Answer key · Unit 29

Seeing and sensing.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 5

Lesson 1 of 5 · About 35 minutes

Lasers

A ruby laser: photons of one energy, a threshold above which a beam appears, stimulated emission that copies photons, and a mirror that leaks too much.

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The 9 steps and their answers

Step 1 of 9
Open this step
One colour
Practice · Calculate
Question
The ruby laser gives out red light with a wavelength of 694.3 nm. What is the energy of one of its photons, in electronvolts?

Correct answer
1.786 eV ± 0.02 eV
Setup, choices and feedback
The Scene at this step
Pump power
4 kW
Output mirror
90 %
Show the photons
Off
Accepted within
± 0.02 eV
Computed from the Scene at this step’s setting
Energy of one photon, hc ÷ λ/nodes/laser/outputs/photonEnergy
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 1.79 eV. That is exactly the energy a chromium ion loses when it falls from the excited level to the ground state, which is why every photon of the laser has the same colour.

Hint
$E = \dfrac{hc}{\lambda}$ ⁠, and 1 eV = 1.602 × 10⁻¹⁹ J.

Worked solution
1. $E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{694.3 \times 10^{-9}}$ ⁠.
2. $E \approx 2.86 \times 10^{-19}$ ⁠ J.
3. Dividing by 1.602 × 10⁻¹⁹ J/eV: about 1.79 eV.
Remember to turn nanometres into metres: 694.3 nm = 6.943 × 10⁻⁷ m.
Step 2 of 9
Open this step
The threshold
Practice · Predict, then look
Question
At a Pump power of 1.5 kW the ruby only glows dimly. What will happen when the pump is raised to 4 kW?

Correct answer
- Correct A narrow, bright beam will appear out of the mirror
Setup, choices and feedback
The Scene at this step
Pump power
1.5 kWfixed
Output mirror
90 %
Show the photons
Off
Choices
- Correct A narrow, bright beam will appear out of the mirror
- Wrong The ruby will glow brighter, in every direction
  Feedback
  The glow does grow, but only up to the threshold. Above it each extra excited ion is drawn straight into the beam, not the glow.
- Wrong Nothing will change
  Feedback
  At 4 kW the pump is already above this mirror’s threshold, about 2.3 kW.
Once the prediction is locked in, the Scene changes
Pump power
4 kWfixed
The Scene then shows: 8.808 W · Power in the beam through the output mirror/nodes/laser/outputs/output
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Above about 2.3 kW more than half the ions are excited, and light passing along the ruby grows instead of being absorbed. At 4 kW about 9 W leaves in the beam.

Worked solution
1. An ion in the ground state absorbs red light, and an excited ion amplifies it by stimulated emission.
2. Below the threshold most ions are in the ground state, so light travelling along the ruby is absorbed more than it grows.
3. Above the threshold, with more than half the ions excited, each round trip between the mirrors amplifies the light more than it loses.
4. The light builds itself into one beam along the axis, and leaves through the output mirror.
Step 3 of 9
Open this step
Copies of a photon
Practice · Choose
Question
What makes a laser’s light different from a lamp’s?

Correct answer
- Correct Most photons are copies of the photon that triggered them: the same direction, colour and phase
Setup, choices and feedback
The Scene at this step
Pump power
4 kW
Output mirror
90 %
Show the photons
On
Choices
Shown in a random order.
- Correct Most photons are copies of the photon that triggered them: the same direction, colour and phase
- Wrong The laser’s lamp is very bright, and the mirrors gather its light into a beam
  Misconception: Thinks a laser is just a very bright lamp whose light the mirrors gather into a beam.
  Feedback from the catalogue
  A lamp, however bright, gives out photons each at its own moment, in its own direction and with its own phase. In a laser most photons are made by stimulated emission: a photon passing an excited ion makes it give out a copy of itself, in the same direction, with the same wavelength and the same phase. The mirrors send the photons back and forth so that they trigger more and more copies, and the beam that leaves is narrow and pure in colour.
- Wrong The ruby filters out every colour but red
  Feedback
  Ruby is coloured, but a filter only takes light away. It cannot make a narrow beam whose photons all keep in step.
Feedback for a right answer
That is what the name LASER says: light amplification by stimulated emission of radiation. The photons in the beam keep in step, so it hardly spreads and can be focused to a tiny spot.

Hint
Choose On under Show the photons and compare the light along the axis with the light leaving sideways.

Worked solution
1. A lamp, and the ruby’s glow, give out light by spontaneous emission: each ion at a random time, in a random direction.
2. A photon passing an excited ion can make it give out an identical copy: stimulated emission.
3. The mirrors send back the photons moving along the axis, and they trigger more copies on every pass.
4. So the beam is made of copies: narrow, one colour, in one phase.
Step 4 of 9
Open this step
The output mirror
Practice · Predict, then look
Question
The Pump power is fixed at 3 kW, and the laser runs with a 98% Output mirror. Swap it for a 60% mirror, which lets far more of the light out. What will happen?

Correct answer
- Correct The beam will vanish
Setup, choices and feedback
The Scene at this step
Pump power
3 kWfixed
Output mirror
98 %fixed
Show the photons
Off
Choices
- Correct The beam will vanish
- Wrong The beam will grow, because more light leaves on each pass
  Feedback
  More light leaves on each pass, but less comes back to amplify itself. For the light to grow on each pass, more ions must now be excited, and that raises the threshold.
- Wrong Nothing will change: the pump has not changed
  Feedback
  The pump is the same, but the threshold has moved. Above the threshold or below it is the difference between a beam and a glow.
Once the prediction is locked in, the Scene changes
Output mirror
60 %fixed
The Scene then shows: 0.0001101 W · Power in the beam through the output mirror/nodes/laser/outputs/output
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With a 60% mirror the threshold rises to about 3.3 kW, above 3 kW, and the beam goes out. A mirror that leaks too much loses more light on each pass than the ruby can make up.

Worked solution
1. The laser works only if on each round trip the gain in the ruby beats the loss at the mirrors.
2. A 60% mirror loses 40% of the light at each bounce, instead of 2%.
3. Making up such a loss takes more excited ions, which takes a stronger pump: the threshold rises to about 3.3 kW.
4. The pump, 3 kW, is below the threshold, so there is no beam.
Step 5 of 9
Open this step
A stream of photons
Practice · Calculate
Question
At a Pump power of 4 kW about 8.8 W leaves in the beam. Each photon carries about 2.86 × 10⁻¹⁹ J. How many photons leave in the beam each second?

Correct answer
30790000000000000000 ± 3%
Setup, choices and feedback
The Scene at this step
Pump power
4 kWfixed
Output mirror
90 %fixed
Show the photons
Off
Accepted within
± 3%
Computed from the Scene at this step’s setting
Photons leaving in the beam each second/nodes/laser/outputs/photonRate
Feedback for a right answer
About $3 \times 10^{19}$ ⁠ photons a second. Even a beam of a few watts is a huge stream of photons, which is why you never see them one by one.

Hint
A watt is a joule a second.

Worked solution
1. Each second 8.8 J leaves.
2. The number of photons: $\dfrac{8.8}{2.86 \times 10^{-19}} \approx 3.1 \times 10^{19}$ ⁠.
A ruby laser is very inefficient: of the 4000 W going into the lamp, only about 9 W leaves in the beam.
Step 6 of 9
Open this step
How fast it swings
Practice · Calculate
Question
The ruby laser’s red light has a wavelength of 694.3 nm, and the speed of light is 3.00 × 10⁸ m/s. What is its frequency?

Correct answer
431.8 THz ± 2%
Setup, choices and feedback
The Scene at this step
Pump power
4 kW
Output mirror
90 %
Show the photons
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
The light's frequency, c ÷ λ/nodes/laser/outputs/frequency
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 432 THz, 4.32 × 10¹⁴ Hz.

Hint
$c = f \lambda$ ⁠. First turn the wavelength into metres.

Worked solution
1. $\lambda = 694.3 \times 10^{-9}$ ⁠ metres.
2. $f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8}{6.943 \times 10^{-7}} \approx 4.32 \times 10^{14}$ ⁠ hertz.
3. That is about 432 THz: more than four hundred million million swings a second.
In a laser every photon has exactly the same frequency. An ordinary lamp gives a mixture of many.
Step 7 of 9
Open this step
Held at threshold
Practice · Predict, then look
Question
The Pump power is at 3 kW, above threshold, and the laser is running. About 54% of the chromium ions are in the upper level. Predict: what will happen to that share when the pump is raised to 6 kW?

Correct answer
- Correct It will stay almost the same
Setup, choices and feedback
The Scene at this step
Pump power
3 kW
Output mirror
90 %fixed
Show the photons
On
Choices
- Wrong It will grow a lot, to almost all the ions, because twice the light arrives
  Misconception: Thinks pumping a running laser harder keeps raising the share of atoms in the upper level.
  Feedback from the catalogue
  Below threshold, pumping harder does excite more ions. But once the laser runs, every extra ion up top strengthens the beam, and the beam brings ions back down by stimulated emission. The excited share stops where the gain balances the losses, and all the extra pump comes out as light.
- Correct It will stay almost the same
- Wrong It will fall, because the ions get hot
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Once the prediction is locked in, the Scene changes
Pump power
6 kW
The Scene then shows: 0.5353 · Share of the chromium ions in the upper laser level/nodes/laser/outputs/excited
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Almost the same, about 54%. The extra pump goes out entirely in the stronger beam.

Hint
Each extra ion in the upper level adds to the gain. What happens to the gain in a laser that is already running?

Worked solution
1. Above threshold, each extra ion in the upper level makes the beam stronger, and the stronger beam brings ions back down by stimulated emission.
2. So the number of excited ions stops exactly where the gain balances the leak through the mirror.
3. All the pump’s extra energy goes out in the beam, not into excited ions: the beam grows, the share stays at about 54%.
In ruby more than half the ions must be up for there to be any gain, because the lower level is the ground state.
Step 8 of 9
Open this step
How much gets out
Practice · Calculate
Question
At a Pump power of 4 kW about 8.8 W leaves in the beam. What percentage of the lamp’s power becomes laser light?

Correct answer
0.2202 ± 3%
Setup, choices and feedback
The Scene at this step
Pump power
4 kWfixed
Output mirror
90 %fixed
Show the photons
Off
Accepted within
± 3%
Computed from the Scene at this step’s setting
Beam power ÷ lamp power/nodes/laser/outputs/efficiency
Wrong answers it recognises
- 0.0022 ± 5%
  Misconception: Gives a share as a decimal where a percentage is asked for.
  Feedback from the catalogue
  A ratio like $\dfrac{8.8}{4000} = 0.0022$ ⁠ is a share of the whole. To write it as a percentage, multiply it by $100$ ⁠: $0.22\%$ ⁠.
Feedback for a right answer
About 0.22%: less than one watt in every four hundred.

Hint
Divide the power in the beam by the lamp’s power, in the same units, and multiply by $100$ ⁠.

Worked solution
1. $\dfrac{8.8}{4000} = 0.0022$ ⁠.
2. As a percentage: $0.22\%$ ⁠.
Most of the lamp’s white light is in colours the ruby does not absorb, and most of what is absorbed becomes heat. That is why ruby lasers need cooling, and why modern lasers are pumped by diodes that shine in exactly the right colour.
Step 9 of 9
Open this step
On the edge
Practice · Reach the target
Question
The Pump power is fixed at 2.5 kW. A leakier mirror lets more of the light out, but raises the threshold. Find the Output mirror at which 2.5 kW is exactly the threshold: the laser only just comes on. Press Check.

Goal
- 2500 ± 30
  At the start of the step: 2304
Setup, choices and feedback
The Scene at this step
Pump power
2.5 kWfixed
Output mirror
90 %
Show the photons
Off
Read from the Scene at
Lamp power at which it starts to lase/nodes/laser/outputs/threshold
One setting that reaches it
Output mirror
81 %
Feedback for a right answer
At a mirror of about 81–82%: any leakier, and the laser goes out.

Hint
Lower the mirror slowly, and watch for the beam to vanish.

Worked solution
1. With a 90% mirror the threshold is about 2.3 kW, and 2.5 kW is enough.
2. The leakier the mirror, the fewer round trips each photon makes between the mirrors, and the more the ions must amplify on each pass. So the threshold rises.
3. At about 81–82% the threshold reaches 2.5 kW, and below that the beam goes out.
Laser designers choose the mirror that gives the most light from the pump they have: not too tight, and not too leaky.

Lesson 2 of 5 · About 25 minutes

Telescopes

A telescope does three things: it gathers light, separates detail and magnifies. The eyepiece sets the magnification, but only the aperture sets how much light and how much detail, because of diffraction.

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The 5 steps and their answers

Step 1 of 5
Open this step
Magnification
Practice · Calculate
Question
The telescope has a focal length of 1000 mm, and the Eyepiece focal length is 25 mm. What is the magnification?

Correct answer
40 ± 0.1
Setup, choices and feedback
The Scene at this step
Aperture
100 mm
Eyepiece focal length
25 mmfixed
Telescope
Refractor
Show the light path
Off
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Objective focal length ÷ eyepiece focal length/nodes/scope/outputs/magnification
Wrong answers it recognises
- 0.025 ± 0.002
  Misconception: Divides the eyepiece's focal length by the objective's to get the magnification.
  Feedback from the catalogue
  The magnification is the focal length of the main lens or mirror divided by the eyepiece’s: $M = \dfrac{f_{\text{obj}}}{f_{\text{eye}}}$ ⁠. A short eyepiece gives a high magnification. If you got a number below 1, you turned the fraction over, and such a telescope would shrink things.
Feedback for a right answer
40 times: the Moon, half a degree across, looks 20° across. A different eyepiece gives a different magnification, so one telescope has many.

Hint
$M = \dfrac{f_{\text{obj}}}{f_{\text{eye}}}$ ⁠.

Worked solution
1. $f_{\text{obj}} = 1000$ ⁠ mm, $f_{\text{eye}} = 25$ ⁠ mm.
2. $M = \dfrac{1000}{25} = 40$ ⁠.
If you got 0.025, you divided the other way round.
Step 2 of 5
Open this step
Gathering light
Practice · Calculate
Question
A dark-adapted eye’s pupil is about 7 mm wide. The telescope’s Aperture is 250 mm. How many times more light does it gather than the eye?

Correct answer
1276 ± 2%
Setup, choices and feedback
The Scene at this step
Aperture
250 mmfixed
Eyepiece focal length
25 mm
Telescope
Refractor
Show the light path
Off
Accepted within
± 2%
Computed from the Scene at this step’s setting
Light gathered, times a dark-adapted eye's 7 mm pupil: (D ÷ 7 mm)²/nodes/scope/outputs/lightGrasp
Wrong answers it recognises
- 35.7 ± 2%
  Misconception: Thinks a telescope gathers light in proportion to its diameter rather than its area.
  Feedback from the catalogue
  A telescope gathers light over the whole area of its lens or mirror, and a circle’s area grows with the square of its diameter. An aperture twice as wide gathers four times the light.
Feedback for a right answer
About 1300 times. That is why such a telescope shows stars a thousand times fainter than the eye sees, down to nearly magnitude 15.

Hint
Light is gathered over an area, and a circle’s area goes with the square of its diameter.

Worked solution
1. The ratio of the diameters: $\dfrac{250}{7} \approx 35.7$ ⁠.
2. The ratio of the areas, and so of the light: $35.7^2 \approx 1276$ ⁠.
If you got 35.7, you compared diameters rather than areas.
Step 3 of 5
Open this step
The diffraction limit
Practice · Calculate
Question
An Aperture of 100 mm, and green light of 550 nm. Diffraction smears each star into a small disc of angular radius $\theta = 1.22\dfrac{\lambda}{D}$ ⁠ in radians. What is that radius in arcseconds? (A radian is about 206,265 arcseconds.)

Correct answer
1.384 arcsec ± 0.03 arcsec
Setup, choices and feedback
The Scene at this step
Aperture
100 mmfixed
Eyepiece focal length
10 mm
Telescope
Refractor
Show the light path
Off
Accepted within
± 0.03 arcsec
Computed from the Scene at this step’s setting
Radius of a star's diffraction disc at 550 nm, 1.22 λ ÷ D, in degrees (1″ = 1/3600°)/nodes/scope/outputs/airyRadius
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About 1.4 arcseconds. It is the same diffraction you saw at a narrow slit, only at a round opening: the wider the aperture, the smaller the disc, and the finer the detail the telescope separates.

Hint
Turn both lengths into metres before dividing.

Worked solution
1. $\lambda = 550 \times 10^{-9}$ ⁠ m, $D = 0.1$ ⁠ m.
2. $\theta = 1.22 \times \dfrac{550 \times 10^{-9}}{0.1} \approx 6.7 \times 10^{-6}$ ⁠ rad.
3. In arcseconds: $6.7 \times 10^{-6} \times 206\,265 \approx 1.38$ ⁠.
A pair of stars closer than this merges into one blob, at any magnification.
Step 4 of 5
Open this step
Empty magnification
Practice · Predict, then look
Question
An Aperture of 50 mm at 100×: each close pair of the Double Double, 2.3″ apart, shows as one smeared star. Switch to a 5 mm eyepiece, 200×. What will you see?

Correct answer
- Correct A bigger blob, still not two stars
Setup, choices and feedback
The Scene at this step
Aperture
50 mmfixed
Eyepiece focal length
10 mmfixed
Telescope
Refractor
Show the light path
Off
Choices
- Correct A bigger blob, still not two stars
- Wrong Two separate stars
  Misconception: Thinks a higher magnification always shows finer detail.
  Feedback from the catalogue
  A lens or mirror $D$ ⁠ wide cannot separate two points closer than $1.22\dfrac{\lambda}{D}$ ⁠, because of diffraction. Magnification only enlarges the image it makes. If two stars have already merged into one blob, more magnification gives a bigger blob, not two stars. Only a wider aperture separates them.
- Wrong Nothing: at that magnification the star leaves the field
  Feedback
  The field does shrink to half, but the telescope is still pointed at the star. The question is what it looks like.
Once the prediction is locked in, the Scene changes
Eyepiece focal length
5 mmfixed
The Scene then shows: 0 · 1 when the eye sees that pair split: resolved and magnified enough, else 0/nodes/scope/outputs/split
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A bigger blob. A 50 mm aperture separates only pairs wider than about 2.3″, and magnification adds no detail the aperture did not gather. With a 100 mm aperture the same eyepiece does split them.

Worked solution
1. The resolving power depends on the aperture: by Dawes, $\dfrac{116}{D}$ ⁠ arcseconds, with $D$ ⁠ in millimetres.
2. At 50 mm: $\dfrac{116}{50} \approx 2.3$ ⁠ arcseconds, right at the limit, and the pair merges.
3. The eyepiece only enlarges the image the lens made, smear and all.
4. So at 200× you see the same blob, twice as big. Such magnification is called empty magnification.
Step 5 of 5
Open this step
Mirror or lens
Practice · Choose
Question
All the world’s large telescopes use a mirror rather than a lens. Why?

Correct answer
- Correct A mirror sends every colour to the same focus, and can be held up from behind
Setup, choices and feedback
The Scene at this step
Aperture
100 mm
Eyepiece focal length
10 mm
Telescope
Reflector
Show the light path
On
Choices
Shown in a random order.
- Correct A mirror sends every colour to the same focus, and can be held up from behind
- Wrong A mirror reflects more light than a lens lets through
  Feedback
  The difference in light is small: an aluminium coating reflects about 90%, and a good lens passes about as much. The real reasons are colour and weight.
- Wrong A mirror magnifies more
  Feedback
  The magnification depends on the focal length and the eyepiece, not on the kind of optics. In the Scene both telescopes have the same focal length and the same magnification.
Feedback for a right answer
A lens bends each colour by a slightly different angle, like a prism, so bright stars get violet fringes. A mirror reflects every colour at the same angle. And a lens can be held only by its rim and sags under its own weight, while a mirror can be supported across its whole back.

Hint
Compare the stars in the two kinds of Telescope: which one has violet fringes?

Worked solution
1. A lens refracts light, and glass’s refractive index depends slightly on colour: violet bends more than red.
2. So each colour has a slightly different focus, and in a wide lens the difference grows.
3. A mirror reflects by the law of reflection, which does not depend on colour at all.
4. A big lens is heavy, can be held only by its rim, and distorts. A mirror is supported from behind. So the largest, metres across, are always mirrors.

Lesson 3 of 5 · About 30 minutes

Cameras and image sensors

A camera counts photons: the aperture and the shutter set how much light reaches each pixel, the ISO only amplifies what was counted, and few photons mean grain.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
The f-number
Practice · Calculate
Question
The lens has a focal length of 50 mm, and the Aperture (f-number) is f/4. What is the diameter of the iris opening?

Correct answer
12.5 mm ± 0.1 mm
Setup, choices and feedback
The Scene at this step
Aperture (f-number)
4fixed
Shutter speed (s)
1/250
ISO
100
Cut to the pixels
Camera
Accepted within
± 0.1 mm
Computed from the Scene at this step’s setting
The iris opening's diameter/nodes/camera/outputs/iris
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 4 mm ± 0.1 mm
  Misconception: Reads the f-number as the aperture's diameter in millimetres.
  Feedback from the catalogue
  The f-number is a ratio, not a length: the focal length divided by the opening’s diameter, $N = \dfrac{f}{D}$ ⁠. On a 50 mm lens at f/4 the opening is $\dfrac{50}{4} = 12.5$ ⁠ mm. That is why a larger f-number is a narrower opening.
Feedback for a right answer
12.5 mm. At f/8 the opening would be half that, and a quarter of the area.

Hint
$N = \dfrac{f}{D}$ ⁠.

Worked solution
1. $N = \dfrac{f}{D}$ ⁠, so $D = \dfrac{f}{N}$ ⁠.
2. $D = \dfrac{50}{4} = 12.5$ ⁠ mm.
The f-number is a ratio, not a diameter. If you wrote 4 mm, you read it as a length.
Step 2 of 7
Open this step
Two stops
Practice · Predict, then look
Question
Close the Aperture (f-number) from f/4 to f/8, with the shutter still at 1/250 s. How much light will reach each pixel?

Correct answer
- Correct A quarter
Setup, choices and feedback
The Scene at this step
Aperture (f-number)
4fixed
Shutter speed (s)
1/250
ISO
100
Cut to the pixels
Camera
Choices
- Correct A quarter
- Wrong Half, because the f-number doubled
  Feedback
  The diameter does halve, but light passes through the opening’s area, and the area falls to a quarter.
- Wrong The same light: the aperture changes only the depth of field
  Feedback
  The aperture does change the depth of field, but first of all it changes how much light gets in.
Once the prediction is locked in, the Scene changes
Aperture (f-number)
8fixed
The Scene then shows: 1024 · Photons a pixel counts from a mid-grey patch/nodes/camera/outputs/photons
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
From about 4100 photons a pixel to about 1000: a quarter, two “stops”. Each stop on the scale f/2.8, f/4, f/5.6, f/8 multiplies the number by $\sqrt2$ ⁠ and halves the light.

Worked solution
1. The opening’s diameter: $D = \dfrac{f}{N}$ ⁠, so at f/8 it is half what it is at f/4.
2. The area goes with $D^2$ ⁠, so it is a quarter.
3. The light let in over a given time is in proportion to the area: a quarter.
Step 3 of 7
Open this step
A faster shutter
Practice · Reach the target
Question
The Shutter speed (s) is fixed at 1/1000, four times faster, to freeze the pinwheel, and the Aperture (f-number) is fixed at f/4. The photo is too dark. Change the ISO until it is correctly exposed, then press Check.

Goal
- 0 ± 0.5
  At the start of the step: -1.966
Setup, choices and feedback
The Scene at this step
Aperture (f-number)
4fixed
Shutter speed (s)
1/1000fixed
ISO
100
Cut to the pixels
Camera
Read from the Scene at
How bright the photo is, in stops from correct/nodes/camera/outputs/exposure
One setting that reaches it
ISO
400
Feedback for a right answer
ISO 400: four times the gain makes up for a quarter of the light. But each pixel still counts only about 1000 photons, so the photo is grainier.

Feedback for a wrong answer
Not yet. The short shutter let in a quarter of the light: which ISO is four times 100?

Hint
The shutter lets in a quarter of the light. How many times must the signal be amplified?

Worked solution
1. From 1/250 s to 1/1000 s the time falls 4 times, and so does the light.
2. For the photo to be as bright as before, the signal must be amplified 4 times.
3. ISO 400 is 4 times ISO 100.
Step 4 of 7
Open this step
Counting photons
Practice · Predict, then look
Question
Go from f/4 at ISO 100 to f/16 at ISO 1600, with the same shutter. The photo is just as bright. How will it look?

Correct answer
- Correct Grainier, with a sharper background
Setup, choices and feedback
The Scene at this step
Aperture (f-number)
4fixed
Shutter speed (s)
1/250
ISO
100fixed
Cut to the pixels
Pixels
Choices
- Correct Grainier, with a sharper background
- Wrong Exactly the same photo: the ISO made up the light
  Misconception: Thinks raising the ISO lets more light reach the sensor, so the photo is as good as one with more light.
  Feedback from the catalogue
  ISO lets in no light. It only amplifies the signal the sensor has already measured, like a volume knob. The number of photons each pixel counted is set only by the aperture and the shutter, and the randomness of the count, $\sqrt{N}$ ⁠, is amplified along with the signal. Fewer photons mean more grain, even when the photo is just as bright.
- Wrong Blurrier, because the opening is small
  Feedback
  The reverse: a narrow opening shrinks the blur disc of whatever is out of focus, so the garden behind is sharper.
Once the prediction is locked in, the Scene changes
Aperture (f-number)
16fixed
ISO
1600fixed
The Scene then shows: 256 · Photons a pixel counts from a mid-grey patch/nodes/camera/outputs/photons
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Each pixel now counts about 256 photons instead of about 4100. The randomness of the count is about $\sqrt{N}$ ⁠, so the relative noise grows four times. The background is sharper because the opening is narrow.

Worked solution
1. From f/4 to f/16 the diameter falls 4 times and the area 16 times: each pixel gets $\tfrac{1}{16}$ ⁠ of the photons.
2. ISO 1600 amplifies 16 times, so the brightness does not change.
3. But photons arrive at random: in a count of $N$ ⁠ photons the noise is about $\sqrt N$ ⁠. At 4096 that is about 1.6%, and at 256 about 6%.
4. The gain amplifies the noise too, so the photo is grainier.
Step 5 of 7
Open this step
Motion blur
Practice · Calculate
Question
The pinwheel turns twice a second. The Shutter speed (s) is fixed at 1/60. Through how many degrees does the pinwheel turn while the shutter is open?

Correct answer
12 ± 0.2
Setup, choices and feedback
The Scene at this step
Aperture (f-number)
4
Shutter speed (s)
1/60fixed
ISO
100
Cut to the pixels
Camera
Accepted within
± 0.2
Computed from the Scene at this step’s setting
Degrees the pinwheel turns while the shutter is open/nodes/camera/outputs/spin
Feedback for a right answer
12°: enough for the blade tips to smear into streaks. At 1/1000 s the pinwheel moves less than a degree, and looks frozen.

Hint
One turn is 360°.

Worked solution
1. In a second the pinwheel turns $2 \times 360^\circ = 720^\circ$ ⁠.
2. The shutter is open for $\dfrac{1}{60}$ ⁠ s.
3. $\dfrac{720}{60} = 12^\circ$ ⁠.
Step 6 of 7
Open this step
One number for two
Practice · Calculate
Question
Aperture (f-number) at f/8, and Shutter speed (s) at 1/250. Photographers mark each such pair with a single exposure value, $\mathrm{EV} = \log_2 \dfrac{N^2}{t}$ ⁠. What is the exposure value here?

Correct answer
13.97 ± 0.1
Setup, choices and feedback
The Scene at this step
Aperture (f-number)
8fixed
Shutter speed (s)
1/250fixed
ISO
100
Cut to the pixels
Camera
Accepted within
± 0.1
Computed from the Scene at this step’s setting
Exposure value of the settings, log₂(N² ÷ t): one more for each halving of the light/nodes/camera/outputs/ev
Feedback for a right answer
About $14$ ⁠: in bright sunshine, at ISO $100$ ⁠, that is a good exposure.

Hint
$N^2 = 64$ ⁠ and $\dfrac{1}{t} = 250$ ⁠. How many times must $2$ ⁠ be multiplied by itself to reach the product?

Worked solution
1. $\dfrac{N^2}{t} = 64 \times 250 = 16\,000$ ⁠.
2. $2^{14} = 16\,384$ ⁠, so $\log_2 16\,000 \approx 13.97$ ⁠, almost $14$ ⁠.
Each step of $1$ ⁠ in EV is half the light. f/4 at the same shutter gives an EV of about $12$ ⁠: two steps, four times the light. Every pair with the same EV gives the same exposure.
Step 7 of 7
Open this step
Counting the noise
Practice · Calculate
Question
f/16, 1/250 and ISO 1600: the photo is as bright as before, but a grey pixel gets only 256 photons. The noise is $\sqrt{N + r^2}$ ⁠, where $r = 3$ ⁠ is the sensor’s read noise. What is the signal-to-noise ratio, $\dfrac{N}{\sqrt{N + r^2}}$ ⁠?

Correct answer
15.73 ± 3%
Setup, choices and feedback
The Scene at this step
Aperture (f-number)
16fixed
Shutter speed (s)
1/250fixed
ISO
1600fixed
Cut to the pixels
Pixels
Accepted within
± 3%
Computed from the Scene at this step’s setting
Signal-to-noise of a mid-grey pixel: photons ÷ √(photons + read noise²)/nodes/camera/outputs/snr
Wrong answers it recognises
- 63.9 ± 3%
  Misconception: Thinks raising the ISO lets more light reach the sensor, so the photo is as good as one with more light.
  Feedback from the catalogue
  ISO lets in no light. It only amplifies the signal the sensor has already measured, like a volume knob. The number of photons each pixel counted is set only by the aperture and the shutter, and the randomness of the count, $\sqrt{N}$ ⁠, is amplified along with the signal. Fewer photons mean more grain, even when the photo is just as bright.
Feedback for a right answer
About $15.7$ ⁠, against $64$ ⁠ before: the photo is as bright as before, but four times as grainy.

Hint
ISO amplifies the signal after the photons have been counted. What sets the noise is $N$ ⁠.

Worked solution
1. $N + r^2 = 256 + 9 = 265$ ⁠.
2. $\sqrt{265} \approx 16.3$ ⁠.
3. $\dfrac{256}{16.3} \approx 15.7$ ⁠.
At f/4 and ISO 100 the pixel got 4096 photons, and the ratio was about $64$ ⁠. Sixteen times less light, four times the grain: the noise goes with the square root.

Lesson 4 of 5 · About 25 minutes

Medical imaging

A CT scanner measures how much X-ray light gets through the body from many directions, and rebuilds a slice from it. Why one reading is not enough, how many directions it takes, and why the lungs show black.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
One reading
Practice · Choose
Question
One detector measures how much of an X-ray beam got through the body. What does that reading tell you?

Correct answer
- Correct How much the beam was absorbed in total along the whole line it crossed
Setup, choices and feedback
The Scene at this step
Slice position
48 cm
Projection angles
24 views
Build the slice
Projections
Choices
Shown in a random order.
- Correct How much the beam was absorbed in total along the whole line it crossed
- Wrong How deep the organ that absorbed it lies
  Misconception: Thinks a single X-ray reading or picture shows how deep each organ lies.
  Feedback from the catalogue
  An X-ray passing through the body is weakened all along its way, and a detector measures only how much of it arrived: the total absorbed along the line. It cannot tell whether the bone that absorbed it was at the front or the back. Only by measuring from many directions and combining the readings can a slice be built.
- Wrong Which organ lies right in front of the detector
  Feedback
  The detector does not see only what is next to it. The beam was weakened everywhere it passed, from the tube to the detector.
Feedback for a right answer
Each detector measures one total: the absorption along the line, $\int \mu\, dl$ ⁠. An ordinary X-ray photo is a set of such totals, which is why organs at different depths pile up on top of each other in it.

Hint
Under Build the slice choose Projections, and follow one ray from the tube to the detector.

Worked solution
1. In each centimetre the beam crosses it loses part of its strength, by the absorption coefficient $\mu$ ⁠ of the tissue there.
2. What reaches the detector is $I = I_0 e^{-\int \mu\, dl}$ ⁠: it depends only on the total along the line.
3. A bone at the front and a bone at the back give the same total, so one reading cannot say where the absorption is.
4. To find out, the same plane is measured from many directions and rebuilt.
Step 2 of 6
Open this step
Half a turn
Practice · Calculate
Question
The scanner takes 24 Projection angles, evenly spaced over half a turn. What is the angle between two neighbouring views?

Correct answer
7.5 ° ± 0.05 °
Setup, choices and feedback
The Scene at this step
Slice position
48 cm
Projection angles
24 viewsfixed
Build the slice
Projections
Accepted within
± 0.05 °
Computed from the Scene at this step’s setting
Angle between neighbouring views: 180° ÷ views/nodes/scanner/outputs/angleStep
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 15 ° ± 0.05 °
  Misconception: Spreads a CT scanner's views over a full turn of 360° rather than half a turn.
  Feedback from the catalogue
  A line through the body seen from one direction and from the opposite one is the same line, and absorbs the same amount. So half a turn, 180°, is enough to see the body from every direction, and the angle between views is $\dfrac{180^\circ}{n}$ ⁠.
Feedback for a right answer
7.5°. A real scanner takes about a thousand views each turn, a fraction of a degree apart.

Hint
Half a turn is 180°.

Worked solution
1. A line measured from one direction and from the opposite one gives the same reading, so half a turn is enough.
2. $\dfrac{180^\circ}{24} = 7.5^\circ$ ⁠.
If you got 15°, you spread the views over a whole turn.
Step 3 of 6
Open this step
More views
Practice · Predict, then look
Question
The slice is built from just 6 Projection angles, and streaks cross it. What will happen when it is built from 180?

Correct answer
- Correct The streaks will fade and the slice will come out sharp
Setup, choices and feedback
The Scene at this step
Slice position
48 cm
Projection angles
6 viewsfixed
Build the slice
Slice
Choices
- Correct The streaks will fade and the slice will come out sharp
- Wrong The slice will only be brighter
  Feedback
  Each tissue’s brightness is set by its $\mu$ ⁠, not by the number of views. What changes is how clean the rebuild is.
- Wrong Nothing: six views already cover every direction
  Feedback
  Six views are 30° apart. Every sharp edge throws a streak along each of the six directions, and between them there is no information.
Once the prediction is locked in, the Scene changes
Projection angles
180 viewsfixed
The Scene then shows: 1 ° · Angle between neighbouring views: 180° ÷ views/nodes/scanner/outputs/angleStep
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Each view is smeared back across the slice along its lines. Few directions leave streaks, and many directions cancel them out, leaving only what is really there.

Worked solution
1. The rebuild smears each view, after filtering, back across the slice along the lines it was measured on.
2. A real organ gets a share from every direction and builds up in its place.
3. Each single direction also leaves a streak along its lines. With six views each sharp edge shows six streaks.
4. With 180 views, a degree apart, the streaks spread in every direction and cancel out.
Step 4 of 6
Open this step
Reading the slice
Practice · Choose
Question
In the chest slice the lungs are almost black. Why?

Correct answer
- Correct They are full of air, and air barely absorbs X-rays
Setup, choices and feedback
The Scene at this step
Slice position
48 cm
Projection angles
180 views
Build the slice
Slice
Choices
Shown in a random order.
- Correct They are full of air, and air barely absorbs X-rays
- Wrong The ribs around them block the rays, which never reach the lungs
  Feedback
  If the ribs blocked the rays, all the detectors behind would read little, and the rebuild would give the ribs a large absorption, not the lungs a small one. The rays get through.
- Wrong The lungs are very dense and absorb all the radiation
  Feedback
  The reverse: in the slice, what absorbs a lot shows white, like bone. Black is little absorption.
Feedback for a right answer
The absorption coefficient of air-filled lung is about a quarter of muscle’s, and bone’s nearly twice muscle’s. The slice shows $\mu$ ⁠ at each point, so air is black, soft tissue grey and bone white.

Hint
In the slice each point is shaded by its $\mu$ ⁠: white absorbs a lot, black little.

Worked solution
1. The rebuild gives each point of the slice its absorption coefficient $\mu$ ⁠.
2. The slice is shaded so that large absorption is white and small absorption black.
3. Most of a lung’s volume is air, about a thousandth as dense as tissue: it absorbs very little.
4. So the lungs are almost black, and the bones around them white.
Step 5 of 6
Open this step
How little gets through
Practice · Calculate
Question
Slice position is locked at Chest. Along the ray straight from front to back through the middle of the slice, $\int \mu \, dl \approx 5.29$ ⁠. What percentage of the X-rays gets through the body? Use $I = I_0 e^{-\int \mu \, dl}$ ⁠.

Correct answer
0.5058 ± 4%
Setup, choices and feedback
The Scene at this step
Slice position
48 cmfixed
Projection angles
24 views
Build the slice
Projections
Accepted within
± 4%
Computed from the Scene at this step’s setting
Share of the X-rays on that ray that get through, e^(−∫μ dl)/nodes/scanner/outputs/transmission
Feedback for a right answer
About 0.51%. What the detector measures is how little got through.

Hint
Work out $e^{-5.29}$ ⁠ on a calculator, and multiply by $100$ ⁠.

Worked solution
1. $\dfrac{I}{I_0} = e^{-5.29} \approx 0.0051$ ⁠.
2. As a percentage: about $0.51\%$ ⁠, less than one in two hundred.
Each unit of $\int \mu \, dl$ ⁠ cuts the beam by a factor of $e \approx 2.7$ ⁠. That is why the machine needs very sensitive detectors, and why any bone in the way shows up clearly.
Step 6 of 6
Open this step
From reading to sum
Practice · Calculate
Question
Slice position is locked at Head. The detector at the back measures that only 1.40% of the X-rays that went in at the front, through the middle of the slice, got through. What is $\int \mu \, dl$ ⁠ along the ray?

Correct answer
4.267 ± 2%
Setup, choices and feedback
The Scene at this step
Slice position
12 cmfixed
Projection angles
24 views
Build the slice
Projections
Accepted within
± 2%
Computed from the Scene at this step’s setting
∫μ dl along the ray straight down through the middle of the slice, front to back/nodes/scanner/outputs/attenuation
Wrong answers it recognises
- 0.986 ± 3%
  Misconception: Takes the share of X-rays stopped for the integral of μ along the ray.
  Feedback from the catalogue
  The beam weakens by multiplying: each short stretch cuts it by a fixed share. So what goes in and what comes out are linked by a power, $\dfrac{I}{I_0} = e^{-\int \mu \, dl}$ ⁠, and the sum is found with a logarithm: $-\ln \dfrac{I}{I_0}$ ⁠, not $1 - \dfrac{I}{I_0}$ ⁠.
Feedback for a right answer
About $4.27$ ⁠: less than through the chest, though the skull is thick, because the head is much narrower.

Hint
Turn $\dfrac{I}{I_0} = e^{-\int \mu \, dl}$ ⁠ round with a natural logarithm.

Worked solution
1. $\dfrac{I}{I_0} = 0.0140$ ⁠.
2. $\int \mu \, dl = -\ln 0.0140 = \ln \dfrac{1}{0.0140} \approx 4.27$ ⁠.
That is all any one detector reading gives: one sum along one line. The scanner gathers thousands of such sums from many angles, and only from all of them together does it build $\mu$ ⁠ at every point.

Lesson 5 of 5 · About 25 minutes

Sonar and echolocation

A sonar maps the seabed with echoes: depth from the echo's time, a beam whose width the wavelength sets, and a frequency that trades detail against range.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
Depth from an echo
Practice · Calculate
Question
The sonar sends a pulse straight down, and the echo returns after 0.0469 s. The speed of sound in seawater is about 1500 m/s. How deep is the water below the ship?

Correct answer
35.15 m ± 0.3 m
Setup, choices and feedback
The Scene at this step
Pulse frequency
200 kHzfixed
Transducer width
12 cmfixed
Ping the sea floor
0.3 (First ping → Whole line mapped)fixed
Other settings
Screen/nodes/sea/props/screen = false
Accepted within
± 0.3 m
Computed from the Scene at this step’s setting
Depth the latest ping reports; 0 when no echo returns/nodes/sea/outputs/measuredDepth
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 70.3 m ± 0.5 m
  Misconception: Uses the whole echo time for the one-way trip, so gets twice the depth.
  Feedback from the catalogue
  The time the sonar measures is until the echo returns: the sound goes down to the seabed and comes back up. The whole path is twice the depth, so $d = \dfrac{vt}{2}$ ⁠.
Feedback for a right answer
About 35 m. Every point of the map drawn under the ship is one such calculation, from one pulse.

Hint
The time measured is there and back.

Worked solution
1. The distance the sound travelled: $vt = 1500 \times 0.0469 \approx 70.4$ ⁠ m.
2. That is the way down and back, so the depth is half of it: $d = \dfrac{vt}{2} \approx 35.2$ ⁠ m.
If you got about 70 m, you forgot that the echo makes the trip twice.
Step 2 of 6
Open this step
The wavelength
Practice · Calculate
Question
The Pulse frequency is 200 kHz, and the speed of sound in water is 1500 m/s. What is the pulse’s wavelength?

Correct answer
7.5 mm ± 0.05 mm
Setup, choices and feedback
The Scene at this step
Pulse frequency
200 kHzfixed
Transducer width
12 cm
Ping the sea floor
Whole line mapped
Accepted within
± 0.05 mm
Computed from the Scene at this step’s setting
Wavelength of the pulse in seawater/nodes/sea/outputs/wavelength
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
7.5 mm. Such a short wave separates details only a few centimetres across, like the wreck’s mast.

Hint
$\lambda = \dfrac{v}{f}$ ⁠, and 200 kHz = 200,000 Hz.

Worked solution
1. $\lambda = \dfrac{v}{f} = \dfrac{1500}{200\,000}$ ⁠.
2. $\lambda = 0.0075$ ⁠ m, that is 7.5 mm.
This is ultrasound: ten times higher than the highest note a human ear can hear.
Step 3 of 6
Open this step
A lower pitch
Practice · Predict, then look
Question
The whole line is mapped at 200 kHz, and the wreck and its mast show. Lower the Pulse frequency to 15 kHz, with the same transducer. What will happen to the map?

Correct answer
- Correct The beam will widen, and the wreck will blur into a hill
Setup, choices and feedback
The Scene at this step
Pulse frequency
200 kHzfixed
Transducer width
12 cmfixed
Ping the sea floor
Whole line mapped
Choices
- Correct The beam will widen, and the wreck will blur into a hill
- Wrong The map will be sharper, because a low note travels better in water
  Feedback
  A low note does reach further, but that is a matter of range, not sharpness. Sharpness depends on the beam’s width, and a long wave’s beam is wide.
- Wrong Nothing will change: the depth is the same depth
  Feedback
  The depth is the same, but the sonar reports the nearest point inside its beam. A wide beam “sees” the top of the mast even when the ship is no longer above it.
Once the prediction is locked in, the Scene changes
Pulse frequency
15 kHzfixed
The Scene then shows: 47.75 ° · Full width of the beam/nodes/sea/outputs/beamAngle
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The beam widens from about 3.6° to about 48°. A transducer $D$ ⁠ wide sends a beam about $\dfrac{\lambda}{D}$ ⁠ radians wide: the same diffraction of a wave through an opening, in sound this time.

Worked solution
1. At 15 kHz the wavelength is $\dfrac{1500}{15\,000} = 0.1$ ⁠ m, 13 times longer than at 200 kHz.
2. The beam’s width is about $\dfrac{\lambda}{D} = \dfrac{0.1}{0.12} \approx 0.83$ ⁠ rad, about 48°.
3. The sonar measures the first echo, from the nearest point in the beam.
4. In a wide beam the nearest point is often the top of the wreck or the canyon wall, so the details smear.
Step 4 of 6
Open this step
A narrow beam
Practice · Reach the target
Question
The Pulse frequency is fixed at 200 kHz. Change the Transducer width until the beam is narrower than 2°, then press Check.

Goal
- 1.25 ± 0.75
  At the start of the step: 3.581
Setup, choices and feedback
The Scene at this step
Pulse frequency
200 kHzfixed
Transducer width
12 cm
Ping the sea floor
Whole line mapped
Read from the Scene at
Full width of the beam/nodes/sea/outputs/beamAngle
One setting that reaches it
Transducer width
22 cm
Feedback for a right answer
A transducer wider than about 22 cm. That is why research ships carry arrays of transducers along the hull, metres wide.

Feedback for a wrong answer
Still too wide. A wider transducer gives a narrower beam.

Hint
The beam is about $\dfrac{\lambda}{D}$ ⁠ radians wide, and 2° is about 0.035 rad.

Worked solution
1. $2^\circ = 2 \times \dfrac{\pi}{180} \approx 0.035$ ⁠ rad.
2. You need $\dfrac{\lambda}{D} < 0.035$ ⁠, with $\lambda = 0.0075$ ⁠ m.
3. $D > \dfrac{0.0075}{0.035} \approx 0.215$ ⁠ m: a transducer of at least 22 cm.
Step 5 of 6
Open this step
Too high a pitch
Practice · Predict, then look
Question
The ship is over the canyon, about 115 m deep. Raise the Pulse frequency from 200 kHz to 600 kHz, for an even narrower beam. What will happen?

Correct answer
- Correct The echoes from the canyon floor will not come back
Setup, choices and feedback
The Scene at this step
Pulse frequency
200 kHzfixed
Transducer width
12 cm
Ping the sea floor
0.7 (First ping → Whole line mapped)fixed
Choices
- Correct The echoes from the canyon floor will not come back
- Wrong The map of the canyon will be sharper
  Misconception: Thinks a higher-frequency sound pulse reaches further, as it is sharper.
  Feedback from the catalogue
  Water absorbs high-frequency sound much more than low: each oscillation loses a little energy, and at a high frequency there are many more oscillations in each metre. That is why submarines and whales use low frequencies to hear far, and a mapping sonar uses a high one, which separates detail, only in shallow water.
- Wrong The echoes will come back sooner
  Feedback
  The speed of sound in water does not depend on the frequency. An echo that returns, returns in the same time.
Once the prediction is locked in, the Scene changes
Pulse frequency
600 kHzfixed
The Scene then shows: 4 · Pings sent so far whose echo never came back/nodes/sea/outputs/lostPings
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At 600 kHz the water absorbs so much that an echo is heard only from about 88 m. Over the canyon the echoes are lost, leaving a hole in the map.

Worked solution
1. With each oscillation the water turns a little of the sound’s energy into heat, and at a high frequency there are more oscillations in each metre.
2. So the absorption grows steeply with frequency: from a few decibels a kilometre at 15 kHz to well over a hundred at 600 kHz.
3. An echo from a floor 115 m down travels over 230 m through water, and fades below the noise.
4. Choosing a frequency is a trade-off: high for detail, low for range.
Step 6 of 6
Open this step
The footprint
Practice · Calculate
Question
Pulse frequency and Transducer width are locked, and the beam is $3.58^\circ$ ⁠ wide in full. The seabed below the ship is 58.7 m deep. How wide is the patch of seabed the beam covers?

Correct answer
3.667 m ± 3%
Setup, choices and feedback
The Scene at this step
Pulse frequency
200 kHzfixed
Transducer width
12 cmfixed
Ping the sea floor
Whole line mapped
Accepted within
± 3%
Computed from the Scene at this step’s setting
Width of the patch of seabed the beam covers straight below the ship/nodes/sea/outputs/footprint
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.833 m ± 3%
  Misconception: Works out only half of a beam’s footprint, from one side of the beam’s middle.
  Feedback from the catalogue
  The beam’s width is the full angle, side to side. The right-angled triangle covers only half of it, from the middle to one side, so after the sum you double it.
Feedback for a right answer
About 3.7 m: details smaller than that merge into one echo.

Hint
Split the beam into two right-angled triangles: in each the angle is half the full width, and the side along the beam’s middle is the depth.

Worked solution
1. Half the angle: $1.79^\circ$ ⁠, and $\tan 1.79^\circ \approx 0.0312$ ⁠.
2. Half the patch: $58.7 \times 0.0312 \approx 1.83$ ⁠ metres.
3. The whole patch: $2 \times 1.83 \approx 3.7$ ⁠ metres.
Everything inside the patch sends back an echo at about the same time, so the sonar cannot pick out details smaller than it. In deep water the patch grows, and the map blurs.

Across the unit · 14 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Lasers

One colour

Question

Correct answer

The Scene at this step

Hint

Worked solution

The threshold

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Copies of a photon

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

The output mirror

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

A stream of photons

Question

Correct answer

The Scene at this step

Hint

Worked solution

How fast it swings

Question

Correct answer

The Scene at this step

Hint

Worked solution

Held at threshold

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Hint

Worked solution

How much gets out

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

On the edge

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Telescopes

Magnification

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Gathering light

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The diffraction limit

Question

Correct answer