RuukkilabraUpper secondary physics
FY7 · at the end of the module

Apply independently

In these tasks nobody tells you which formula to use. Decide yourself which phenomena and models belong to the situation, solve the task on paper and justify your solution, as in the matriculation examination.

Five tasks bring the module together: the motion of a charged particle in electric and magnetic fields, a generator and induction, the transmission of electrical energy and the force between current-carrying conductors, thermal radiation, the path of light in an optical fibre, interference in a thin film, and the interference of light at a grating. In each one you decide which models fit the situation and justify your choice.

Solve the task yourself first, on paper or on a computer. You may use a calculator, a data booklet and graphing software. Write out your reasoning, formulas, substitutions and answers with units.

Ion implanter 16 points

Made-up example data · The dimensions and voltages of the device are made up but of the same order of magnitude as in a small ion implanter. Masses of the boron atoms: AME2020 atomic mass evaluation (Wang et al., 2021); share of boron-10 in natural boron: IUPAC (2013).

In the manufacture of microchips, boron ions are shot into a silicon wafer so that the electrical conductivity of the silicon changes in the desired way. In an ion implanter, positive ions with a charge of one elementary charge come from the ion source and are accelerated by a voltage, pass through a velocity selector and are guided through an analysing magnet to the wafer. In the field of the magnet the ions travel along a semicircle, and only ions whose path hits a slit in the wall of the chamber reach the wafer. About $19.9\,\%$ of natural boron is the isotope boron-10 and the rest is boron-11. Only boron-11 ions are wanted at the wafer.

Data:

  • mass of a boron-11 ion $m_{11} = 1.8281 \cdot 10^{-26}\,\text{kg}$ and of a boron-10 ion $m_{10} = 1.6627 \cdot 10^{-26}\,\text{kg}$
  • charge of an ion $e = 1.602 \cdot 10^{-19}\,\text{C}$
  • accelerating voltage $U = 30.0\,\text{kV}$; the kinetic energy of the ions before acceleration is negligibly small
  • distance between the plates of the velocity selector $d = 2.00\,\text{cm}$ and magnetic field $B_\text{s} = 0.200\,\text{T}$
  • radius of the path of the boron-11 ions in the analysing magnet $r = 40.0\,\text{cm}$.

In the figure both magnetic fields point into the page. The effect of gravity on the ions is negligibly small.

$\text{source}$
$U$
$\text{velocity selector}$
$\overline{B}_\text{s}$
$d$
$\overline{B}$
$r$
$\text{silicon wafer}$
$\text{analysing magnet}$
  1. a) The voltage between the plates of the selector is adjusted so that the boron-11 ions pass through the selector in a straight line. Which forces act on an ion travelling straight through the selector, in which directions do they point and how do their magnitudes compare? Which plate is positive? Justify your answer. You may illustrate your answer with a force diagram. 3 p
  2. b) What is the speed of the boron-11 ions after the acceleration? 3 p
  3. c) What voltage must be connected between the plates of the selector? 3 p
  4. d) How strong must the magnetic field of the analysing magnet be? If boron-10 ions could also pass through the selector at the same speed, how far from the slit of the boron-11 ions would they hit the wall of the chamber? 4 p
  5. e) In reality the kinetic energy of the ions after the acceleration varies by as much as $10\,\%$ because the conditions in the ion source and the accelerating voltage vary. An engineer suggests: “The analysing magnet is unnecessary. All the ions are accelerated by the same voltage, so the boron-10 ions are faster and do not get through the selector.” Evaluate the suggestion. 3 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Find

the forces acting on a boron-11 ion travelling straight, and which plate of the selector is positive

Principle

In the selector the ion is acted on by a magnetic force, because it moves in a magnetic field, and by an electric force, because there is an electric field between the plates. Gravity is negligibly small. The ion travels straight at constant velocity only if the sum of the forces is zero, that is, the forces are equal in magnitude and opposite in direction. The magnetic force is perpendicular to both the velocity and the field, and its direction is given by the right-hand rule. The electric force on a positive charge is in the direction of the electric field.

$+$
$-$
$\overline{E}$
$\overline{B}_\text{s}$
$\overline{F}_\text{m}$
$\overline{F}_\text{s}$
$\overline{v}$

Model answer

The ion moves to the right and the magnetic field points into the page. By the right-hand rule, the magnetic force $\overline{F}_\text{m}$ on a positive ion points upwards.
- For the ion to travel straight, the electric force $\overline{F}_\text{s}$ must point downwards and be equal in magnitude to the magnetic force. The force arrows are drawn equally long.
- The velocity $\overline{v}$ is drawn separately in its own colour, because it is not a force.

The electric force on a positive ion is in the direction of the electric field, so the electric field $\overline{E}$ points downwards. The electric field points from the positive plate to the negative one, so the upper plate is positive. The choice does not depend on the mass of the ion: the same polarity suits all positive ions.

b)

Given

accelerating voltage $U = 30.0\,\text{kV}$, charge of the ion $e = 1.602 \cdot 10^{-19}\,\text{C}$ and mass $m_{11} = 1.8281 \cdot 10^{-26}\,\text{kg}$; the kinetic energy of the ion before acceleration is negligibly small

Find

the speed $v$ of the ion after the acceleration

Principle

The electric field in the accelerating gap does work $W = eU$ on the ion. Because the kinetic energy of the ion before acceleration is negligibly small and there are no other forces, the work becomes entirely the kinetic energy of the ion (the energy principle).

Formula

$eU = \tfrac{1}{2} m_{11} v^2$, so $v = \sqrt{\dfrac{2eU}{m_{11}}}$

Conversions

$U$$= 30.0\,\text{kV}$
$= 30.0 \cdot 10^{3}\,\text{V}$

Substitution

$v$$= \sqrt{\dfrac{2 \cdot 1.602 \cdot 10^{-19}\,\text{C} \cdot 30.0 \cdot 10^{3}\,\text{V}}{1.8281 \cdot 10^{-26}\,\text{kg}}}$
$= \sqrt{5.258 \cdot 10^{11}\,\tfrac{\text{m}^2}{\text{s}^2}}$
$= 7.251 \cdot 10^{5}\,\text{m/s}$

Answer

$v \approx 7.25 \cdot 10^{5}\,\text{m/s}$

Three significant figures like the accelerating voltage.

Reasonableness

The speed is about $0.24\,\%$ of the speed of light, so the classical expression for kinetic energy is sufficient. The kinetic energy of the ion, $4.8 \cdot 10^{-15}\,\text{J}$, is small, but the ion is very light, so the speed is high.

c)

Given

$v = 7.251 \cdot 10^{5}\,\text{m/s}$ (part b), $B_\text{s} = 0.200\,\text{T}$ and the distance between the plates $d = 2.00\,\text{cm}$

Find

the voltage $U_\text{s}$ between the plates

Principle

The ion travels straight when the electric force $eE$ and the magnetic force $evB_\text{s}$ are equal in magnitude (part a). The electric field between the plates is uniform, so its strength is the voltage between the plates divided by the distance between the plates.

Formula

$eE = evB_\text{s}$, so $E = vB_\text{s}$. When $E = U_\text{s} : d$, we get $U_\text{s} = Ed = vB_\text{s}d$.

Conversions

$d$$= 2.00\,\text{cm}$
$= 0.0200\,\text{m}$

Substitution

$E$$= 7.251 \cdot 10^{5}\,\text{m/s} \cdot 0.200\,\text{T}$
$= 1.450 \cdot 10^{5}\,\text{V/m}$
$U_\text{s}$$= 1.450 \cdot 10^{5}\,\text{V/m} \cdot 0.0200\,\text{m}$
$= 2900\,\text{V}$

Answer

$U_\text{s} \approx 2.90\,\text{kV}$

Three significant figures like the data.

Reasonableness

The units are consistent, because

$1\,\tfrac{\text{m}}{\text{s}} \cdot 1\,\text{T}$$= 1\,\tfrac{\text{V}}{\text{m}}$

The voltage is about a tenth of the accelerating voltage, which is easy to achieve with a plate separation of $2\,\text{cm}$.

d)

Given

$v = 7.251 \cdot 10^{5}\,\text{m/s}$, $m_{11} = 1.8281 \cdot 10^{-26}\,\text{kg}$, $m_{10} = 1.6627 \cdot 10^{-26}\,\text{kg}$, $e = 1.602 \cdot 10^{-19}\,\text{C}$ and the radius of the path of the boron-11 ions $r = 40.0\,\text{cm}$

Find

the magnetic flux density $B$ of the analysing magnet and the distance of the point of impact of the boron-10 ions from the slit of the boron-11 ions

Principle

The magnetic force is perpendicular to the velocity, so it does not change the speed but turns the direction of motion. In a uniform field the ion moves in uniform circular motion, and the magnetic force is its centripetal force. After the selector all the ions have the same velocity, so the radius of the path is directly proportional to the mass of the ion. After the semicircle the ion hits the wall at the far end of a diameter $2r$ from the point of entry, so the distance between the points of impact of the two kinds of ion is the difference of their diameters.

Formula

$evB = \dfrac{m v^2}{r}$, so $B = \dfrac{m_{11} v}{e r}$. At the same speed and in the same field

$r_{10}$$= \dfrac{m_{10} v}{eB}$
$= \dfrac{m_{10}}{m_{11}}\,r$

and the distance between the points of impact is

$\Delta x$$= 2r - 2r_{10}$
$= 2(r - r_{10})$

Conversions

$r = 40.0\,\text{cm} = 0.400\,\text{m}$

Substitution

$mv$$= 1.8281 \cdot 10^{-26}\,\text{kg} \cdot 7.251 \cdot 10^{5}\,\text{m/s}$
$= 1.3256 \cdot 10^{-20}\,\text{kg} \cdot \text{m/s}$

so

$B$$= \dfrac{1.3256 \cdot 10^{-20}\,\text{kg} \cdot \text{m/s}}{1.602 \cdot 10^{-19}\,\text{C} \cdot 0.400\,\text{m}}$
$= 0.2069\,\text{T}$

The radius of the boron-10 ions is

$r_{10}$$= \dfrac{1.6627}{1.8281} \cdot 0.400\,\text{m}$
$= 0.3638\,\text{m}$

so

$\Delta x$$= 2 \cdot (0.400\,\text{m} - 0.3638\,\text{m})$
$= 0.0724\,\text{m}$

Answer

$B \approx 0.207\,\text{T}$

And $\Delta x \approx 7.2\,\text{cm}$: the boron-10 ions would hit the wall about $7\,\text{cm}$ closer to the point of entry of the ions than the slit. The field has three significant figures like the data; the distance is the difference of two almost equal numbers, so it is left with two.

Reasonableness

The field is of the same order of magnitude as that of an ordinary electromagnet. The masses differ by about $9\,\%$, so the radii also differ by about $9\,\%$, that is, about $3.6\,\text{cm}$, and the diameters by twice that. The difference is much larger than the size of the slit, so the magnet certainly separates the isotopes.

e)

Find

whether the analysing magnet can be left out when the kinetic energy of the ions varies by as much as $10\,\%$ after the acceleration

Principle

In the balance of forces on an ion passing through the selector, $eE = evB_\text{s}$, the charge cancels and the mass does not appear at all. The selector therefore lets straight through every ion whose speed is $E : B_\text{s}$. The speed after the acceleration depends on both the mass and the kinetic energy.

Model answer

Without variation in energy. A boron-10 ion that has received the full acceleration is lighter, so its speed is greater: $v_{10} = \sqrt{\dfrac{2eU}{m_{10}}}$. Substituting the values of part b and the mass $m_{10} = 1.6627 \cdot 10^{-26}\,\text{kg}$ gives $v_{10} \approx 7.60 \cdot 10^{5}\,\text{m/s}$. The magnetic force at that speed is greater than the electric force, so the ion curves upwards in the selector and does not get through. Here the engineer is right.

With varying energy. A boron-10 ion passes through the selector if its speed is the same as that of the boron-11 ions, that is, $7.25 \cdot 10^{5}\,\text{m/s}$. Its kinetic energy is then $\tfrac{1}{2} m_{10} v^2$, while that of a boron-11 ion is $\tfrac{1}{2} m_{11} v^2$, so the ratio is $m_{10} : m_{11} = 1.6627 : 1.8281 \approx 0.91$. A boron-10 ion whose kinetic energy is about $9\,\%$ smaller than the full value therefore passes straight through. When the energy varies by as much as $10\,\%$, there are certain to be such ions.

Conclusion. The suggestion does not work. The selector selects speed, not mass, and when the energy varies, boron-10 ions also get through it. Only the analysing magnet separates ions travelling at the same speed according to their mass, because the radius of the path is then directly proportional to the mass.

Also accepted

  • The speed in part b can also be calculated through the kinetic energy: first $W = eU = 4.806 \cdot 10^{-15}\,\text{J}$ and then $v = \sqrt{2W : m_{11}}$.
  • The distance in part d can also be calculated by first finding the radius of the boron-10 ions directly from the equation $r_{10} = \dfrac{m_{10} v}{eB}$. The answer $7.2$–$7.3\,\text{cm}$ is accepted.
  • In part e it is also enough to give a justification that calculates the accelerating voltage at which a boron-10 ion would get the right speed: $0.91 \cdot 30.0\,\text{kV} \approx 27.3\,\text{kV}$.

Common mistakes

  • The magnetic force is drawn in the direction of the velocity or the field, or its direction is deduced with the rule for a negative charge.
  • The electric force is drawn regardless of the plates, and from the upper plate it is concluded that the electric field points towards the positive plate.
  • The velocity is drawn in the force diagram as a force or in the same colour as the forces.
  • The kilovolts of the accelerating voltage are forgotten when converting to volts, so the speed is about $32$ times too small.
  • In part c the voltage is calculated with the formula $U = Ed$ without taking into account that $E$ must first be solved from the balance of forces.
  • In part d the difference of the radii is given as the distance between the points of impact instead of the difference of the diameters.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

Names the forces acting on the ion: the magnetic force $\overline{F}_\text{m}$ and the electric force $\overline{F}_\text{s}$. The weight is left out because it is negligibly small. 1 p
Justifies the directions and magnitudes: the magnetic force is perpendicular to the velocity and the field and on a positive ion points upwards, the electric force points downwards, and the forces are equal in magnitude because the ion travels straight at constant velocity. 1 p
Concludes that the electric field points in the direction of the electric force on the positive ion, that is, downwards, so the upper plate is positive. 1 p

b)3 p

Justifies with the energy principle: the work $W = eU$ done on the ion by the electric field becomes the kinetic energy of the ion. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $eU = \tfrac{1}{2} m_{11} v^2$, so $v = \sqrt{\dfrac{2eU}{m_{11}}}$ 1 p
The answer has the correct unit and a precision that matches the given data. $v \approx 7.25 \cdot 10^{5}\,\text{m/s}$ 1 p

c)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $eE = evB_\text{s}$ and in a uniform field $E = U_\text{s} : d$, so $U_\text{s} = vB_\text{s}d$. 1 p
The numerical values are substituted with their units, and the calculation is correct. $E = vB_\text{s} \approx 1.45 \cdot 10^{5}\,\text{V/m}$ 1 p
The answer has the correct unit and a precision that matches the given data. $U_\text{s} \approx 2.90\,\text{kV}$ 1 p

d)4 p

Uses the magnetic force as the centripetal force of the circular motion: $evB = \dfrac{mv^2}{r}$, so at the same speed the radius of the path is directly proportional to the mass. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $B = \dfrac{m_{11}v}{er}$ and $r_{10} = \dfrac{m_{10}}{m_{11}}\,r$ 1 p
The answer has the correct unit and a precision that matches the given data. $B \approx 0.207\,\text{T}$ 1 p
The numerical values are substituted with their units, and the calculation is correct. The ions hit the wall at the far end of a diameter from the point of entry, so the distance between the points of impact is $2(r - r_{10}) \approx 7.2\,\text{cm}$. 1 p

e)3 p

States that the selector lets straight through every ion whose speed is $E : B_\text{s}$, regardless of its mass and charge: it selects speed, not mass. 1 p
Calculates that a boron-10 ion passes through the selector if its kinetic energy is $m_{10} : m_{11} \approx 0.91$ of the full value, that is, about $9\,\%$ smaller, which is possible when the energy varies by as much as $10\,\%$. 1 p
The conclusion answers the question and is based on the data. The suggestion does not work: without the variation in energy the engineer would be right, because a boron-10 ion that has received the full acceleration is faster, but when the energy varies, boron-10 ions also get through the selector, and only the magnet separates ions travelling at the same speed according to their mass. 1 p

Self-assessed points: 0 / 16 (16 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Hub dynamo 16 points

Made-up example data · The data of the dynamo and the graph of the voltage are made up but typical: a bicycle dynamo light is usually $6\,\text{V}$ and $3\,\text{W}$, and a hub dynamo often has $28$ magnetic poles.

In a dynamo built into the hub of a bicycle's front wheel, a ring of permanent magnets with $28$ magnetic poles, alternately north and south poles, rotates with the wheel. The stationary coil has an iron core, and the magnetic flux through it changes direction each time the next pole passes it. The voltage therefore has $14$ cycles for every revolution of the wheel. The diameter of the wheel with its tyre is $0.70\,\text{m}$. A light with a rated voltage of $6.0\,\text{V}$ and a power of $3.0\,\text{W}$ can be connected to the dynamo.

A cyclist measured the voltage of the dynamo with an oscilloscope while the light was off and the cyclist rode at a steady speed. The graph is below.

$10$
$20$
$30$
$40$
$50$
$60$
$70$
$80$
$90$
$-12$
$-8$
$-4$
$0$
$4$
$8$
$12$
$16$
$t\ (\text{ms})$
$u\ (\text{V})$
$0$
  1. a) From the graph, determine the frequency and the effective value of the voltage. 3 p
  2. b) At what speed was the cyclist riding? 3 p
  3. c) The light shines properly when the effective value of the voltage of the dynamo is at least the rated voltage of the light. Estimate the lowest speed at which this is so, assuming that the light receives the same voltage as was measured with no load. 3 p
  4. d) The cyclist claims: “When the light is on, pedalling is harder, even though nothing in the dynamo rubs against anything else. When the light is off, the dynamo does not brake the wheel at all.” Evaluate both parts of the claim. 3 p
  5. e) When the light is on, the efficiency of the dynamo is $60\,\%$. How large an extra force must the cyclist overcome at the speed of part b? Does it matter, when air resistance and rolling resistance at this speed total about $10\,\text{N}$? 4 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

the graph of the voltage of the dynamo with the light off

Find

the frequency $f$ and the effective value $U$ of the voltage

Principle

The voltage varies sinusoidally, so the period is the duration of one complete oscillation and the frequency is its reciprocal. The effective value is the direct voltage that would produce the same average power in a resistor. For a sinusoidal voltage it is the peak value divided by $\sqrt{2}$.

Reading the graph

The voltage starts from zero and returns to zero, changing in the same direction, at $30\,\text{ms}$, $60\,\text{ms}$ and $90\,\text{ms}$: three cycles last $90\,\text{ms}$, so $T = 90\,\text{ms} : 3 = 30\,\text{ms}$. Reading three cycles is more accurate than reading one. The peaks are at $7.5\,\text{ms}$, $37.5\,\text{ms}$ and $67.5\,\text{ms}$ on the grid line $14\,\text{V}$, so $e_0 = 14\,\text{V}$.

Formula

$f = \dfrac{1}{T}$ and $U = \dfrac{e_0}{\sqrt{2}}$

Conversions

$T = 30\,\text{ms} = 0.030\,\text{s}$

Substitution

$f$$= \dfrac{1}{0.030\,\text{s}}$
$= 33.3\,\text{Hz}$
$U$$= \dfrac{14\,\text{V}}{\sqrt{2}}$
$= 9.90\,\text{V}$

Answer

$f \approx 33\,\text{Hz}$

And $U \approx 9.9\,\text{V}$, two significant figures like the values read from the graph.

Reasonableness

The frequency is lower than the $50\,\text{Hz}$ of the mains, which suits a wheel that turns only a few revolutions per second. The effective value is about $70\,\%$ of the peak value, as it should be for a sinusoidal voltage.

b)

Given

$f = 33.3\,\text{Hz}$ (part a), $14$ cycles of the voltage per revolution of the wheel and the diameter of the wheel $D = 0.70\,\text{m}$

Find

the speed $v$ of the cyclist

Principle

The magnet ring has $14$ pairs of north and south poles. When one pair of poles passes the coil, the flux through the coil makes one complete oscillation, so the voltage has $14$ cycles per revolution of the wheel. The rotation frequency of the wheel is therefore $14$ times smaller than the frequency of the voltage. When the wheel rolls without slipping, in one revolution it advances the length of its circumference $\pi D$.

Formula

$f_\text{w}$$= \dfrac{f}{14}$
$v$$= \pi D f_\text{w}$
$= \pi D \cdot \dfrac{f}{14}$

Conversions

The answer is converted at the end: $1\,\text{m/s} = 3.6\,\text{km/h}$.

Substitution

$f_\text{w}$$= \dfrac{33.3\,\text{Hz}}{14}$
$= 2.381\,\text{Hz}$
$v$$= \pi \cdot 0.70\,\text{m} \cdot 2.381\,\tfrac{1}{\text{s}}$
$= 5.236\,\text{m/s}$
$= 5.236 \cdot 3.6\,\text{km/h}$
$= 18.8\,\text{km/h}$

Answer

$v \approx 5.2\,\text{m/s} \approx 19\,\text{km/h}$

Two significant figures like the diameter and the period.

Reasonableness

The speed is a normal commuting cycling speed. The wheel turns only about two and a half revolutions per second, but $28$ poles still make the frequency of the voltage more than $30\,\text{Hz}$.

c)

Given

the effective voltage with no load $U = 9.90\,\text{V}$ at the speed $v = 5.236\,\text{m/s}$ (parts a and b), rated voltage of the light $U_\text{lamp} = 6.0\,\text{V}$

Find

the lowest speed $v_\text{min}$ at which the effective voltage is at least $6.0\,\text{V}$

Principle

The peak value of the voltage of a generator is $e_0 = NBA \cdot 2\pi f$. The number of turns of the coil, the field of the magnets and the area stay the same, so the peak value, and with it the effective value, is directly proportional to the frequency. The frequency in turn is directly proportional to the rotation frequency of the wheel and to the speed. The effective voltage is therefore directly proportional to the speed: the more slowly one rides, the more slowly the magnets pass the coil and the more slowly the flux changes.

Formula

$\dfrac{v_\text{min}}{v}$$= \dfrac{U_\text{lamp}}{U}$

so

$v_\text{min}$$= v \cdot \dfrac{U_\text{lamp}}{U}$

Conversions

The answer is converted at the end to kilometres per hour: $1\,\text{m/s} = 3.6\,\text{km/h}$.

Substitution

$v_\text{min}$$= 5.236\,\text{m/s} \cdot \dfrac{6.0\,\text{V}}{9.90\,\text{V}}$
$= 3.17\,\text{m/s}$
$= 3.17 \cdot 3.6\,\text{km/h}$
$= 11.4\,\text{km/h}$

Answer

$v_\text{min} \approx 3.2\,\text{m/s} \approx 11\,\text{km/h}$

Two significant figures like the data.

Reasonableness

The result is slow cycling but brisker than walking, so at walking speed the light stays dim. The assumption is rough, however: when the light is on, a current flows in the coil and part of the voltage is left in the coil itself, so the light gets less than the no-load voltage. The real lowest speed is therefore somewhat greater.

d)

Find

whether the dynamo brakes the wheel when the light is on and when it is off

Principle

When the coil is in a closed circuit, the changing flux produces an induced current in it. By Lenz's law, the magnetic field of the induced current opposes the change that produces it. By conservation of energy, the energy radiated by the light must come from somewhere. A changing magnetic field also produces eddy currents in conducting metal parts.

Model answer

Light on. An induced current flows in the coil, and its magnetic field opposes the change in flux: it pushes an approaching magnetic pole away and pulls a receding one back. The rotating magnet ring is therefore acted on by a braking magnetic force, even though nothing touches anything else. The same follows from conservation of energy: the light radiates energy that comes from the work done by the cyclist. If there were no braking force, the light would get energy from nothing. The first part of the claim is true.

Light off. The circuit is open, so no current flows in the coil and the coil does not brake. The changing field of the magnet ring does, however, produce eddy currents in the metal parts of the dynamo, for example in the iron core of the coil and in the metal hub. The eddy currents heat the metal and brake the rotation in accordance with Lenz's law. In addition, the magnetisation of the iron core reverses continuously, which takes a little energy.

Conclusion. The second part of the claim is not exactly true: the dynamo brakes the wheel a little even when the light is off. The braking is then much smaller than when the light is on, because the iron core is made of thin insulated sheets, in which the eddy currents stay small.

e)

Given

power of the light $P_\text{lamp} = 3.0\,\text{W}$, efficiency of the dynamo $\eta = 60\,\% = 0.60$, speed $v = 5.236\,\text{m/s}$ (part b) and air resistance and rolling resistance together about $10\,\text{N}$

Find

the extra braking force $F$ of the dynamo and its significance

Principle

The dynamo converts the cyclist's work into electrical energy. The light receives the fraction of the power taken by the dynamo given by the efficiency, so the power taken by the dynamo is $P_\text{lamp} : \eta$. When the braking force acts on a wheel moving at speed $v$, the power of the work it does is $P = Fv$, as in an eddy-current brake.

Formula

$P_\text{in}$$= \dfrac{P_\text{lamp}}{\eta}$

$P_\text{in} = Fv$, so $F = \dfrac{P_\text{lamp}}{\eta v}$

Conversions

Not needed: the power is in watts and the speed in metres per second.

Substitution

$P_\text{in}$$= \dfrac{3.0\,\text{W}}{0.60}$
$= 5.0\,\text{W}$
$F$$= \dfrac{5.0\,\text{W}}{5.236\,\text{m/s}}$
$= 0.955\,\text{N}$

Answer

$F \approx 0.95\,\text{N}$

Two significant figures like the power of the light and the efficiency.

Reasonableness

Air resistance and rolling resistance at this speed total about $10\,\text{N}$, so the light adds about a tenth to the resistance. The difference can be felt, but it is small; for example, a $1\,\%$ uphill gradient adds about $9\,\text{N}$ for a $90\,\text{kg}$ cyclist and bicycle. When riding more slowly the same power requires a larger force, so the light is felt most on an uphill.

Also accepted

  • In part a the period can also be read from a single cycle, for example between two successive peaks: $37.5\,\text{ms} - 7.5\,\text{ms} = 30\,\text{ms}$.
  • The answers to parts b and c can be given in metres per second; in part c, $11$–$12\,\text{km/h}$ is accepted.
  • In part c the peak values can be compared: the light needs a peak value of $6.0\,\text{V} \cdot \sqrt{2} = 8.5\,\text{V}$, and $v_\text{min} = v \cdot 8.5\,\text{V} : 14\,\text{V}$.

Common mistakes

  • The period is read as the interval between zero crossings, which is only half a cycle, so the frequency is twice too large.
  • The effective value is calculated by multiplying the peak value by $\sqrt{2}$ instead of dividing by it.
  • In part b the frequency of the voltage is used as the rotation frequency of the wheel, so the speed comes out at over $250\,\text{km/h}$.
  • The circumference is taken to be the product of the radius and $\pi$, or the product of the diameter and $2\pi$.
  • In part d the braking effect is explained by friction, or it is said that the magnet attracts iron all the time in the same direction.
  • In part e the power of the light is multiplied by the efficiency instead of divided by it.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. Three cycles last $90\,\text{ms}$, so $T = 30\,\text{ms}$; the peak value $e_0 = 14\,\text{V}$ is read at the peak. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $f = \dfrac{1}{T}$ and $U = \dfrac{e_0}{\sqrt{2}}$ 1 p
The answer has the correct unit and a precision that matches the given data. $f \approx 33\,\text{Hz}$ and $U \approx 9.9\,\text{V}$ 1 p

b)3 p

Justifies that the wheel turns $14$ times more slowly than the voltage oscillates, and that in one revolution the wheel advances the length of its circumference $\pi D$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $v = \pi D \cdot \dfrac{f}{14}$ 1 p
The answer has the correct unit and a precision that matches the given data. $v \approx 5.2\,\text{m/s} \approx 19\,\text{km/h}$ 1 p

c)3 p

Justifies that the peak value of the voltage is directly proportional to the rotation frequency and therefore to the speed, because the faster the cyclist rides, the faster the magnets pass the coil and the faster the flux changes. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\dfrac{v_\text{min}}{v} = \dfrac{U_\text{lamp}}{U}$, so $v_\text{min} = v \cdot \dfrac{U_\text{lamp}}{U}$ 1 p
The answer has the correct unit and a precision that matches the given data. $v_\text{min} \approx 3.2\,\text{m/s} \approx 11\,\text{km/h}$ 1 p

d)3 p

Explains that the first part of the claim is true: when the light is on, an induced current flows in the coil, and its magnetic field opposes the change that produces it, that is, it brakes the rotating magnet ring. 1 p
Justifies with conservation of energy: the energy radiated by the light comes from the cyclist's work, so some force brakes the rotation even though nothing rubs. 1 p
Explains that the second part of the claim is not exactly true: when the light is off, no current flows in the coil, but the changing magnetic field produces eddy currents in the metal parts of the dynamo, such as the iron core and the hub, so a little braking remains. 1 p

e)4 p

Connects power and force: the power of the braking force $P = Fv$ is the power taken by the dynamo, which is the power of the light divided by the efficiency. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $F = \dfrac{P_\text{lamp}}{\eta v}$ 1 p
The answer has the correct unit and a precision that matches the given data. $F \approx 0.95\,\text{N}$ 1 p
The reasonableness of the result has been assessed. The extra force is about a tenth of the sum of air resistance and rolling resistance, $10\,\text{N}$: the difference can be felt, but it is small. 1 p

Self-assessed points: 0 / 16 (16 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Electricity supply for a mine 17 points

Made-up example data · The mine, the line and the powers are made up. The resistivity and density of aluminium are table values.

A mining company opens a new mine $25\,\text{km}$ from the nearest substation. A power of about $6.0\,\text{MW}$ is transmitted to the mine continuously. From the substation it is possible to build either a $20\,\text{kV}$ line straight to the mine or a $110\,\text{kV}$ line, at the end of which a transformer is built at the mine. The transformer steps the voltage down to $20\,\text{kV}$ for the mine's own network.

In both options the line has two aluminium conductors with a cross-sectional area of $150\,\text{mm}^2$. The resistivity of aluminium is $2.82 \cdot 10^{-8}\,\Omega\,\text{m}$ and its density is $2700\,\text{kg/m}^3$.

Real power lines are three-phase, but a simple model is enough here: the substation sends a power of $6.0\,\text{MW}$ into the line, the power is calculated as for direct current, the voltages and currents are effective values, and the transformer is ideal.

  1. a) How much power is converted into heat in the line if the $20\,\text{kV}$ line is used? What percentage of the transmitted power is that? 3 p
  2. b) How large is the power loss if the $110\,\text{kV}$ line is used? 3 p
  3. c) The equipment of the mine works correctly when the voltage of the network deviates from the rated value $20\,\text{kV}$ by at most $10\,\%$. What voltage is left at the mine end of the $20\,\text{kV}$ line? Is it acceptable? 3 p
  4. d) A municipal councillor suggests: “Let us build the cheaper $20\,\text{kV}$ line without a transformer and make the conductors thicker, so that the losses are as small as with the $110\,\text{kV}$ line.” Evaluate the suggestion. 4 p
  5. e) The two conductors of the line run side by side on the poles at a distance of $1.5\,\text{m}$ from each other. Do the conductors attract or repel each other? How large is the force on each metre of conductor in the $20\,\text{kV}$ line in normal operation and in a short circuit near the substation, when the current briefly rises to $8.0\,\text{kA}$? Does the force matter? The permeability of vacuum is $\mu_0 = 4\pi \cdot 10^{-7}\,\tfrac{\text{Vs}}{\text{Am}}$. 4 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

transmitted power $P = 6.0\,\text{MW}$, transmission voltage $U = 20\,\text{kV}$, length of the line $25\,\text{km}$, two conductors, cross-sectional area $A = 150\,\text{mm}^2$ and resistivity $\rho = 2.82 \cdot 10^{-8}\,\Omega\,\text{m}$

Find

the power loss $P_\text{h}$ of the line and its share of the transmitted power

Principle

The current flows to the mine along one conductor and returns along the other, so the length of the conductors in the circuit is $2 \cdot 25\,\text{km}$. The resistance is obtained from the resistivity, the length and the cross-sectional area. The transmitted power is the product of the voltage and the current, so the current is the power divided by the voltage. The power converted into heat in the resistance of the line is the product of the resistance and the square of the current.

Formula

$R = \rho \dfrac{l}{A}$, $I = \dfrac{P}{U}$ and $P_\text{h} = R I^2$

Conversions

$l$$= 2 \cdot 25\,\text{km}$
$= 5.0 \cdot 10^{4}\,\text{m}$
$A$$= 150\,\text{mm}^2$
$= 150 \cdot 10^{-6}\,\text{m}^2$

$P = 6.0 \cdot 10^{6}\,\text{W}$ and $U = 20 \cdot 10^{3}\,\text{V}$

Substitution

$R$$= 2.82 \cdot 10^{-8}\,\Omega\,\text{m} \cdot \dfrac{5.0 \cdot 10^{4}\,\text{m}}{150 \cdot 10^{-6}\,\text{m}^2}$
$= 9.40\,\Omega$
$I$$= \dfrac{6.0 \cdot 10^{6}\,\text{W}}{20 \cdot 10^{3}\,\text{V}}$
$= 300\,\text{A}$
$P_\text{h}$$= 9.40\,\Omega \cdot (300\,\text{A})^2$
$= 846\,000\,\text{W}$

The share is

$846\,\text{kW} : 6000\,\text{kW}$
$= 0.141$
$= 14.1\,\%$

Answer

$P_\text{h} \approx 850\,\text{kW}$

That is, about $14\,\%$ of the transmitted power, two significant figures like the power and the voltage.

Reasonableness

The losses of the whole electricity grid in Finland are only a few per cent in total, so a loss of $14\,\%$ in a single line is far too large. At a low voltage a large current is needed, and the loss grows as the square of the current.

b)

Given

the data of part a and the resistance $R = 9.40\,\Omega$, transmission voltage $U = 110\,\text{kV}$

Find

the power loss $P_\text{h}$

Principle

The same power is transmitted at $110 : 20 = 5.5$ times the voltage, so the current is $5.5$ times smaller. The power loss is proportional to the square of the current, so it decreases by a factor of $5.5^2 \approx 30$. The conductors are the same, so the resistance is the same as in part a.

Formula

$P_\text{h}$$= R I^2$
$= R\left(\dfrac{P}{U}\right)^2$

Conversions

$U$$= 110\,\text{kV}$
$= 110 \cdot 10^{3}\,\text{V}$

Substitution

$I$$= \dfrac{6.0 \cdot 10^{6}\,\text{W}}{110 \cdot 10^{3}\,\text{V}}$
$= 54.5\,\text{A}$
$P_\text{h}$$= 9.40\,\Omega \cdot (54.5\,\text{A})^2$
$= 27\,970\,\text{W}$

The share is $28.0\,\text{kW} : 6000\,\text{kW} \approx 0.47\,\%$.

Answer

$P_\text{h} \approx 28\,\text{kW}$

That is, about $0.5\,\%$, two significant figures like the power.

Reasonableness

$846\,\text{kW} : 28.0\,\text{kW} \approx 30$, as the square of the ratio of the voltages predicted. The loss is now a fraction of a per cent, which is usual in electricity transmission.

c)

Given

in the $20\,\text{kV}$ line the current $I = 300\,\text{A}$ and the resistance $R = 9.40\,\Omega$ (part a); permitted deviation $10\,\%$

Find

the voltage $U_\text{mine}$ left at the mine and whether it is acceptable

Principle

The substation maintains a voltage of $20\,\text{kV}$ at the start of the line. There is a voltage drop across the resistance of the conductors, which is the product of the resistance and the current. What is left at the mine is the difference between the transmission voltage and the voltage drop.

Formula

$U_\text{mine} = U - RI$

Conversions

$U = 20\,\text{kV} = 20\,000\,\text{V}$

Substitution

$RI$$= 9.40\,\Omega \cdot 300\,\text{A}$
$= 2820\,\text{V}$
$U_\text{mine}$$= 20\,000\,\text{V} - 2820\,\text{V}$
$= 17\,180\,\text{V}$

The deviation is $2820\,\text{V} : 20\,000\,\text{V} \approx 14\,\%$.

Answer

$U_\text{mine} \approx 17.2\,\text{kV}$

Which is about $14\,\%$ below the rated value. It is not acceptable, because the deviation is more than $10\,\%$. Three significant figures, so that the deviation can be compared with the limit.

Reasonableness

The product of the voltage drop and the current, $2820\,\text{V} \cdot 300\,\text{A} = 846\,\text{kW}$, is equal to the power loss of part a, as it should be. In the $110\,\text{kV}$ line the voltage drop would be only $9.40\,\Omega \cdot 54.5\,\text{A} \approx 0.51\,\text{kV}$, that is, less than $0.5\,\%$.

d)

Find

whether the loss of the $20\,\text{kV}$ line can be made as small by using thicker conductors

Principle

The power loss is $P_\text{h} = RI^2$. At a voltage of $20\,\text{kV}$ the current is always $300\,\text{A}$, so the only way to reduce the loss is to reduce the resistance. The resistance is inversely proportional to the cross-sectional area, and the mass of the conductors is directly proportional to their volume.

Model answer

Required thickness. For the loss to be the same as in the $110\,\text{kV}$ line, the resistance must be $846\,\text{kW} : 28.0\,\text{kW} \approx 30$ times smaller. The cross-sectional area must therefore be $30.25 \cdot 150\,\text{mm}^2 \approx 4500\,\text{mm}^2$. The diameter of a solid round conductor would be $2\sqrt{4.5 \cdot 10^{-3}\,\text{m}^2 : \pi} \approx 7.6\,\text{cm}$.

Amount of aluminium. The mass of the present conductors is $2700\,\text{kg/m}^3 \cdot 5.0 \cdot 10^{4}\,\text{m} \cdot 150 \cdot 10^{-6}\,\text{m}^2 \approx 20\,\text{t}$. The mass of the thick conductors would be $30$ times as much, about $610\,\text{t}$. The poles would have to carry all this, and the aluminium would cost many times the price of a transformer.

Why it pays to raise the voltage. Raising the voltage reduces the current in the same ratio, and the loss depends on the square of the current: $5.5$ times the voltage reduces the loss by a factor of about $30$. A thicker conductor reduces the loss only in the same ratio as the cross-sectional area increases, so to get the same result the cross-sectional area of the conductor has to be made $30$ times as large.

Conclusion. The suggestion would work in principle, but the line would be enormously heavy and expensive. That is why long transmission is done at high voltage and the voltage is stepped down with a transformer only near the place of use.

e)

Given

distance between the conductors $r = 1.5\,\text{m}$, current in normal operation $I = 300\,\text{A}$ (part a) and in a short circuit $I_\text{sc} = 8.0\,\text{kA}$, $\mu_0 = 4\pi \cdot 10^{-7}\,\tfrac{\text{Vs}}{\text{Am}}$, cross-sectional area of a conductor $A = 150\,\text{mm}^2$ and density of aluminium $\rho_\text{Al} = 2700\,\text{kg/m}^3$

Find

whether the conductors attract or repel each other, the force per metre of conductor $F : l$ in both situations and its significance

Principle

The current flows to the mine along one conductor and returns along the other, so currents of equal magnitude flow in the conductors in opposite directions. Each conductor is a long straight conductor, and its magnetic field at the other conductor is obtained from the formula for the field of a straight conductor, because the conductors are much longer than the distance between them. The field is perpendicular to the other conductor, so a force $F = IlB$ acts on it. By the right-hand rule, currents in opposite directions repel each other. With alternating current the currents change direction at the same time, so they are always in opposite directions and the conductors repel each other all the time. The weight of a metre of conductor is obtained from its mass, which is the product of the density and the volume.

Formula

$B = \dfrac{\mu_0 I}{2\pi r}$ and $F = IlB$, so

$\dfrac{F}{l}$$= \dfrac{\mu_0 I^2}{2\pi r}$

For the weight, $\dfrac{G}{l} = \rho_\text{Al} A g$.

Conversions

$I_\text{sc}$$= 8.0\,\text{kA}$
$= 8.0 \cdot 10^{3}\,\text{A}$
$A$$= 150\,\text{mm}^2$
$= 150 \cdot 10^{-6}\,\text{m}^2$

Substitution

In normal operation

$B$$= \dfrac{4\pi \cdot 10^{-7}\,\tfrac{\text{Vs}}{\text{Am}} \cdot 300\,\text{A}}{2\pi \cdot 1.5\,\text{m}}$
$= 4.0 \cdot 10^{-5}\,\text{T}$
$\dfrac{F}{l}$$= 300\,\text{A} \cdot 4.0 \cdot 10^{-5}\,\text{T}$
$= 0.012\,\text{N/m}$

In a short circuit

$B$$= \dfrac{4\pi \cdot 10^{-7}\,\tfrac{\text{Vs}}{\text{Am}} \cdot 8.0 \cdot 10^{3}\,\text{A}}{2\pi \cdot 1.5\,\text{m}}$
$= 1.067 \cdot 10^{-3}\,\text{T}$
$\dfrac{F}{l}$$= 8.0 \cdot 10^{3}\,\text{A} \cdot 1.067 \cdot 10^{-3}\,\text{T}$
$= 8.53\,\text{N/m}$

The weight of the conductor per metre is

$\dfrac{G}{l}$$= 2700\,\tfrac{\text{kg}}{\text{m}^3} \cdot 150 \cdot 10^{-6}\,\text{m}^2 \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}$
$= 3.97\,\text{N/m}$

Answer

The conductors repel each other. The force is about $0.012\,\text{N}$ in normal operation and about $8.5\,\text{N}$ in a short circuit for each metre of conductor, two significant figures like the distance and the short-circuit current. In normal operation the force does not matter, but in a short circuit it is about twice the weight of the conductor.

Reasonableness

In normal operation the force is only about $0.3\,\%$ of the weight of the conductor, about $4.0\,\text{N}$ per metre. In a short circuit the current is about $27$ times as large and the force grows as the square of the current, so it is about $710$ times as large: $8.53 : 3.97 \approx 2.1$ times the weight of the conductor. Conductors hanging between poles then swing apart and may swing back and slap together. That is why the spacing of the conductors is designed for a short circuit and the short circuit is interrupted in a fraction of a second. A current this large is possible only near the substation, where the resistance of the line up to the fault is small.

Also accepted

  • Part a can be calculated directly with the formula $P_\text{h} = \dfrac{R P^2}{U^2}$, and part b from the result of part a with the ratio $P_\text{h,110} = P_\text{h,20} \cdot (20 : 110)^2$.
  • In part c the voltage drop can be compared directly with the permitted deviation $0.10 \cdot 20\,\text{kV} = 2.0\,\text{kV}$.
  • In part d an estimate based on the cross-sectional area and the mass having to be about $30$ times as large is enough.
  • The short-circuit force of part e can be calculated from the force in normal operation using the square of the ratio of the currents: $0.012\,\text{N/m} \cdot (8000 : 300)^2 \approx 8.5\,\text{N/m}$.

Common mistakes

  • The length of the line is taken to be $25\,\text{km}$ although the circuit has two conductors, so the resistance and the losses are half of what they should be.
  • Square millimetres of the cross-sectional area are converted to square metres with the factor $10^{-3}$ instead of $10^{-6}$.
  • The power loss is calculated with the formula $U^2 : R$ using the transmission voltage, although the transmission voltage is not the voltage across the resistance of the line.
  • In part b it is assumed that the current stays the same when the voltage increases.
  • In part d it is assumed that the loss decreases in proportion to the thickness, that is, the diameter, although the resistance depends on the cross-sectional area.
  • In part e it is concluded that the conductors attract each other, although the currents in them flow in opposite directions.
  • In part e the current is not squared, although the field of the other conductor and the force on the conductor are each directly proportional to the current.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

Calculates the resistance of the conductors from the length of the whole circuit, $2 \cdot 25\,\text{km}$: $R = \rho\dfrac{l}{A} \approx 9.4\,\Omega$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $I = \dfrac{P}{U}$ and $P_\text{h} = RI^2$ 1 p
The answer has the correct unit and a precision that matches the given data. $P_\text{h} \approx 850\,\text{kW}$, that is, about $14\,\%$ of the transmitted power 1 p

b)3 p

Justifies that the same power at $5.5$ times the voltage means a current $5.5$ times smaller, so the loss decreases by a factor of $5.5^2 \approx 30$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $P_\text{h} = R\left(\dfrac{P}{U}\right)^2$ 1 p
The answer has the correct unit and a precision that matches the given data. $P_\text{h} \approx 28\,\text{kW}$, that is, about $0.5\,\%$ 1 p

c)3 p

Justifies that part of the transmission voltage is left across the resistance of the conductors: the voltage drop is $RI$, and what is left at the mine is the difference between the transmission voltage and the voltage drop. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $U_\text{mine} = U - RI$ 1 p
The conclusion answers the question and is based on the data. $U_\text{mine} \approx 17.2\,\text{kV}$ is about $14\,\%$ below the rated value, so it is not acceptable; in the $110\,\text{kV}$ line the voltage drop is only about $0.5\,\text{kV}$, that is, $0.5\,\%$. 1 p

d)4 p

Calculates that the resistance must be about $30$ times smaller, so the cross-sectional area must be about $4500\,\text{mm}^2$, which corresponds to a solid conductor about $7.6\,\text{cm}$ thick. 1 p
Estimates the amount of aluminium: the mass of the conductors would grow from about $20\,\text{t}$ to about $610\,\text{t}$. 1 p
Explains why raising the voltage is more effective: the loss depends on the square of the current and the current decreases in the same ratio as the voltage increases, but a thicker conductor reduces the loss only in the same ratio as the cross-sectional area increases. 1 p
The conclusion answers the question and is based on the data. The suggestion would work in principle, but the line would be enormously heavy and expensive. That is why long transmission is done at high voltage. 1 p

e)4 p

Justifies that currents of equal magnitude flow in the conductors in opposite directions, because the current returns to the substation along the other conductor, and that currents in opposite directions repel each other. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. The magnetic field of the other conductor is $B = \dfrac{\mu_0 I}{2\pi r}$ and the force is $F = IlB$, so $\dfrac{F}{l} = \dfrac{\mu_0 I^2}{2\pi r}$. 1 p
The answer has the correct unit and a precision that matches the given data. In normal operation $\dfrac{F}{l} \approx 0.012\,\text{N/m}$ and in a short circuit about $8.5\,\text{N/m}$. 1 p
The reasonableness of the result has been assessed. Compares with the weight of the conductor, about $4.0\,\text{N}$ per metre: in normal operation the force is negligible, but in a short circuit it is about twice the weight, so the conductors swing apart and may slap together. 1 p

Self-assessed points: 0 / 17 (17 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Pyrometer in a rolling mill 17 points

Values calculated by a model · The spectrum has been calculated from the black-body radiation model. The data of the steel, the fibre and the thermocouple are made-up examples.

In the hot rolling mill of a steelworks, a glowing steel slab is flattened into a plate. The temperature of the slab is monitored with a pyrometer, which measures the spectrum of the radiation coming from the slab at wavelengths of about $0.5$–$2.5\,\mu\text{m}$ without touching the slab. The light passes through a protective window near the rolls and is guided along an optical fibre to a protected measuring instrument, because the instrument cannot withstand the heat.

The figure shows the spectrum of the radiation emitted by the slab over a wider range of wavelengths, assuming that the slab is a black body.

Data:

  • constant $b = 2.898 \cdot 10^{-3}\,\text{m} \cdot \text{K}$
  • refractive index of the core of the fibre $1.44$ and of the cladding $1.42$, of air $1.00$
  • refractive index of the glass of the protective window $1.50$; on the surface of the window there is a thin film of magnesium fluoride with a refractive index of $1.38$.

Assume that the refractive indices are the same at all wavelengths.

$1$
$2$
$3$
$4$
$5$
$6$
$0.2$
$0.4$
$0.6$
$0.8$
$1.0$
$0$
$\lambda\ (\mu\text{m})$
$\text{relative intensity}$
$\text{visible}$
  1. a) Determine the temperature of the surface of the slab in degrees Celsius. 3 p
  2. b) The roller operator wonders: “According to the graph the slab radiates almost everything as infrared. Why does it then glow orange-red?” Explain. 3 p
  3. c) Light that hits the end of the fibre continues along the fibre only if it is totally internally reflected at the interface between the core and the cladding. What is the critical angle at this interface? What is the greatest angle of incidence allowed for light entering the end face of the fibre from air? 4 p
  4. d) The temperature of the same slab was measured as $1210\,{}^\circ\text{C}$ with a thermocouple pushed into a hole drilled in it. The foreman suspects that the pyrometer reads wrongly. Evaluate whether the result of part a conflicts with the reading of the thermocouple and what the difference may be due to. 4 p
  5. e) The film of the protective window is designed so that the window reflects as little as possible at the wavelength of the peak of the spectrum, $2.0\,\mu\text{m}$, when the light arrives perpendicularly. How thick is the thinnest film that works in this way? An installer notices that greenish light is reflected from the window in daylight and suspects that the film is wrong. Evaluate the suspicion. 3 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Given

the spectrum of the radiation of the slab (graph) and $b = 2.898 \cdot 10^{-3}\,\text{m} \cdot \text{K}$

Find

the temperature of the surface of the slab in degrees Celsius

Principle

The spectrum of the thermal radiation of a hot object is continuous, and its peak shifts to shorter wavelengths as the temperature increases. For a black body, the product of the wavelength of the peak of the spectrum and the absolute temperature is a constant (Wien's displacement law). The slab is assumed to be a black body, as the graph does.

Reading the graph

The peak of the curve is on the vertical grid line at $2.0\,\mu\text{m}$, so $\lambda_\text{max} = 2.0\,\mu\text{m}$. The peak is flat, so its position can be read to an accuracy of about $\pm 0.05\,\mu\text{m}$ (part d).

Formula

$\lambda_\text{max} T = b$, so $T = \dfrac{b}{\lambda_\text{max}}$. In degrees Celsius the numerical value of the temperature is $273$ smaller than in kelvins.

Conversions

$\lambda_\text{max}$$= 2.0\,\mu\text{m}$
$= 2.0 \cdot 10^{-6}\,\text{m}$

Substitution

$T$$= \dfrac{2.898 \cdot 10^{-3}\,\text{m} \cdot \text{K}}{2.0 \cdot 10^{-6}\,\text{m}}$
$= 1449\,\text{K}$

so

$t$$= (1449 - 273)\,{}^\circ\text{C}$
$\approx 1176\,{}^\circ\text{C}$

Answer

$T \approx 1450\,\text{K}$

That is, $t \approx 1180\,{}^\circ\text{C}$, three significant figures because the peak is on a grid line; the effect of the reading accuracy is estimated in part d.

Reasonableness

Steel to be hot rolled is typically at a temperature of about $1100$–$1250\,{}^\circ\text{C}$, and it glows orange-red, as the result implies. The temperature is clearly lower than the melting point of steel, about $1500\,{}^\circ\text{C}$, as it should be because the slab is solid.

b)

Find

why the slab glows orange-red although the peak of the spectrum is in the infrared

Principle

The spectrum of thermal radiation is continuous: the object radiates at all wavelengths, but a different amount at each. The eye sees only wavelengths of about $0.38$–$0.75\,\mu\text{m}$, and the colour of the glow depends on how the radiation is distributed in this range.

Model answer

In the graph the curve begins to rise already in the range of visible light, although it looks almost zero there. Visible light is therefore only a small part of the total radiation, but it is there. In the visible range the intensity increases steeply with increasing wavelength, so there is much more red and orange light than green or blue. That is why the glow looks orange-red and not white.

As a whole, the hot slab radiates with an enormous power, so even a small visible fraction is bright to the eye. The peak of the spectrum of an object at room temperature is at about $10\,\mu\text{m}$, and its share of visible light is vanishingly small, so it does not glow. If the slab were heated further, the peak would shift to shorter wavelengths and the glow would become yellower and brighter.

c)

Given

refractive index of the core $n_1 = 1.44$, of the cladding $n_2 = 1.42$ and of air $1.00$

Find

the critical angle $\alpha_\text{r}$ at the interface between the core and the cladding and the greatest angle of incidence $\theta$ at the end face of the fibre

Principle

The light passes through two interfaces. At the end face it enters the core from the air and is refracted towards the normal to an angle $\beta$ (the law of refraction). At the side surface of the core the light is totally reflected if its angle of incidence is at least the critical angle. The end face is perpendicular to the side surface, so the normal of the end face is parallel to the side surface and the angle of incidence of the ray at the side surface is $90^\circ - \beta$. The greater $\theta$ is, the greater $\beta$ is and the smaller the angle of incidence at the side surface, so at the greatest allowed angle of incidence the angle of incidence at the side surface is exactly the critical angle.

$\text{cladding } n_2$
$\text{cladding } n_2$
$\text{core } n_1$
$\text{air}$
$\theta$
$\beta$
$\alpha_\text{r}$

Formula

$n_1 \sin\alpha_\text{r} = n_2 \sin 90^\circ$, so

$\sin\alpha_\text{r}$$= \dfrac{n_2}{n_1}$

At the end face $1.00 \cdot \sin\theta = n_1 \sin\beta$, where $\beta = 90^\circ - \alpha_\text{r}$, so $\sin\theta = n_1 \sin(90^\circ - \alpha_\text{r})$.

Conversions

Not needed: refractive indices have no units, and the angles are in degrees.

Substitution

$\sin\alpha_\text{r}$$= \dfrac{1.42}{1.44}$
$= 0.9861$

so $\alpha_\text{r} = 80.44^\circ$. Then

$\beta$$= 90^\circ - 80.44^\circ$
$= 9.56^\circ$
$\sin\theta$$= 1.44 \cdot \sin 9.56^\circ$
$= 1.44 \cdot 0.1661$
$= 0.2392$

so $\theta = 13.84^\circ$.

Answer

$\alpha_\text{r} \approx 80.4^\circ$

And $\theta \approx 13.8^\circ$, three significant figures like the refractive indices.

Reasonableness

The refractive indices of the core and the cladding are almost equal, so the critical angle is close to $90^\circ$ and only light that arrives almost parallel to the fibre gets into it. The pyrometer therefore has to look at the slab along the direction of the fibre, and a lens is needed in front of the fibre to gather the light into a narrow cone. The angle is of the same order of magnitude as in telecommunications fibres.

d)

Find

whether the result of part a conflicts with the thermocouple reading $1210\,{}^\circ\text{C}$ and what the difference may be due to

Principle

Two measurements conflict only if the difference between them is greater than their uncertainty. In addition, one must consider whether the methods measure the same thing and whether the assumptions of the model hold.

Model answer

Uncertainty. The peak of the spectrum is flat, so its position can be read from the graph to an accuracy of about $\pm 0.05\,\mu\text{m}$. The temperature is inversely proportional to the wavelength of the peak, so the relative uncertainty is the same: $0.05 : 2.0 = 2.5\,\%$, that is, $0.025 \cdot 1449\,\text{K} \approx 36\,\text{K}$. The difference from the thermocouple is $1210\,{}^\circ\text{C} - 1176\,{}^\circ\text{C} = 34\,\text{K}$, so it is within the reading accuracy.

What the methods measure. The pyrometer measures the radiation of the surface, the thermocouple the temperature of the interior of the slab. The surface cools faster than the interior through radiation and through the effect of the air and the rolls, so the surface may really be tens of degrees cooler.

Assumptions of the model. The pyrometer assumes the slab is a black body. If the steel radiated at every wavelength the same fraction of the radiation of a black body, the spectrum would simply be lower with the peak in the same place, and the result would not change. However, the emissivity of real steel and of the scale layer on its surface depends on the wavelength, so the shape of the spectrum and the position of the peak differ from the spectrum of a black body. The temperature given by the model is therefore only approximate.

Conclusion. The results do not conflict: the difference is within the reading accuracy and is also explained by the cooling of the surface. The pyrometer cannot be said to read wrongly. For a more accurate comparison the pyrometer should be calibrated with an object at a known temperature.

e)

Given

refractive index of the film $n = 1.38$, of the glass $1.50$, of air $1.00$ and wavelength $\lambda = 2.0\,\mu\text{m}$; the light arrives perpendicularly

Find

the thickness $d$ of the thinnest anti-reflection film and whether the greenish reflection is a sign of a wrong film

Principle

The waves reflected from the upper and lower surfaces of the film interfere. When the light arrives perpendicularly, the wave reflected from the lower surface travels an extra distance $2d$ in the film, where the wavelength is $\lambda : n$, so the path difference is $2nd$ measured in wavelengths in air. Both reflections take place against a material with a higher refractive index ($1.00 \to 1.38 \to 1.50$), so each reverses the wave and the reversals cancel each other. The waves then weaken each other when $2nd = (k + \tfrac{1}{2})\lambda$ and reinforce each other when $2nd = k\lambda$. The thinnest weakening film is obtained with $k = 0$.

Formula

$2nd = \tfrac{1}{2}\lambda$, so $d = \dfrac{\lambda}{4n}$

Conversions

$\lambda$$= 2.0\,\mu\text{m}$
$= 2.0 \cdot 10^{-6}\,\text{m}$

Substitution

$d$$= \dfrac{2.0 \cdot 10^{-6}\,\text{m}}{4 \cdot 1.38}$
$= 3.62 \cdot 10^{-7}\,\text{m}$
$= 362\,\text{nm}$

Answer

The thinnest film is $d \approx 0.36\,\mu\text{m}$, that is, about $360\,\text{nm}$ thick, two significant figures like the wavelength. A greenish reflection is not a sign of a wrong film; it is to be expected with a film designed for the infrared.

Reasonableness

The film is a quarter of the wavelength in the film: $2000\,\text{nm} : 1.38 = 1449\,\text{nm}$, and a quarter of that is $362\,\text{nm}$, as it should be for an anti-reflection film. The path difference is

$2nd$$= 2 \cdot 1.38 \cdot 362\,\text{nm}$
$= 1000\,\text{nm}$

In the visible range it is twice $500\,\text{nm}$, so greenish light of $500\,\text{nm}$ is reinforced ($2nd = 2\lambda$). Red light of $667\,\text{nm}$ ($2nd = 1.5\lambda$) and violet of $400\,\text{nm}$ ($2nd = 2.5\lambda$) are weakened. That is why the window reflects greenish light, and the installer's suspicion is unfounded.

Also accepted

  • In part a the temperature can be converted to degrees Celsius by subtracting $273$; the answer $1180\,{}^\circ\text{C}$ or $1176\,{}^\circ\text{C}$ is accepted.
  • In part c the angle $\beta$ can first be calculated from the condition $\cos\beta = \sin\alpha_\text{r}$, or directly from the equation $\sin\theta = \sqrt{n_1^2 - n_2^2}$, which follows from the same laws.
  • In part d a justified estimate of the reading accuracy between $\pm 0.05$ and $0.1\,\mu\text{m}$ is accepted.

Common mistakes

  • The temperature is calculated in degrees Celsius from Wien's law, or the wavelength is not converted to metres.
  • The wavelength of the peak is read in the range of visible light because the slab glows.
  • In part b it is claimed that infrared radiation is seen as red.
  • In part c the angle of refraction at the end face is used directly as the angle of incidence at the side surface, or the critical angle is calculated for the interface between air and the core.
  • In part c the refractive indices are swapped in the expression for the critical angle, so the value of the sine comes out greater than $1$.
  • In part d it is stated that the pyrometer reads wrongly without comparing the difference with the uncertainty of the measurement.
  • In part d it is said that a radiation weaker than that of a black body shifts the peak, although the peak shifts only if the emissivity depends on the wavelength.
  • In part e it is assumed that only one reflection reverses the wave, as in a soap film, so the thickness comes out twice as large, $\lambda : (2n) \approx 725\,\text{nm}$.
  • In part e the refractive index is forgotten, so the thickness comes out as $\lambda : 4 = 500\,\text{nm}$.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)3 p

The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. The peak is on a grid line: $\lambda_\text{max} = 2.0\,\mu\text{m}$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\lambda_\text{max} T = b$, so $T = \dfrac{b}{\lambda_\text{max}}$ 1 p
The answer has the correct unit and a precision that matches the given data. $T \approx 1450\,\text{K}$, that is, about $1180\,{}^\circ\text{C}$ 1 p

b)3 p

Explains that the spectrum is continuous: a small part of the radiation is in the range of visible light, even though the peak is in the infrared. 1 p
Explains that in the visible range the intensity increases with increasing wavelength, so there is much more red and orange light than blue. 1 p
States that the hot slab radiates with such a high power that even a small visible fraction is enough for the eye; an object at room temperature does not glow, because there is practically no visible light in its spectrum. 1 p

c)4 p

Sees the geometry of the two interfaces: at the end face the light is refracted to an angle $\beta$, and since the end face is perpendicular to the side surface of the core, the angle of incidence on the side surface is $90^\circ - \beta$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\sin\alpha_\text{r} = \dfrac{n_2}{n_1}$ and $1.00 \cdot \sin\theta = n_1\sin\beta$, where $\beta = 90^\circ - \alpha_\text{r}$ 1 p
The numerical values are substituted with their units, and the calculation is correct. $\beta \approx 9.56^\circ$ and $\sin\theta \approx 0.239$ 1 p
The answer has the correct unit and a precision that matches the given data. $\alpha_\text{r} \approx 80.4^\circ$ and $\theta \approx 13.8^\circ$ 1 p

d)4 p

The uncertainty of the measurement has been estimated. The peak is flat, so its position can be read to an accuracy of about $\pm 0.05\,\mu\text{m}$; in temperature this corresponds to about $\pm 36\,\text{K}$. 1 p
The main sources of error are named, and it is explained how they affect the result. The surface cools faster than the interior of the slab, which the thermocouple measures, through radiation and the effect of the rolls and the air, so the surface may really be cooler. 1 p
The assumption is named, and its effect on the result is justified. The emissivity of steel and of the scale layer on its surface depends on the wavelength, so the shape of the spectrum and the position of the peak differ from the spectrum of a black body, and the temperature given by the model is only approximate. Merely a radiation weaker than that of a black body, by the same fraction at every wavelength, would not shift the peak. 1 p
The conclusion answers the question and is based on the data. The difference, about $34\,\text{K}$, is within the reading accuracy and is also explained by the cooling of the surface, so the pyrometer cannot be said to read wrongly. 1 p

e)3 p

Justifies that both reflections, from air to film and from film to glass, take place against a material with a higher refractive index, so the phase shifts cancel each other and the reflected waves weaken each other when $2nd = (k + \tfrac{1}{2})\lambda$; for the thinnest film $k = 0$. 1 p
The answer has the correct unit and a precision that matches the given data. $d = \dfrac{\lambda}{4n} \approx 362\,\text{nm} \approx 0.36\,\mu\text{m}$ 1 p
Explains the green reflection: the path difference $2nd = 1000\,\text{nm}$ is twice $500\,\text{nm}$, so green light is reinforced in the reflection ($2nd = k\lambda$, $k = 2$). The film is therefore not wrong but designed for the infrared. 1 p

Self-assessed points: 0 / 17 (17 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).

Spectrum of a hydrogen lamp 17 points

Simulated example data · Wavelengths: NIST Atomic Spectra Database (2024), wavelengths in air. The angles have been calculated from the model of the interference of light for an invented grating and rounded to the reading accuracy of the goniometer, $0.1^\circ$, so they contain variation consistent with that accuracy.

A group of students studied the light of a hydrogen lamp with a grating spectrometer. The light passes through a narrow slit and a lens and strikes the grating perpendicularly, and the directions of the maxima are measured with a rotating telescope. The line density of the grating had been marked on the device, but the marking had worn illegible. The group measured the directions $\theta$ of the maxima of four spectral lines of hydrogen relative to the direction of the incident light. The wavelengths $\lambda$ of the lines are from a data book.

line $\lambda\ (\text{nm})$ order $\theta$
red $656.28$ $1$ $23.2^\circ$
blue-green $486.13$ $1$ $17.0^\circ$
blue $434.05$ $1$ $15.1^\circ$
violet $410.17$ $1$ $14.2^\circ$
red $656.28$ $2$ $52.0^\circ$
blue-green $486.13$ $2$ $35.7^\circ$

The group also noticed a yellow line in the spectrum that does not belong to hydrogen. Its first-order maximum was at an angle of $20.7^\circ$. The reading accuracy of the goniometer is $0.1^\circ$.

  1. a) From the measurements, form a graph such that, according to the model, the same straight line through the origin fits all six points. 5 p
  2. b) From the graph, determine the line density of the grating, that is, the number of lines per millimetre. 3 p
  3. c) What is the wavelength of the yellow line? 3 p
  4. d) According to the data book, the yellow line of helium has a wavelength of $587.56\,\text{nm}$ and the yellow lines of sodium have wavelengths of $588.99\,\text{nm}$ and $589.59\,\text{nm}$. The group concludes that the lamp contains sodium as an impurity. Evaluate the conclusion. How could the measurement be improved so that helium and sodium could be told apart? 3 p
  5. e) A student remarks: “In a glass prism violet light is deviated from its direction the most, but in a grating red light. One of the observations must be wrong.” Evaluate the claim. 3 p
Show the model answer and marking Open only when your own solution is ready.

Model answer

a)

Find

a graph such that the same straight line through the origin fits all six points

Graph

At a grating a maximum forms in the direction in which the path difference $d\sin\theta$ of the waves from adjacent slits is a multiple of the wavelength: $d\sin\theta = k\lambda$. From this, $\sin\theta = \dfrac{1}{d} \cdot k\lambda$. When $k\lambda$ is chosen for the horizontal axis and $\sin\theta$ for the vertical axis, the points of both the first and the second order are on the same straight line through the origin, whose gradient is $\tfrac{1}{d}$. The wavelength alone on the horizontal axis would give two different lines.

line $k$ $k\lambda\ (\text{nm})$ $\sin\theta$
red $1$ $656$ $0.394$
blue-green $1$ $486$ $0.292$
blue $1$ $434$ $0.261$
violet $1$ $410$ $0.245$
red $2$ $1313$ $0.788$
blue-green $2$ $972$ $0.584$

For example, for the second-order red line $k\lambda = 2 \cdot 656.28\,\text{nm} = 1313\,\text{nm}$ and $\sin 52.0^\circ = 0.788$.

$400$
$800$
$1200$
$0.2$
$0.4$
$0.6$
$0.8$
$0$
$k\lambda\ (\text{nm})$
$\sin\theta$
$(1300;\ 0.780)$

Model answer

The horizontal axis shows $k\lambda$ in nanometres and the vertical axis the dimensionless $\sin\theta$, both with evenly spaced scales starting from zero. Each measurement is a point of its own. A straight line through the origin is fitted to the points, because according to the model $\sin\theta = 0$ when $k\lambda = 0$ (the central maximum). The line passes through all six points within the reading accuracy, so the model fits the data and there is no deviation at the second-order points.

b)

Given

the graph of part a and its line

Find

the line density of the grating, that is, the number of lines per millimetre

Principle

According to the model, the gradient of the line is $\tfrac{1}{d}$, where $d$ is the grating spacing, that is, the distance between adjacent slits. The gradient is read from the line using two points far from each other; one of them can be the origin, because the line passes through it. The line density is a millimetre divided by the grating spacing.

Reading the graph

The point $(1300\,\text{nm}$; $0.780)$, which is far from the origin, is read from the line. The gradient is $0.780 : 1300\,\text{nm} = 6.00 \cdot 10^{-4}\,\tfrac{1}{\text{nm}}$.

Formula

$\dfrac{1}{d} = \text{gradient}$, so $d = \dfrac{1}{\text{gradient}}$, and the line density is $N = \dfrac{1\,\text{mm}}{d}$.

Conversions

$1\,\text{mm} = 10^{6}\,\text{nm}$

Substitution

$d$$= \dfrac{1}{6.00 \cdot 10^{-4}\,\tfrac{1}{\text{nm}}}$
$= 1667\,\text{nm}$
$N$$= \dfrac{10^{6}\,\text{nm}}{1667\,\text{nm}}$
$= 600$

Answer

$d \approx 1.67\,\mu\text{m}$

So the grating has about $600$ lines per millimetre, three significant figures like the point read from the graph.

Reasonableness

Gratings for teaching use typically have $100$–$600$ lines per millimetre, so the result is a common one. The grating spacing is about three times the wavelength of visible light, so the first-order maxima are at angles of about $15$–$25^\circ$, as measured.

c)

Given

the first-order maximum of the yellow line at the angle $\theta = 20.7^\circ$ and the grating spacing $d = 1667\,\text{nm}$ (part b)

Find

the wavelength $\lambda$ of the yellow line

Principle

The same grating and the same condition for maxima apply to all light, so an unknown wavelength is obtained from the measured angle when the grating spacing is known.

Formula

$d\sin\theta = k\lambda$, so when $k = 1$, $\lambda = d\sin\theta$.

Conversions

Not needed: the grating spacing is in nanometres, so the wavelength is obtained in nanometres.

Substitution

$\lambda$$= 1667\,\text{nm} \cdot \sin 20.7^\circ$
$= 1667\,\text{nm} \cdot 0.3535$
$= 589.1\,\text{nm}$

Answer

$\lambda \approx 589\,\text{nm}$

Three significant figures like the angle.

Reasonableness

The angle is between those of the red ($23.2^\circ$) and blue-green ($17.0^\circ$) lines, and the wavelength is correspondingly between theirs, in the range of yellow light. The same result is obtained from the graph: with $\sin 20.7^\circ \approx 0.354$, $k\lambda \approx 590\,\text{nm}$ is read from the line.

d)

Find

whether the conclusion about sodium is justified and how the measurement could be improved

Principle

Two alternatives can be told apart by a measurement only if the difference between them is greater than the uncertainty of the measurement. The uncertainty of the angle is carried over to the wavelength through the grating equation.

Model answer

Uncertainty. The reading accuracy of the angle is $0.1^\circ$. We calculate the wavelengths from the angles $20.6^\circ$ and $20.8^\circ$: $1667\,\text{nm} \cdot \sin 20.6^\circ \approx 586.5\,\text{nm}$ and $1667\,\text{nm} \cdot \sin 20.8^\circ \approx 592.0\,\text{nm}$. The uncertainty of the wavelength is therefore about $\pm 3\,\text{nm}$, that is, $\lambda = (589 \pm 3)\,\text{nm}$. The grating spacing also has a small uncertainty, which only adds to this.

Evaluating the conclusion. The helium line $587.56\,\text{nm}$ and the sodium lines $588.99\,\text{nm}$ and $589.59\,\text{nm}$ all lie between $586$ and $592\,\text{nm}$. The difference between helium and sodium, about $1.4\,\text{nm}$, is smaller than the uncertainty. In terms of angles the difference is only $20.64^\circ$ against $20.70^\circ$, that is, about $0.05^\circ$, which is below the reading accuracy of the goniometer. The measurement therefore does not distinguish sodium from helium, and the group's conclusion is not justified by this measurement. The lamp may contain sodium, but that does not follow from this measurement.

Improvements. From the condition for maxima $\sin\theta = \dfrac{k\lambda}{d}$ we see that when the wavelength changes, the sine of the angle changes $k$ times as much. In the second order the maxima of helium and sodium are at $44.84^\circ$ and $44.98^\circ$, so the difference is about $0.14^\circ$, almost three times that in the first order, and the uncertainty in the wavelength is about $\pm 1\,\text{nm}$. The second order would make it easier to tell them apart, but on its own it is barely sufficient: the angular difference is only slightly larger than the reading accuracy of $0.1^\circ$, and $\pm 1\,\text{nm}$ is almost as large as the $1.4\,\text{nm}$ separation of the lines. A grating with more lines, in which the angles are larger, or a more accurate goniometer is therefore also needed. In addition, the angle can be measured on both sides of the central maximum and half of the angle between them used, which removes the error caused by tilting of the grating.

e)

Find

whether either observation is wrong when in a prism violet and in a grating red is deviated the most

Principle

A grating and a prism separate colours by different phenomena. In a grating the direction is determined by the interference of waves, in a prism by refraction, which depends on the refractive index.

Model answer

Grating. A maximum forms in the direction in which the waves from adjacent slits meet in the same phase, that is, their path difference $d\sin\theta$ is a multiple of the wavelength. A longer wavelength requires a larger path difference and therefore a larger angle, so red light is deviated the most and violet the least, as in the table.

Prism. Light is refracted when it enters the glass and when it leaves it. The refractive index of glass depends on the wavelength (dispersion): it is greater for violet light than for red, because violet light travels more slowly in glass. Violet is therefore refracted the most.

Conclusion. The claim is wrong: both observations are correct. The colours are separated by different phenomena, in the grating by interference and in the prism by refraction, so the order of the colours can be the opposite.

Also accepted

  • In part a the horizontal axis can be chosen to show $k\lambda$ in metres or micrometres; the gradient is then in the corresponding unit.
  • The gradient in part b can also be read from two other points of the line far from each other. The answer $590$–$610$ lines per millimetre is accepted.
  • The wavelength in part c can also be read from the graph, once $\sin 20.7^\circ$ has first been calculated for the vertical axis.
  • The uncertainty in part d can also be estimated from the difference of the angles: the difference between the maxima of helium and sodium is only about $0.05^\circ$, which is below the reading accuracy.

Common mistakes

  • Only the wavelength is chosen for the horizontal axis of the graph, so the second-order points lie on a different line.
  • The angles are plotted in the graph in degrees, although the model is linear in the sine of the angle.
  • The gradient is read from a single measured point instead of the fitted line.
  • The line density is confused with the grating spacing, or nanometres are converted incorrectly to millimetres.
  • In part d the result $589\,\text{nm}$ is accepted as sodium without comparing the difference between helium and sodium with the uncertainty of the measurement.
  • In part e it is claimed that in a grating, too, the refractive index causes the separation of the colours.

Marking and self-assessment

Compare your solution with the model answer and, for each item, choose the points you think your solution earns. Part marks are possible, and any other physically correct method is accepted too.

a)5 p

Justifies the quantities of the graph from the condition for maxima $d\sin\theta = k\lambda$: when $k\lambda$ is on the horizontal axis and $\sin\theta$ on the vertical axis, the points of both orders are on the same straight line, whose gradient is $\tfrac{1}{d}$. 1 p
The numerical values are substituted with their units, and the calculation is correct. The sines of the angles and the products $k\lambda$ have been calculated, for example for the second-order red line $\sin 52.0^\circ \approx 0.788$ and $k\lambda \approx 1313\,\text{nm}$. 1 p
The axes show the correct quantities and units, and the scales are evenly spaced. 1 p
Each measurement is marked as a separate point in the right place. 1 p
The model is fitted to the points it suits, and the points have not been joined into a broken line. A straight line through the origin fits all six points. 1 p

b)3 p

The slope or other quantity is read visibly from the graph, using two points on the line that are far apart. The gradient $\approx 6.00 \cdot 10^{-4}\,\tfrac{1}{\text{nm}}$ has been read from a point of the line far from the origin, for example $(1300\,\text{nm}$; $0.780)$. 1 p
The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. The gradient is $\tfrac{1}{d}$, so $d = \tfrac{1}{\text{gradient}}$, and the line density is $1\,\text{mm} : d$. 1 p
The answer has the correct unit and a precision that matches the given data. $d \approx 1.67\,\mu\text{m}$, so there are about $600$ lines per millimetre. 1 p

c)3 p

The starting equation is correct, and the quantity asked for is solved from it before the numbers are substituted. $\lambda = d\sin\theta$ when $k = 1$ 1 p
The numerical values are substituted with their units, and the calculation is correct. $\lambda = 1667\,\text{nm} \cdot \sin 20.7^\circ$ 1 p
The answer has the correct unit and a precision that matches the given data. $\lambda \approx 589\,\text{nm}$ 1 p

d)3 p

The uncertainty of the measurement has been estimated. A reading accuracy of $0.1^\circ$ corresponds in the first order to an uncertainty of about $\pm 3\,\text{nm}$ in the wavelength. 1 p
The conclusion answers the question and is based on the data. The result fits sodium but also helium, because the difference between them, about $1.4\,\text{nm}$, is smaller than the uncertainty: the conclusion is not justified by this measurement. 1 p
Proposes a justified improvement: in the second order $\sin\theta = \dfrac{k\lambda}{d}$ changes twice as much with the wavelength and the difference in angle grows to about $0.14^\circ$, but this is only slightly larger than the reading accuracy, so a more accurate goniometer or a grating with more lines is also needed. 1 p

e)3 p

Explains that in a grating a maximum forms in the direction where the path difference of the waves coming from adjacent slits is a multiple of the wavelength; a longer wavelength requires a larger angle, so red is deviated the most. 1 p
Explains that in a prism light is refracted and the refractive index of glass is greater for violet light than for red, so violet is refracted the most. 1 p
The conclusion answers the question and is based on the data. The claim is wrong: both observations are correct, because the colours are separated by different phenomena, in the grating by interference and in the prism by refraction. 1 p

Self-assessed points: 0 / 17 (17 items not assessed)

The points are your own estimate, not a checked grade. The marking is Ruukkilabra's own, written according to the features of a good matriculation examination answer, and it is not official marking by the Matriculation Examination Board (YTL).