . .

• Multiplication and division: constants factor out of relative errors

Z =kA×B × . . .

C ×D × . . .

∆Z

Z=

√

(

∆A

A

)2

+

(

∆B

B

)2

+

(

∆C

C

)2

+

(

∆D

D

)2

. . .

• Functions of one variable: if the quantity A is measured with uncertainty ∆A and is then used tocompute F (A), then the uncertainty ∆F in the value of F (A) is given by

∆F =

(

dF

dA

)

∆A

Function F (A) Derivative, dFdA Error equation

An nAn−1 ∆FF = n∆A

A

logeA A−1 ∆F = ∆AA

exp(A) exp(A) ∆FF = ∆A

sin(A) cos(A) ∆F = cos(A)∆A

cos(A) − sin(A) ∆F = − sin(A)∆A

tan(A) sec(A)2 ∆F = sec(A)2∆A

All trigonometric functions and the errors in the angle variables are evaluated in radians

51

52 APPENDIX B. ERROR PROPAGATION RULES

How to derive an error equation

Let’s use the change of variable method to determine the error equation for the following expression:

y =M

m

√

0.5 kx (1− sin θ) (B.1)

• Begin by rewriting Equation B.1 as a product of terms:

y = M ∗ m−1 ∗ [ 0.5 ∗ k ∗ x ∗ (1− sin θ)] 1/2 (B.2)

= M ∗ m−1 ∗ 0.51/2 ∗ k1/2 ∗ x1/2 ∗ (1− sin θ)1/2 (B.3)

• Assign to each term in Equation B.3 a new variable name A,B,C, . . . , then express v in terms ofthese new variables,

y = A ∗ B ∗ C ∗ D ∗ E ∗ F (B.4)

• With ∆(y) representing the error or uncertainty in the magnitude of y, the error expression for y iseasily obtained by applying Rule 4 to the product of terms Equation B.4:

∆(y)

y=

√

(

∆(A)

A

)2

+

(

∆(B)

B

)2

+

(

∆(C)

C

)2

+

(

∆(D)

D

)2

+

(

∆(E)

E

)2

+

(

∆(F )

F

)2

(B.5)

• Select from the table of error rules an appropriate error expression for each of these new variables asshown below. Note that F requires further simplification since there are two terms under the squareroot, so we equate these to a variable G:

A = M , ∆(A) = ∆(M)

B = m−1,∆(B)B = −1

∆(m)m = −∆(m)

m

C = 0.51/2,∆(C)C = 1

2∆(0.5)|0.5| = 0

D = k1/2,∆(D)D = 1

2∆(k)k

=∆(k)2k

E = x1/2,∆(E)E = 1

2∆(x)x =

∆(x)2x

F = G1/2,∆(F )F = 1

2∆(G)G =

∆(G)2G

G = 1− sin θ, ∆(G) =√

(∆(1))2 + (∆(sin θ))2 = cos θ∆θ

• Finally, replace the error terms into the original error Equation B.5, simplify and solve for ∆(y) bymultiplying both sides of the equation with y:

∆(y) = y

√

(

∆M

M

)2

+

(

∆m

m

)2

+

(

∆k

2k

)2

+

(

∆x

2x

)2

+

(

cos θ∆θ

2− 2 sin θ

)2

(B.6)

Appendix C

Graphing techniques

Figure C.1: Proper scaling of axes

Figure C.2: Improper scaling of axes

A mathematical function y = f(x) describes the one to onerelationship between the value of an independent variable xand a dependent variable y. During an experiment, we analysesome relationship between two quantities by performing a se-ries of measurements. To perform a measurement, we set somequantity x to a chosen value and measure the correspondingvalue of the quantity y. A measurement is thus representedby a coordinate pair of values (x, y) that defines a point on atwo dimensional grid.

The technique of graphing provides a very effective methodof visually displaying the relationship between two variables.By convention, the independent variable x is plotted along thehorizontal axis (x-axis) and the dependent variable y is plottedalong the vertical axis (y-axis) of the graph. The graph axesshould be scaled so that the coordinate points (x, y) are welldistributed across the graph, taking advantage of the maxi-mum display area available. This point is especially impor-tant when results are to be extracted directly form the datapresented in the graph. The graph axes do not have to startat zero.

Scale each axis with numbers that represent the range ofvalues being plotted. Label each axis with the name and unitof the variable being plotted. Include a title above the graph-ing area that clearly describes the contents of the graph beingplotted. Refer to Figure C.1 and Figure C.2.

The line of best fit

Suppose there is a linear relationship between x and y, so thaty = f(x) is the equation of a straight line y = mx + b where m is the slope of the line and b is the valueof y at x = 0. Having plotted the set of coordinate points (x, y) on the graph, we can now extract a valuefor m and b from the data presented in the graph.

Draw a line of ’best fit’ through the data points. This line should approximate as well as possible thetrend in your data. If there is a data point that does not fit in with the trend in the rest of the data, youshould ignore it.

53

54 APPENDIX C. GRAPHING TECHNIQUES

The slope of a straight line

Figure C.3: Slope of a line

The slope m of a straight-line graph is determined bychoosing two points, P1 = (x1, y1) and P2 = (x2, y2),on the line of best fit, not from the original data, andevaluating Equation C.1. Note that these two pointsshould be as far apart as possible.

m =rise

run

m =∆y

∆x=

y2 − y1x2 − x1

(C.1)

Error bars

Figure C.4: Error Bars for Point (x, y)

All experimental values are uncertain to some degree dueto the limited precision in the scales of the instrumentsused to set the value of x and to measure the result-ing value of y. This uncertainty σ of a measurement isgenerally determined from the physical characteristics ofthe measuring instrument, i.e. the graduations of a scale.When plotting a point (x, y) on a graph, these uncertain-ties σ(x) and σ(y) in the values of x and y are indicatedusing error bars.

For any experimental point (x± σ(x), y ± σ(y)), theerror bars will consist of a pair of line segments of length2σ(x) and 2σ(y), parallel to the x and y axes respec-tively and centered on the point (x, y). The true value lies within the rectangle formed by using the errorbars as sides. The rectangle is indicated by the dotted lines in Figure C.3. Note that only the error bars,and not the rectangle are drawn on the graph.

The uncertainty in the slope

Figure C.5: Determining slope error

Figure C.5 shows a set of data points for a linear rela-tionship. The slope is that of line 2, the line of best fitthrough these points. The uncertainty in this slope istaken to be one half the difference between the line ofmaximum slope line 1 and the line of miinimum slope,line 3:

σ(slope) =slopemax − slopemin

2(C.2)

The lines of maximum and minimum slope should gothrough the diagonally opposed vertices of the rectanglesdefined by the error bars of the two endpoints of thegraph, as in Figure C.5.

55

Logarithmic graphs

In science courses you will encounter a great number of functions and relationships, both linear and non-linear. Linear functions are distinguished by a proportional change in the value of the function with a changein value of one of the variables, and can be analyzed by plotting a graph of y versus x to obtain the slope mand vertical intercept b. Non-linear functions do not exhibit this behaviour, but can be analyzed in a similarmanner with some modification. For example, a commonly occuring function is the exponential function,

Figure C.6: The exponential function y = aebx.

y = aebx, (C.3)

where e = 2.71828 . . ., and a and b are constants.Plotted on linear (i.e. regular) graph paper, thefunction y = aebx appears as in Figure C.6. Tak-ing the natural logarithm of both sides of equa-tion (C.3) gives

ln y = ln(

aebx)

ln y = ln a+ ln(

ebx)

ln y = ln a+ bx ln e

ln y = ln a+ bx (since ln e = 1)

Equation (C.4) is the equation of a straight linefor a graph of ln y versus x, with ln a the verticalintercept, and b the slope. Plotting a graph of ln yversus x (semilogarithmic, i.e. logarithmic on the vertical axis only) should result in a straight line, whichcan be analyzed.

There are two ways to plot semilogarithmic data for analysis:

1. Calculate the natural logarithms of all the y values, and plot ln y versus x on linear scales. The slopeand vertical intercept can then be determined after plotting the line of best fit.

2. Use semilogarithmic graph paper. On this type of paper, the divisions on the horizontal axis areproportional to the number plotted (linear), and the divisions on the vertical axis are proportional tothe logarithm of the number plotted (logarithmic). This method is preferable since only the naturallogarithms of the vertical coordinates used to determine the slope of the lines best fit, minimum andmaximum slope need to be calculated.

Semilogarithmic graph paper

The horizontal axis is linear and the vertical axis is logarithmic. The vertical axis is divided into a seriesof bands called decades or cycles.

• Each decade spans one order of magnitude, and is labelled with numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 1.

• the second “1” represents 10× what the first “1” does, the third 10× the second, et cetera, and

• there is no zero on the logarithmic axis since the logarithm of zero does not exist.

A logarithmic axis often has more than one decade, each representing higher powers of 10. In Figure C.7,the axis has 3 decades representing three consecutive orders of magnitude. For instance, if the data to beplotted covered the range 1 → 1000, the lowest decade would represent 1 → 10 (divisions 1, 2, 3, . . . , 9),

56 APPENDIX C. GRAPHING TECHNIQUES

Figure C.7: A 3-Decade Logarithmic Scale.

the second decade 10 → 100 (divisions 10, 20, 30, . . . , 90) and the third decade 100 → 1000 (divisions 100,200, 300, . . . 900, 1000).

Another advantage of using a logarithmic scale is that it allows large ranges of data to be plotted. Forinstance, plotting 1 → 1000 on a linear scale would result in the data in the lower range (e.g. 1 → 100)being compressed into a very small space, possibly to the point of being unreadable. On a logarithmicscale this does not occur.

Calculating the slope on semilogarithmic paper

The slope of a semilogarithmic graph is calculated in the usual manner:

m = slope

=rise

run

=∆(vertical)

∆(horizontal).

For ∆(vertical) it is necessary to calculate the change in the logarithm of the coordinates, not the changein the coordinates themselves. Using points (x1, y1) and (x2, y2) from a line on a semilogarithmic graph ofy versus x and Equation C.4, the slope of the line is obtained.

m =rise

run

m =ln y2 − ln y1x2 − x1

m =ln (y2/y1)

x2 − x1(C.4)

Note that the units for m will be (units of x)−1 since ln y results in a pure number.

Analytical determination of slope

There are analytical methods of determining the slope m and intercept b of a straight line. The advantageof using an analytical method is that the analysis of the same data by anyone using the same analyticalmethod will always yield the same results. Linear Regression determines the equation of a line of best fitby minimizing the total distance between the data points and the line of best fit.

To perform “Linear Regression” (LR), one can use the preprogrammed function of a scientific calculatoror program a simple routine using a spreadsheet program. Based on the x and y coordinates given to it, aLR routine will return the slope m and vertical intercept b of the line of best fit as well as the uncertaintiesσ(m) and σ(b) in these values. Be aware that performing a LR analysis on non-linear data will producemeaningless results. You should first plot the data points and determine visually if a LR analysis is indeedvalid.

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