Corrected exercises: lab measurements, uncertainties and graphical analysis (203-SN3-RE)

A vernier caliper graduated to fiftieths has fifty vernier divisions spanning forty-nine millimetres of the main scale. Each vernier division is therefore worth $\dfrac{49}{50}=0.98$ mm, which is $0.02$ mm less than a millimetre: the offset builds up from one division to the next, and the line that lines up tells you the number of hundredths. The reading uncertainty is the smallest readable division, $0.02$ mm. Unlike a graduated ruler, you do NOT take half: a vernier gives a clean reading, not an interpolation by eye.

Thirty divisions span twenty-nine half-degrees, that is $29\times 30'=870'$. Each vernier division is worth $\dfrac{870}{30}=29'$, against $30'$ for a division of the circle: the difference is $1'$ per division. The smallest readable division is therefore ONE ARCMINUTE, that is $\dfrac{1}{60}$ of a degree. A goniometer in an optics lab thus reads thirty times more finely than its circle is engraved.

$54'=\dfrac{54}{60}=0.90^{\circ}$, so $40^{\circ}54'=40.90^{\circ}$. For the uncertainty: $2'=\dfrac{2}{60}=0.03333^{\circ}$, and in radians $0.03333\times\dfrac{\pi}{180}=5.82\times 10^{-4}$ rad. The conversion to radians is not cosmetic: any propagation through the derivative of a sine or a cosine requires it, and forgetting it multiplies the result by fifty-seven.

The uncertainty is rounded to TWO significant figures when it starts with a $1$, so $\Delta\lambda=1.3$ nm. The value is aligned on the same decimal place, the tenth of a nanometre: $\lambda=(545.6\pm 1.3)$ nm. The four decimals from the calculator are calculation noise; writing them would suggest a measurement ten thousand times better than it is.

Because the eye cannot separate two lines closer than about a tenth of a millimetre, and because an engraving fifty times finer would be both impossible to make cleanly and unreadable without a microscope. The vernier gets around the problem: it engraves nothing finer, it converts a small DISTANCE into a COINCIDENCE, and the eye is excellent at judging whether two lines are aligned. It is the same principle as the Wheatstone bridge in electricity or the zero of a beam balance: a difficult absolute reading is replaced by an easy comparison.

$221^{\circ}00'-139^{\circ}12'$. The arcminutes cannot be subtracted directly, since $00'-12'$ is negative: you BORROW $60'$ from the degrees, exactly as you borrow $60$ seconds from a minute. So you calculate $220^{\circ}60'-139^{\circ}12'$, which gives $48'$ and $220-139=81$ degrees: the difference is $81^{\circ}48'$.

The difference of the two readings is $2\theta$, since the line is deflected by $\theta$ on one side and by $\theta$ on the other. So $\theta=\dfrac{81^{\circ}48'}{2}=40^{\circ}54'=40.90^{\circ}$.

The two readings are independent and each carries $2'$: a difference adds the absolute uncertainties, $\Delta(2\theta)=2'+2'=4'$. Dividing by two, $\Delta\theta=\dfrac{4'}{2}=2'$, that is $0.0333^{\circ}$, or $5.82\times 10^{-4}$ rad. Result: $\theta=(40.900\pm 0.033)^{\circ}$, a relative uncertainty of only $0.08\,\%$.

With the zero offset by $+15'$, the two readings become $139^{\circ}27'$ and $221^{\circ}15'$. Their difference: $221^{\circ}15'-139^{\circ}27'=81^{\circ}48'$, unchanged, so $\theta=40.90^{\circ}$ again. The offset is added to BOTH readings and disappears in the subtraction. That is the whole point of the method: a SYSTEMATIC setting error, which no amount of repetition would have corrected, is eliminated by the geometry of the measurement itself. Measuring on one side only, assuming the zero is right, would have given $41^{\circ}15'$, that is $0.35^{\circ}$ too much, ten times the stated uncertainty.

Because the division by two applies to EVERYTHING, to the quantity as well as to its uncertainty. The difference did add the uncertainties, $4'$ instead of $2'$, but it also DOUBLED the measured quantity, which is $2\theta$ and not $\theta$: the relative uncertainty has not moved. The classic trap of the chapter, a difference of two nearby quantities, does not apply here: the two readings are very FAR apart, $82$ degrees, so the subtraction destroys nothing. Remember the exact wording: a difference is dangerous when its result is SMALL compared with the terms, not simply because it is a difference.

$\mu=\dfrac{m}{L}=\dfrac{2.50\times 10^{-3}}{1.200}=2.083\times 10^{-3}$ kg/m. This is a quotient, so the relative uncertainties add: $\dfrac{0.01}{2.50}=0.40\,\%$ for the mass and $\dfrac{0.002}{1.200}=0.17\,\%$ for the length, that is $0.57\,\%$ in all.

$v=\sqrt{\dfrac{12.00}{2.083\times 10^{-3}}}=\sqrt{5760}=75.89$ m/s. Order-of-magnitude check: a sonometer string stretched by about ten newtons carries waves at a few tens of metres per second, more slowly than sound in air, $343$ m/s; the two speeds are unrelated anyway, since the speed on the string depends only on its tension and its linear mass density. The figure is plausible.

The exponent of $T$ is $+\dfrac{1}{2}$ and that of $\mu$ is $-\dfrac{1}{2}$, so each relative uncertainty is HALVED: $\dfrac{\Delta v}{v}=\dfrac{1}{2}\dfrac{\Delta T}{T}+\dfrac{1}{2}\dfrac{\Delta\mu}{\mu}$. In detail: the tension contributes $\dfrac{1}{2}\times\dfrac{0.05}{12.00}=0.21\,\%$, the mass $\dfrac{1}{2}\times 0.40=0.20\,\%$ and the length $\dfrac{1}{2}\times 0.17=0.08\,\%$. Total: $0.49\,\%$. None of the three dominates, and that is precisely the effect of the root: it has flattened the differences between the contributions.

$\Delta v=75.89\times 0.0049=0.37$ m/s, rounded to $0.4$ m/s. Result: $v=(75.9\pm 0.4)$ m/s.

Without the root, the three relative uncertainties would enter in full: $0.42+0.40+0.17=0.98\,\%$, exactly DOUBLE. What to take from this fits in one sentence: an exponent is an amplifier of uncertainty, in both directions. A square doubles, a cube triples, a square root halves, a cube root divides by three. That is why a quantity obtained through a root is, for equal measurements, more reliable than one obtained through a product; and it is why a squared diameter, in a wire cross-section or a beam area, is almost always the dominant term in an uncertainty budget.

$n=\dfrac{\sin 60.0^{\circ}}{\sin 35.0^{\circ}}=\dfrac{0.8660}{0.5736}=1.510$. Check: the ray bends toward the normal on entering, so the second medium is the more refractive one and $n>1$. If the calculation had given $0.66$, the quotient would have been inverted.

The sine is INCREASING on $[0, 90^{\circ}]$, so $\sin r$ grows with $r$. Since $\sin r$ is in the DENOMINATOR, the index is largest when $\sin r$ is smallest, that is for $r$ at its MINIMUM, $34.5^{\circ}$, and simultaneously for $i$ at its maximum, $60.5^{\circ}$. The argument rests on the monotonicity of the sine: over a range containing $90^{\circ}$ it would no longer hold.

$n_{max}=\dfrac{\sin 60.5^{\circ}}{\sin 34.5^{\circ}}=\dfrac{0.8704}{0.5664}=1.537$ and $n_{min}=\dfrac{\sin 59.5^{\circ}}{\sin 35.5^{\circ}}=\dfrac{0.8616}{0.5807}=1.484$. The central value $1.510$ does lie between the two.

$\Delta n=\dfrac{1.537-1.484}{2}=\dfrac{0.053}{2}=0.026$, rounded to $0.03$. Result: $n=(1.51\pm 0.03)$, a relative uncertainty of $1.8\,\%$. The refractive index is a number with NO UNIT: write nothing after it, neither a degree sign nor an invented unit.

No. The interval $[1.48, 1.54]$ contains both $1.52$ and $1.49$: both materials are CONSISTENT with the measurement, so the experiment does not decide. That is the right conclusion to write, and not “it is crown glass since $1.51$ is closer to $1.52$”: comparing central values when the intervals overlap has no value as proof. To tell them apart you would have to bring the uncertainty below $0.015$, therefore divide $\Delta i$ and $\Delta r$ by two at least, that is read the angles to a quarter of a degree or better, which is precisely what a vernier goniometer does.

$d=\dfrac{1}{600}$ mm $=1.667\times 10^{-3}$ mm $=1.667$ µm $=1.6667\times 10^{-6}$ m. Then $\lambda=\dfrac{1.6667\times 10^{-6}\times\sin 40.90^{\circ}}{2}=\dfrac{1.6667\times 10^{-6}\times 0.6547}{2}=5.456\times 10^{-7}$ m, that is $545.6$ nm. The order-of-magnitude check is immediate: a visible wavelength lies between $400$ and $700$ nm, and $545$ nm is indeed in the green.

$\dfrac{\partial\lambda}{\partial\theta}=\dfrac{d\cos\theta}{m}=\dfrac{1.6667\times 10^{-6}\times 0.7559}{2}=6.299\times 10^{-7}$ m/rad. The uncertainty MUST be in radians: $\Delta\theta=0.03333^{\circ}=5.818\times 10^{-4}$ rad. Contribution: $6.299\times 10^{-7}\times 5.818\times 10^{-4}=3.66\times 10^{-10}$ m, that is $0.37$ nm. Keeping degrees would have given $21$ nm, fifty-seven times too much: this is the costliest error of the whole chapter, because the result stays plausible.

The spacing $d$ enters to the power $1$, so its relative contribution passes through unchanged: $\dfrac{\Delta d}{d}=\dfrac{1}{600}=0.167\,\%$, since an uncertainty of one line out of six hundred carries over identically to their spacing. Contribution: $545.6\times 0.00167=0.91$ nm.

$\Delta\lambda=0.37+0.91=1.28$ nm, rounded to $1.3$ nm. Result: $\lambda=(545.6\pm 1.3)$ nm, interval $[544.3, 546.9]$ nm. The green mercury line, $546.1$ nm, lies inside it: the measurement is CONSISTENT, and the experiment identifies the line correctly.

It is now the GRATING that limits: its contribution, $0.91$ nm, is two and a half times that of the angle. Reading the angles even more finely would hardly help; you would need a grating whose spacing is better known, or to calibrate it on a line of known wavelength, which is standard practice. As for the order: going from $m=1$ to $m=2$ doubles $\sin\theta$, so it takes the angle from $19^{\circ}$ to $41^{\circ}$, where the cosine is smaller. Now $\dfrac{\partial\lambda}{\partial\theta}$ is proportional to $\dfrac{\cos\theta}{m}$: the two effects work in the same direction and divide the contribution of the angle by about $2.5$. Higher orders are therefore more precise, until $\sin\theta$ exceeds $1$ and the line disappears.

The sum is $376$ counts, so $\bar N=\dfrac{376}{8}=47.0$ counts per ten seconds.

The deviations from the mean are $0$; $+5$; $-8$; $-2$; $+11$; $-4$; $+3$; $-5$, which sum to zero. The sum of squares is $0+25+64+4+121+16+9+25=264$, so $s=\sqrt{\dfrac{264}{7}}=\sqrt{37.71}=6.14$ counts.

$\sqrt{\bar N}=\sqrt{47.0}=6.86$ counts. The measured standard deviation, $6.14$, is within $10\,\%$ of it, which is all you can hope for with only eight values: the uncertainty on a standard deviation estimated from $N$ values is of the order of $\dfrac{1}{\sqrt{2(N-1)}}$, that is $27\,\%$ here. The conclusion is a strong one: the observed scatter is the one predicted by the root-$N$ law, SO YOU DO NOT NEED TO REPEAT to know the uncertainty of a count. This is a unique situation in the whole chapter: elsewhere, only repetition reveals the scatter.

$\dfrac{s}{\sqrt{8}}=\dfrac{6.14}{2.828}=2.17$ counts, that is $\dfrac{2.17}{47.0}=4.6\,\%$. You would write $\bar N=(47.0\pm 2.2)$ counts per ten seconds.

A single $80$ s count accumulates the same decays: about $376$ counts, whose uncertainty is $\sqrt{376}=19.4$ counts, that is $\dfrac{19.4}{376}=5.2\,\%$. This is PRACTICALLY THE SAME result as the eight separate counts, and it is no accident: what fixes the precision of a count is the TOTAL NUMBER OF COUNTS ACCUMULATED, not the way they are split into sessions. The only thing the eight counts bring in addition is the check in c), that is, the proof that the source is stable and the counter does not drift. Practical rule: to halve the relative uncertainty, you need FOUR times as many counts, hence four times as much time.

You plot $(2n-1)$ on the horizontal axis, that is $1$, $3$, $5$, $7$, and the length $L$ on the vertical axis. The relation $L=(2n-1)\dfrac{\lambda}{4}-e$ is then of the form $Y=aX+b$ with $a=\dfrac{\lambda}{4}$ and $b=-e$. Plotting simply $n$ would also work, but the slope would be $\dfrac{\lambda}{2}$ and the intercept $-e-\dfrac{\lambda}{4}$: the two unknowns would be mixed, whereas the choice of $(2n-1)$ separates them cleanly.

With $\bar X=4$ and $\bar L=0.6575$ m, we form $\sum(X_{i}-\bar X)(L_{i}-\bar L)=3.35$ and $\sum(X_{i}-\bar X)^{2}=20$, so $a=\dfrac{3.35}{20}=0.1675$ m. Hence $\lambda=4a=0.670$ m. Immediate check: the gaps between successive resonances are all $0.335$ m, that is HALF a wavelength, which gives $\lambda=0.670$ m again.

$v=\lambda f=0.670\times 512=343.0$ m/s. This is exactly the value expected in a room at ambient temperature.

$b=\bar L-a\bar X=0.6575-0.1675\times 4=0.6575-0.6700=-0.0125$ m. Since $b=-e$, the end correction is $e=0.0125$ m $=1.25$ cm. Theory gives $e\approx 0.6\,d=0.6\times 2.0=1.2$ cm: the agreement is excellent. The intercept is therefore NOT here a mere check that you hope will be zero, it is a PHYSICAL RESULT, and this is the only way to obtain it. A student who had forced the line through the origin would have distorted the slope, hence the speed of sound, while thinking they were doing the right thing.

$\theta=\dfrac{343.0-331}{0.6}=20.0$ °C, the temperature of a lab room. This check does not measure the temperature, it VALIDATES the measurement chain: it verifies that the frequency of the tuning fork, the reading of the lengths and the graphical processing are consistent with one another. Had it given $-40$ °C or $95$ °C, you would have had to look for the error, for example a misidentified tuning fork or a skipped resonance. A good lab report always ends with a check of this kind, which confronts the result with something known independently.

FALSE. The counter counts exactly, but what it counts FLUCTUATES: decay is random, and a second count of the same duration will give a different integer. The uncertainty is $\sqrt{N}=\sqrt{47}=6.9$ counts, that is $15\,\%$. Correct version: “the count is exact, the quantity is not; the uncertainty of a count is the square root of the number of counts, and it is predicted without repeating”.

FALSE, and this is the costliest error of the chapter because the result stays plausible. A derivative such as $\dfrac{\partial\lambda}{\partial\theta}=\dfrac{d\cos\theta}{m}$ is valid only in RADIANS, since the derivative of the sine equals the cosine only in that unit. You must therefore convert: $0.03333^{\circ}=5.818\times 10^{-4}$ rad. Keeping degrees multiplies the contribution by $\dfrac{180}{\pi}=57.3$. Correct version: “any angular uncertainty entering a propagation through a derivative is first converted to radians”.

FALSE. The background, cosmic rays and ambient radioactivity, is present all the time and ALWAYS in the same direction: it is a systematic error by excess. Averaging it only measures it more precisely, it does not remove it. It is SUBTRACTED, and that subtraction has a cost, since it adds the uncertainties of the two counts. Correct version: “the background is measured with the source removed, then subtracted from the total count; repeating only improves the knowledge of the background, never its absence”.

FALSE. The gap, $0.5$ nm, is SMALLER than the uncertainty, $1.3$ nm: the tabulated value lies inside the interval $[544.3, 546.9]$ nm, so the measurement is CONSISTENT and the experiment succeeded. Declaring a failure because two numbers differ is ignoring what an uncertainty means. Correct version: “two values are consistent when their intervals overlap; the $0.5$ nm gap is $0.4$ times the uncertainty, an excellent agreement”.

FALSE. The square root halves each relative contribution, which flattens the differences: the tension contributes $0.21\,\%$, the mass $0.20\,\%$ and the length $0.08\,\%$. None dominates, and improving the tension alone would gain at best a fifth of the budget. Correct version: “you find the dominant contribution by calculation, not by the place of the quantity in the formula; and when none dominates, you must improve several measurements at once or change method”.

Isolating the voltage: $V_{s}=\dfrac{hc}{e}\cdot\dfrac{1}{\lambda}$. This is of the form $Y=aX$ with $X=\dfrac{1}{\lambda}$ and $Y=V_{s}$, so you plot the INVERSE OF THE WAVELENGTH on the horizontal axis and the SLOPE IS $\dfrac{hc}{e}$. Plotting $\lambda$ directly would give a hyperbola, from which no tangent can be read with a ruler.

Expressing $\dfrac{1}{\lambda}$ in µm$^{-1}$: $1.5748$; $1.7094$; $1.9048$; $2.1277$; $2.4691$. The regression gives $\bar X=1.9572$ µm$^{-1}$, $\bar V=2.344$ V, and $a=1.278$ V·µm. Rough check with the two extreme points: $\dfrac{3.00-1.85}{2.4691-1.5748}=\dfrac{1.15}{0.8943}=1.286$, the right order.

$a=\dfrac{hc}{e}$, so $h=\dfrac{ae}{c}$. Converting the slope to volt-metres, $a=1.278\times 10^{-6}$ V·m: $h=\dfrac{1.278\times 10^{-6}\times 1.602\times 10^{-19}}{3.00\times 10^{8}}=6.83\times 10^{-34}$ J·s. The conversion of micrometres to metres is the only delicate step: forgetting it gives $10^{-28}$, a million times too much.

$\dfrac{6.83-6.626}{6.626}=0.031$, that is $+3.1\,\%$. The deviation is POSITIVE, and it always is with this method.

$\Delta a=\dfrac{1.32-1.24}{2}=0.04$ V·µm, a relative uncertainty of $\dfrac{0.04}{1.278}=3.1\,\%$ on the slope, hence on $h$: $\Delta h=6.83\times 0.031=0.21$, and $h=(6.83\pm 0.21)\times 10^{-34}$ J·s. The interval $[6.62, 7.04]\times 10^{-34}$ contains the accepted value, just barely: the measurement is CONSISTENT. What to conclude is more interesting than the verdict. The diode method carries a SYSTEMATIC BIAS upward, because the “threshold” is not a sharp point: the diode already emits a little below it, and the eye or the voltmeter declares it lit a little too late; the ohmic drop in the junction adds to this. The bias here is three percent, and it happens to be of the same order as the uncertainty, which makes it invisible. An honest uncertainty therefore saved the conclusion, but did not prove it: if you really wanted to measure $h$, you would need a method without a fuzzy threshold, such as the photoelectric effect with a stopping potential.

You subtract $20$ counts from each count: $800$; $610$; $464$; $354$; $270$. The subtraction is done BEFORE the logarithm because the logarithm is not a linear operation: $\ln(N_{tot}-N_{bg})$ has nothing to do with $\ln N_{tot}-\ln N_{bg}$, which equals $\ln\dfrac{N_{tot}}{N_{bg}}$. More deeply, the exponential law describes the SOURCE alone; the background is a constant additive term laid on top of it, and it has to be removed to recover the quantity the law describes.

The logarithms are $6.6846$; $6.4135$; $6.1399$; $5.8693$; $5.5984$. Their successive differences, $-0.2711$; $-0.2736$; $-0.2706$; $-0.2709$, are practically constant: the decay is indeed exponential, and this is the check to make before any calculation. The regression gives $\bar t=120$ s, $\overline{\ln N}=6.1411$, $\sum(t_{i}-\bar t)(\ln N_{i}-\overline{\ln N})=-163.0$ and $\sum(t_{i}-\bar t)^{2}=36{,}000$, so $\lambda=-a=4.528\times 10^{-3}$ s$^{-1}$.

$t_{1/2}=\dfrac{\ln 2}{\lambda}=\dfrac{0.6931}{4.528\times 10^{-3}}=153.1$ s, that is $\dfrac{153.1}{60}=2.55$ min. The agreement with the tabulated value is exact to the hundredth of a minute.

$t_{1/2}=\dfrac{0.6931}{4.334\times 10^{-3}}=159.9$ s, that is $2.67$ min. The relative deviation is $\dfrac{2.67-2.55}{2.55}=4.5\,\%$, and it is POSITIVE: forgetting the background always makes the source look slower than it is. This is not a random error that repetition would correct, it is a systematic bias, of the same kind as the offset zero of a goniometer.

A constant background of $20$ counts represents $2.4\,\%$ of the first count but $6.9\,\%$ of the last: its RELATIVE weight grows as the source dies away. The curve of totals therefore falls less and less quickly on a logarithmic scale, it flattens, and would tend to the horizontal once the source was exhausted, since only the background would remain. It is a valuable signal: ON A SEMI-LOGARITHMIC PLOT, AN UPWARD CURVATURE BETRAYS A FORGOTTEN ADDITIVE TERM. The same diagnosis applies elsewhere: a background of stray light in an absorption measurement, a dark current in a photodiode, a leak in parallel on a capacitor. And the opposite curvature, downward, generally betrays the presence of a second isotope with a shorter life.