Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal

Revision sheet: lab measurements, uncertainties and graphical analysis (203-SN3-RE)

The waves and modern physics lab report is marked as much on the uncertainty as on the value, and the traps are not those of mechanics. Here the instrument reads angles on a vernier, the quantity sought is a millionth of a millimetre, and half of the measurements concern a phenomenon that fluctuates all by itself.

This sheet gives what is really lost: the angle unit forgotten in a propagation, the uncertainty of a count assumed to be zero because the counter shows a whole number, the consistency conclusion written the wrong way round. The corrected exercises of the same chapter practise each of these moves on a real setup.

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The thread of the chapter

In this course, the interesting quantities are either tiny, a wavelength or a frequency shift, or random by nature, a count of decays. The uncertainty is therefore no longer read off the instrument, it is PREDICTED, by the square root of the number of counts or by the derivative of a sine. And you never get a small quantity by subtracting two large ones: you measure it directly, or you count long enough for it to emerge from the noise.

This chapter is part of 203-SN3-RE, Waves and Modern Physics

The essentials

Reading a vernier: the line that lines up

  • • Vernier caliper graduated to fiftieths: 5050 vernier divisions span 4949 mm, each division is worth 0.980.98 mm, that is 0.020.02 mm less than a millimetre.
  • • The vernier zero gives the whole millimetres; the vernier line that LINES UP with a line of the main scale gives the fiftieths: seventh line, 7×0.02=0.147\times 0.02=0.14 mm.
  • • The reading is clean: the uncertainty is the smallest readable division, 0.020.02 mm, without taking half as on a graduated ruler.
101520250123main scale (mm)vernier12 mm + 7 × 0.02 mm = 12.14 mmuncertainty: 0.02 mm, not half of it
The vernier zero falls after 12 mm; its seventh line lines up with a line of the main scale: 12.14 mm, read to within 0.02 mm.

Propagating, linearizing, and the angle unit that decides everything

  • • Vernier: the smallest readable division is the GAP between one vernier division and one division of the main scale, 0.020.02 mm for fiftieths, the arcminute on a goniometer with thirty divisions. You do NOT take half, the reading is clean.
  • • Sum or difference: ABSOLUTE uncertainties. Product, quotient, power: relative, each multiplied by its EXPONENT. A square doubles, a square root halves.
  • • Differential method: Δz≈∣∂z∂x∣Δx+…\Delta z\approx\left|\dfrac{\partial z}{\partial x}\right|\Delta x+\ldots, and any ANGULAR uncertainty is converted first to RADIANS, otherwise the result is multiplied by 57.357.3.
  • • Linearizing means turning the law into a line: ln⁡u\ln u linear in tt for an exponential decay, II linear in cos⁡2θ\cos^{2}\theta for Malus's law, VsV_{s} linear in 1λ\dfrac{1}{\lambda} for a diode, LL linear in (2n−1)(2n-1) for a sound tube.
  • • The INTERCEPT is not an ornament: zero when the law passes through the origin, otherwise it is a physical result, the end correction of a tube or the stray light of a detector. A curvature on a semi-logarithmic plot betrays a forgotten additive term.
0.511.522.530.511.522.533.51/λ (µm⁻¹)Vs (V)slope = hc/e
Threshold voltage of five diodes against the inverse of the wavelength: the slope gives hce\dfrac{hc}{e}, hence Planck's constant.

The uncertainty of a SLOPE does not come out of the calculator: you draw the two extreme lines that still cross all the uncertainty bars, and take half the spread of their slopes.

Counting, subtracting a background, and knowing what you really showed

  • • The uncertainty of a count of NN events is N\sqrt{N}, so the relative uncertainty is 1N\dfrac{1}{\sqrt{N}}. It is PREDICTED without repeating, because the fluctuation comes from the phenomenon and not from the instrument.
  • • What fixes the precision is the TOTAL NUMBER of counts accumulated, not the way they are split into sessions. Halving the relative uncertainty takes four times as many counts, hence four times as much time.
  • • The background is a SYSTEMATIC error by excess: it is SUBTRACTED, not averaged. And the subtraction is done BEFORE the logarithm, since the logarithm is not linear.
  • • Grating: dsin⁡θ=mλd\sin\theta=m\lambda, and resolving power R=λΔλmin=mNR=\dfrac{\lambda}{\Delta\lambda_{min}}=mN with NN the number of ILLUMINATED lines, not engraved ones.
  • • SEPARATING two lines and LOCATING a line are two different qualities: resolution is gained by widening the beam or going up in order, accuracy is bought by calibrating.
1234567891011123035404550556065mean 47+ √N− √Ncount number
Eight ten-second counts on the same source: their standard deviation, 6.16.1 counts, falls on the square root of the mean, 6.96.9 counts, as the statistics predict.

Consistency: two values are consistent when their intervals overlap, and a gap is expressed as a NUMBER OF UNCERTAINTIES. A gap of 0.50.5 nm for an uncertainty of 1.31.3 nm is an excellent agreement, not a failure.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Leaving an angular uncertainty in degrees in a propagation

3 points, and an uncertainty fifty-seven times too large

What not to write

“∣∂λ∂θ∣Δθ=6.30×10−7×0.0333=21\left|\dfrac{\partial\lambda}{\partial\theta}\right|\Delta\theta=6.30\times 10^{-7}\times 0.0333=21 nm.”

What to write

“Δθ=0.0333∘=5.82×10−4\Delta\theta=0.0333^{\circ}=5.82\times 10^{-4} rad, so the contribution is 0.370.37 nm.”

1 rad = 57.3°1° = 0.01745 rad
The derivative of the sine assumes radians, and one radian is 57.3°: leaving 0.0333 in degrees in the derivative amounts to counting 0.0333 rad, that is 1.9°, 57 times too much.

Why: The derivative of the sine equals the cosine only in RADIANS. It is the costliest error of the chapter because it leaves a plausible result: 2121 nm on a wavelength of 546546 nm shocks nobody, whereas the measurement is in fact fifty times better. The check: any angular uncertainty is converted BEFORE it enters a derivative.

2. Believing a whole-number count is an exact measurement

2 points, and the whole rest of the calculation is left without uncertainty

What not to write

“The counter shows 4747 counts, a whole number: this measurement has no uncertainty.”

What to write

“ΔN=47=6.9\Delta N=\sqrt{47}=6.9 counts, that is 15 %15\,\%: the counter is exact, the quantity is not.”

Why: Decay is a RANDOM phenomenon: a second count of the same duration will give another integer. It is the only situation in the chapter where the uncertainty is predicted without repeating, and that is an advantage, not an exemption: you still have to write it down.

3. Averaging the background instead of subtracting it

3 points, and a half-life overestimated by several percent

What not to write

“If we repeat the background count long enough, it cancels in the average.”

What to write

“The background is measured with the source removed, then SUBTRACTED from each count: 820−20=800820-20=800 counts.”

1503004506007509002.533.544.555.566.57background forgotten: the curve bends upbackground subtracted: a straight lineln 20t (s)ln N
With a background of 20 counts, the logarithm of the total counts bends toward ln 20; once the background is subtracted, ln N is a straight line again, with slope − λ.

Why: Cosmic rays and ambient radioactivity are always present and always in the same direction: it is a systematic error. Repeating only improves the knowledge of the background. Forgotten on a decay, it makes the source look slower, 2.672.67 min instead of 2.552.55 min, and it bends the semi-logarithmic plot upward.

4. Taking the logarithm before subtracting the background

2 points, and a slope with no connection to reality

What not to write

“ln⁡(Ntot)−ln⁡(Nbg)\ln(N_{tot})-\ln(N_{bg}) gives the logarithm of the net count.”

What to write

“You first calculate Ntot−NbgN_{tot}-N_{bg}, THEN its logarithm.”

Why: ln⁡(a−b)\ln(a-b) has nothing to do with ln⁡a−ln⁡b\ln a-\ln b, which equals ln⁡ab\ln\dfrac{a}{b}. More deeply, the exponential law describes the SOURCE alone, and the background is an additive term laid on top: you have to remove it to recover the quantity the law describes.

5. Confusing resolving power and accuracy

2 points, often on the conclusion identifying a line

What not to write

“The grating separates 0.0910.091 nm, so it measures the wavelength to within 0.0910.091 nm.”

What to write

“It SEPARATES to 0.0910.091 nm and LOCATES to 1.31.3 nm: calibration is the limit, not the optics.”

5865875885895905915925930.20.40.60.811.21.41.6position: ± 1.3 nmtwo sharp lines, 0.6 nmλ (nm)intensity
The sodium doublet, 589.0 and 589.6 nm, appears as two separate lines, although each is located only to within ± 1.3 nm: separating and locating are two different things.

Why: Resolution is a discriminating power, R=mNR=mN, which tells you whether two neighbouring lines appear separated. Accuracy tells you where they are on the scale, and it depends on the grating spacing, hence on the calibration. Here you separate fourteen times more finely than you locate: this is the normal case for a spectrometer, and confusing the two makes you buy the wrong instrument.

6. Taking both upper bounds in the extreme-values method

2 points: the value obtained is neither the maximum nor the minimum

What not to write

“nmax=sin⁡60.5∘sin⁡35.5∘n_{max}=\dfrac{\sin 60.5^{\circ}}{\sin 35.5^{\circ}}, with both angles at their maximum.”

What to write

“sin⁡r\sin r is in the DENOMINATOR and the sine increases: nmax=sin⁡60.5∘sin⁡34.5∘n_{max}=\dfrac{\sin 60.5^{\circ}}{\sin 34.5^{\circ}}.”

n = 1.521n = 1.484 minn = 1.537 maxn = 1.49959.560.5i (°)34.535.5r (°)
n increases with i and decreases with r, since sin r is in the denominator: the maximum is at the corner (i max, r min). Both upper bounds give 1.499, in the middle.

Why: Each bound is chosen according to the DIRECTION OF VARIATION of the result, and here that direction goes through the monotonicity of the sine on [0,90∘][0, 90^{\circ}]. Check: the central value must fall between zminz_{min} and zmaxz_{max}, roughly in the middle.

7. Declaring an experiment failed on a gap smaller than the uncertainty

3 points, and a wrong conclusion about the whole experiment

What not to write

“Measured 545.6545.6 nm, table 546.1546.1 nm: the experiment failed.”

What to write

“546.1546.1 lies in [544.3,546.9][544.3, 546.9]: consistent, the gap is 0.40.4 uncertainty.”

543544545546547548λ (nm)measurementtable: 546.1 nm−3−2−10123gap counted in uncertainties0.4 uncertainty
546.1 nm falls inside the interval of the measurement: the gap is 0.4 uncertainty, well inside the green zone of one uncertainty, an excellent agreement.

Why: Two different numbers can designate the same quantity: that is precisely what the uncertainty allows. You compare the INTERVALS, and you express the gap as a number of uncertainties. Statistically, a gap of the order of one uncertainty is expected; finding three times less is even a little lucky.

8. Claiming a weak source is active when its signal drowns

3 points, and it is the reasoning that manufactures false discoveries

What not to write

“520520 counts against 480480 of background: the sample does emit 4040 more counts.”

What to write

“Net =(40±45)=(40\pm 45) counts, consistent with zero: this measurement shows nothing, it is too short.”

−1000100200300400500600net (counts)3,000 s: (400 ± 141) counts300 s: (40 ± 45) countszero net
Over 300 s, the interval of the net contains zero: nothing is shown. Over 3,000 s, the net is ten times larger, its uncertainty only √10 times.

Why: A difference adds the uncertainties, here 520+480=45\sqrt{520}+\sqrt{480}=45 counts, and the net is 4040: the interval contains zero. The signal-to-noise ratio grows as the SQUARE ROOT of the time, so gaining a factor of ten costs a hundred times more time. Counting 30003000 s instead of 300300 s, the net becomes (400±141)(400\pm 141) counts and a conclusion becomes possible.

Which method to choose

Which propagation rule, according to the form of the relation

The form of the formula that links the result to the measured quantities

x ± Δxy ± Δy± Δx± (Δx + Δy)sum or difference: x ± yx · yxyx · Δyy · Δxproduct: x · y
Difference of two angle readings or of two counts: the ABSOLUTE uncertainties add. Product, like d·sin θ: the two strips, so the RELATIVE uncertainties add.
  • If a sum or a difference → add the ABSOLUTE uncertainties

    Example: 221∘00′−139∘12′=81∘48′221^{\circ}00'-139^{\circ}12'=81^{\circ}48' with Δ=4′\Delta=4'

  • If a product, a quotient or a power → add the relative uncertainties, each multiplied by its EXPONENT

    Example: v=Tμv=\sqrt{\dfrac{T}{\mu}}: the root halves each relative uncertainty

  • If a count of decays → the uncertainty is N\sqrt{N}, with no repetition at all

    Example: N=376N=376 counts gives ΔN=19.4\Delta N=19.4, that is 5.2 %5.2\,\%

  • If a trigonometric function of a measured angle → DIFFERENTIAL method, uncertainty converted to radians

    Example: dcos⁡θm×5.82×10−4=0.37\dfrac{d\cos\theta}{m}\times 5.82\times 10^{-4}=0.37 nm

  • If a formula that is monotonic in each variable, but complicated → EXTREME-VALUES method, each bound chosen according to the direction of variation

    Example: Δn=1.537−1.4842=0.026\Delta n=\dfrac{1.537-1.484}{2}=0.026

  • If the law is not linear → LINEARIZE, then regress and read the slope AND the intercept

    Example: L=(2n−1)λ4−eL=(2n-1)\dfrac{\lambda}{4}-e: the slope gives λ\lambda, the intercept gives ee

A mixed formula is handled in two steps. For a net count divided by a duration, you first apply the ABSOLUTE rule to the difference of the two counts, then the RELATIVE rule to the quotient. Applying a single rule to the whole formula is the mistake that comes back most often.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Presenting a result from a waves or modern physics lab

When to use it: The report asks for a quantity measured with a spectrometer, a counter or an optics setup, and its comparison with a tabulated value.

  1. 1 Give each measurement with its uncertainty AND say where it comes from: vernier, grating ruling, counting statistics, observed scatter.
  2. 2 Write the relation, name the propagation rule chosen, convert the angles to radians, and detail the contributions one by one.
  3. 3 Round the uncertainty to one significant figure, two if it starts with 11, then align the value on the same decimal place.
  4. 4 Conclude on CONSISTENCY by comparing the intervals, express the gap as a number of uncertainties, and say which quantity dominates the budget.

Concluding sentence

“λ=(545.6±1.3)\lambda=(545.6\pm 1.3) nm. The green mercury line, 546.1546.1 nm, lies in [544.3,546.9][544.3, 546.9] nm: the measurement is consistent, the gap being 0.40.4 uncertainty. The grating spacing supplies 71 %71\,\% of the uncertainty, it is what should be calibrated.”

The trap: Concluding “the result is good, the gap is only 0.1 %0.1\,\%” without mentioning the intervals: the marker expects a criterion, not an impression. And for a count, forgetting to write the uncertainty on the grounds that the display shows a whole number costs just as much.

Marking: Typically 1 point for the uncertainties of the measurements, 2 for the propagation, 1 for the rounded writing, 1 for the justified conclusion.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The typical problem, taken apart

A wavelength from a grating, from the goniometer reading to the identification of the line

On the spectrometer, a line appears on both sides of the direct beam. Readings: θL=(139∘12′±2′)\theta_{L}=(139^{\circ}12'\pm 2') and θR=(221∘00′±2′)\theta_{R}=(221^{\circ}00'\pm 2').

The grating has 600600 lines per millimetre, known to within one line, and the line is observed at order m=2m=2. Formula: dsin⁡θ=mλd\sin\theta=m\lambda.

collimatorgratingdirect beamθR = 221°00′θL = 139°12′θθ
The line is read on both sides of the direct beam: half the difference of the two readings gives the deflection angle and cancels the zero offset of the graduated circle.

Step 1

Angle: θ=221∘00′−139∘12′2=81∘48′2=40.90∘\theta=\dfrac{221^{\circ}00'-139^{\circ}12'}{2}=\dfrac{81^{\circ}48'}{2}=40.90^{\circ}.

Why

Measuring on both sides eliminates the zero of the graduated circle: an offset is added to both readings and disappears in the difference. Measuring on one side only with a zero offset by 15′15' would have given 0.35∘0.35^{\circ} too much.

Step 2

Uncertainty: Δ(2θ)=2′+2′=4′\Delta(2\theta)=2'+2'=4', so Δθ=2′=0.0333∘=5.82×10−4\Delta\theta=2'=0.0333^{\circ}=5.82\times 10^{-4} rad.

Why

The difference adds the absolute uncertainties, but the division by two applies to the quantity as well as to its uncertainty. The conversion to radians is done now, not later.

Step 3

Wavelength: λ=1.6667×10−6×sin⁡40.90∘2=545.6\lambda=\dfrac{1.6667\times 10^{-6}\times\sin 40.90^{\circ}}{2}=545.6 nm.

Why

The grating spacing is the inverse of the number of lines per unit length. Order 22 doubles the angle, which makes the measurement more precise than order 11.

Step 4

Contribution of the angle: dcos⁡θmΔθ=6.30×10−7×5.82×10−4=0.37\dfrac{d\cos\theta}{m}\Delta\theta=6.30\times 10^{-7}\times 5.82\times 10^{-4}=0.37 nm.

Why

Differential method, because the angle enters through a sine. Keeping degrees would have given 2121 nm, fifty-seven times too much, without the result looking absurd.

Step 5

Contribution of the grating: 1600=0.167 %\dfrac{1}{600}=0.167\,\%, that is 0.910.91 nm. Total Δλ=1.3\Delta\lambda=1.3 nm.

Why

The spacing enters to the power 11, so its relative uncertainty passes through unchanged. It now dominates: reading the angles more finely would hardly help.

The conclusion, written out

“λ=(545.6±1.3)\lambda=(545.6\pm 1.3) nm. The green mercury line, 546.1546.1 nm, lies in [544.3,546.9][544.3, 546.9] nm: the measurement is consistent, the gap being 0.40.4 uncertainty. The grating spacing supplies 71 %71\,\% of the uncertainty budget.”

The classic mistake on this problem: Keeping Δθ\Delta\theta in degrees in front of the derivative, or measuring on one side of the direct beam only while assuming the zero of the circle is right.

Learn by heart

  • • Vernier: the smallest readable division is the gap, not half of it.
  • • Sum or difference: absolute uncertainties added.
  • • Product, quotient, power: relative uncertainties added, each multiplied by its exponent.
  • • A square doubles the relative uncertainty, a square root halves it.
  • • Any angular uncertainty is converted to radians before a derivative.
  • • Count of N events: uncertainty root N, predicted without repeating.
  • • Background: subtracted, not averaged, and before the logarithm.
  • • Resolving power R=mNR=mN, with N the number of ILLUMINATED lines.
  • • Separating two lines and locating a line are two different qualities.
  • • Consistency: overlapping intervals, gap expressed as a number of uncertainties.

Frequently asked questions

Why must an angle uncertainty be converted to radians?

Because the derivative of the sine equals the cosine only if the angle is measured in radians. As soon as an angular uncertainty enters a propagation through a derivative, it must therefore be converted first, by multiplying by pi and dividing by one hundred and eighty. Keeping degrees multiplies the contribution by about 57.3. It is the costliest error of the chapter, because it leaves a perfectly plausible result: an uncertainty of twenty nanometres on a wavelength of five hundred and forty-six nanometres shocks nobody, whereas the measurement is in fact fifty times better.

Why is the uncertainty of a count equal to the square root of the number of counts?

Because radioactive decay is a random phenomenon: each nucleus decays independently of the others, with a small probability, and the statistics of such a process give a scatter equal to the square root of the number of events. This has a practical consequence unique in the whole chapter: you know the uncertainty without having to repeat the measurement. That said, what matters is the total number of counts accumulated, not the way they are split into sessions, and halving the relative uncertainty takes four times as many counts.

Should the background be subtracted or averaged?

Subtracted, always. The background comes from cosmic rays and ambient radioactivity: it is present all the time and always in the same direction, so it is a systematic error, not a fluctuation. Averaging it only measures it more precisely. And the subtraction is done before taking a logarithm, because the logarithm of a difference has nothing to do with the difference of the logarithms. Forgotten on a decay, a constant background makes the source look slower and bends the semi-logarithmic plot upward.

What is the difference between resolving power and accuracy?

Resolution tells you how finely the instrument separates two neighbouring lines, accuracy tells you how precisely it places a line on the wavelength scale. A school spectrometer commonly separates a tenth of a nanometre but locates only to within a nanometre, because the grating spacing is not known exactly. To improve the resolution, you widen the beam or go up in order; to improve the accuracy, you calibrate on a known line. Confusing the two leads to buying the wrong instrument.

How do you know whether a weak source is really active?

You compare the net count, that is the count with the source minus the background, with its uncertainty, which is the sum of the square roots of the two counts. If the net is smaller than its uncertainty, the interval contains zero and the measurement shows nothing: it does not prove that the sample is inert either, it is simply too short. The ratio of the net to its uncertainty grows as the square root of the time, so counting ten times longer multiplies it by about three. That is also why you prefer to reduce the background, with a lead shield, rather than count longer.

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