Answer key · Unit 14

Light and optics.

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 6

Lesson 1 of 6 · About 30 minutes

Reflection and refraction

Angles are measured from the normal, light entering glass bends towards it, Snell's law, and light that cannot get out of the glass: the critical angle and total internal reflection.

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The 5 steps and their answers

Step 1 of 5
Open this step
An angle from the normal
Practice · Calculate
Question
The ray strikes the glass block at an angle of $50^\circ$ ⁠ to the surface of the glass. What is the angle of incidence?

Correct answer
40 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Angle of incidence
40 °hidden
Refractive index
1.5fixed
Light travels
Into glassfixed
Show normals and angles
Off
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
Incidence/nodes/disc/props/incidence
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 50 ° ± 0.5 °
  Misconception: Measures the angle of incidence, refraction or the critical angle from the surface instead of from the normal.
  Feedback from the catalogue
  In refraction and reflection every angle is measured from the normal, the line square to the surface where the ray strikes, not from the surface itself. An angle from the surface is the complement to $90^\circ$ ⁠: a ray at $50^\circ$ ⁠ to the surface is at $40^\circ$ ⁠ to the normal. Only the angle from the normal gives the right answer in Snell’s law.
Feedback for a right answer
$90^\circ - 50^\circ = 40^\circ$ ⁠. Choose On under Show normals and angles: the Scene shows the dashed normal and the angle 40.0° measured from it.

Hint
The normal is square to the surface, at $90^\circ$ ⁠ to it. How much is missing from $50^\circ$ ⁠ to make $90^\circ$ ⁠?

Worked solution
1. The angle of incidence is measured from the normal, and the normal is at $90^\circ$ ⁠ to the surface.
2. The angle between the ray and the surface and the angle between the ray and the normal add up to $90^\circ$ ⁠.
3. So the angle of incidence is $90^\circ - 50^\circ = 40^\circ$ ⁠.
If you answered $50^\circ$ ⁠, you measured from the surface. On the Scene’s marked disc the zero sits on the dashed normal, so it reads $40^\circ$ ⁠.
Step 2 of 5
Open this step
Towards the normal
Practice · Predict, then look
Question
The ray passes from air into glass. When it is tilted so that the angle of incidence is $40^\circ$ ⁠, which way will it turn inside the glass?

Correct answer
- Correct Towards the normal: the angle of refraction will be less than $40^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Angle of incidence
0 °
Refractive index
1.5fixed
Light travels
Into glassfixed
Show normals and angles
On
Choices
- Correct Towards the normal: the angle of refraction will be less than $40^\circ$ ⁠
- Wrong Away from the normal: the angle of refraction will be more than $40^\circ$ ⁠
  Misconception: Thinks light entering glass from air bends away from the normal (or that it leaving bends towards it).
  Feedback from the catalogue
  Light is slower in glass, so glass has the larger refractive index. In Snell’s law, $n_1 \sin\theta_1 = n_2 \sin\theta_2$ ⁠: when $n_2$ ⁠ is larger than $n_1$ ⁠, the sine of the angle of refraction is smaller, and so is the angle. Light entering a slower material bends towards the normal, and light leaving it bends away from the normal.
- Wrong It will not turn: the angle of refraction will be $40^\circ$ ⁠
  Feedback
  A ray carries straight on only when it strikes head on, along the normal. At a slant one side of its wavefront enters the glass and slows before the other, and the ray turns.
Once the prediction is locked in, the Scene changes
Angle of incidence
40 °
The Scene then shows: 25.4 ° · Angle of the refracted ray from the normal where incidence is measured: asin(sin θ / n) going into a block, prism or drop; asin(n sin θ) out of a half-cylinder's flat face; NaN past the critical angle and for lenses/nodes/disc/outputs/refractionAngle
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Light slows down in glass, so it bends towards the normal: the angle of refraction is about $25.4^\circ$ ⁠, less than the angle of incidence.

Worked solution
1. In air $n_1 = 1$ ⁠, and in glass $n_2 = 1.5$ ⁠.
2. Snell’s law: $\sin\theta_2 = \dfrac{n_1 \sin\theta_1}{n_2} = \dfrac{\sin 40^\circ}{1.5} \approx \dfrac{0.643}{1.5} \approx 0.429$ ⁠.
3. $\theta_2 = \sin^{-1}(0.429) \approx 25.4^\circ$ ⁠, less than $40^\circ$ ⁠.
Light slows down in glass, so it bends towards the normal. Leaving the glass it bends exactly the opposite way, away from the normal.
Step 3 of 5
Open this step
Snell's law
Practice · Calculate
Question
A ray of light in air strikes glass with a refractive index of $1.5$ ⁠, at an angle of incidence of $60^\circ$ ⁠. What is the angle of refraction?

Correct answer
35.31 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Angle of incidence
60 °fixed
Refractive index
1.5fixed
Light travels
Into glassfixed
Show normals and angles
Off
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
Angle of the refracted ray from the normal where incidence is measured: asin(sin θ / n) going into a block, prism or drop; asin(n sin θ) out of a half-cylinder's flat face; NaN past the critical angle and for lenses/nodes/disc/outputs/refractionAngle
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 40 ° ± 0.4 °
  Misconception: Applies the refractive index to the angle itself (θ₂ = θ₁ / n) instead of to its sine.
  Feedback from the catalogue
  Snell’s law links the sines of the angles, not the angles themselves: $\sin\theta_2 = \dfrac{n_1 \sin\theta_1}{n_2}$ ⁠. First take the sine of the angle of incidence, divide by the refractive index, and only at the end find the angle with $\sin^{-1}$ ⁠. Dividing the angle itself comes close at small angles and goes wrong at large ones.
- 54.7 ° ± 0.4 °
  Misconception: Measures the angle of incidence, refraction or the critical angle from the surface instead of from the normal.
  Feedback from the catalogue
  In refraction and reflection every angle is measured from the normal, the line square to the surface where the ray strikes, not from the surface itself. An angle from the surface is the complement to $90^\circ$ ⁠: a ray at $50^\circ$ ⁠ to the surface is at $40^\circ$ ⁠ to the normal. Only the angle from the normal gives the right answer in Snell’s law.
- 0.577 ° ± 3%
  Misconception: Stops at the sine (a ratio between −1 and 1) and gives it as the angle.
  Feedback from the catalogue
  The number that comes out of Snell’s law or of $\sin\theta_c = \dfrac{1}{n}$ ⁠ is the sine of the angle, a ratio without units, always between $0$ ⁠ and $1$ ⁠ here. To get the angle itself takes one more step: $\theta = \sin^{-1}(\ldots)$ ⁠, in degrees.
Feedback for a right answer
$\sin\theta_2 = \dfrac{\sin 60^\circ}{1.5} \approx 0.577$ ⁠, so $\theta_2 \approx 35.3^\circ$ ⁠. Choose On under Show normals and angles to see it on the disc.

Hint
Snell’s law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ ⁠, with $n_1 = 1$ ⁠ for air. Find $\sin\theta_2$ ⁠ first, then the angle with $\sin^{-1}$ ⁠.

Worked solution
1. Given: $n_1 = 1$ ⁠ (air), $\theta_1 = 60^\circ$ ⁠, $n_2 = 1.5$ ⁠.
2. Snell’s law: $\sin\theta_2 = \dfrac{n_1 \sin\theta_1}{n_2}$ ⁠.
3. Substitute: $\sin\theta_2 = \dfrac{\sin 60^\circ}{1.5} \approx \dfrac{0.866}{1.5} \approx 0.577$ ⁠.
4. $\theta_2 = \sin^{-1}(0.577) \approx 35.3^\circ$ ⁠.
If you got $40^\circ$ ⁠, you divided the angle itself by $1.5$ ⁠; Snell’s law divides its sine. And if you got $54.7^\circ$ ⁠, that is the angle from the surface, not from the normal.
Step 4 of 5
Open this step
Out of the glass
Practice · Predict, then look
Question
The light leaves the glass ( $n = 1.5$ ⁠) for air. At an angle of incidence of $30^\circ$ ⁠ it comes out at about $48.5^\circ$ ⁠. What will happen when the angle of incidence grows to $50^\circ$ ⁠?

Correct answer
- Correct No light will come out: all of it will reflect back inside
Setup, choices and feedback
The Scene at this step
Angle of incidence
30 °
Refractive index
1.5fixed
Light travels
Out of glassfixed
Show normals and angles
On
Choices
- Wrong The light will come out at an even larger angle, close to the surface
  Misconception: Thinks light leaving glass always gets out, only bent further, however steep the angle.
  Feedback from the catalogue
  Leaving the glass, $\sin\theta_2 = n \sin\theta_1$ ⁠, and a sine cannot be larger than $1$ ⁠. Past the critical angle, where $n \sin\theta_c = 1$ ⁠, no angle of refraction satisfies Snell’s law, so no light gets out: all of it reflects back inside. This is total internal reflection.
- Wrong The light will come out at a smaller angle, close to the normal
  Feedback
  Leaving the glass the angle of refraction is always larger than the angle of incidence, so when the angle of incidence grows, it grows too.
- Correct No light will come out: all of it will reflect back inside
Once the prediction is locked in, the Scene changes
Angle of incidence
50 °
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$1.5 \sin 50^\circ \approx 1.15$ ⁠, more than $1$ ⁠, so Snell’s law has no solution: there is no refracted ray, and all the light reflects back inside. The Scene shows the caption Total internal reflection.

Worked solution
1. Leaving the glass: $n_1 = 1.5$ ⁠, $n_2 = 1$ ⁠, so $\sin\theta_2 = 1.5 \sin\theta_1$ ⁠.
2. At $30^\circ$ ⁠: $\sin\theta_2 = 1.5 \cdot 0.5 = 0.75$ ⁠, and $\theta_2 \approx 48.6^\circ$ ⁠ (in the Scene’s red light $n$ ⁠ is a little under $1.5$ ⁠, and it shows 48.5°).
3. At $50^\circ$ ⁠: $\sin\theta_2 = 1.5 \cdot 0.766 \approx 1.15$ ⁠.
4. No angle has a sine larger than $1$ ⁠, so there is no refracted ray: all the light reflects back inside, at an angle of reflection of $50^\circ$ ⁠.
This happens from about $41.8^\circ$ ⁠ on, the angle at which $1.5 \sin\theta_1 = 1$ ⁠.
Step 5 of 5
Open this step
The critical angle
Practice · Calculate
Question
The refractive index of water is $1.33$ ⁠. What is the critical angle for light going from water into air?

Correct answer
48.84 ° ± 0.4 °
Setup, choices and feedback
The Scene at this step
Angle of incidence
30 °
Refractive index
1.33fixed
Light travels
Out of glassfixed
Show normals and angles
On
Accepted within
± 0.4 °
Computed from the Scene at this step’s setting
The critical angle of the glass (or the drop), asin(1/n), for light inside it meeting air/nodes/disc/outputs/criticalAngle
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 41.2 ° ± 0.4 °
  Misconception: Measures the angle of incidence, refraction or the critical angle from the surface instead of from the normal.
  Feedback from the catalogue
  In refraction and reflection every angle is measured from the normal, the line square to the surface where the ray strikes, not from the surface itself. An angle from the surface is the complement to $90^\circ$ ⁠: a ray at $50^\circ$ ⁠ to the surface is at $40^\circ$ ⁠ to the normal. Only the angle from the normal gives the right answer in Snell’s law.
- 0.752 ° ± 3%
  Misconception: Stops at the sine (a ratio between −1 and 1) and gives it as the angle.
  Feedback from the catalogue
  The number that comes out of Snell’s law or of $\sin\theta_c = \dfrac{1}{n}$ ⁠ is the sine of the angle, a ratio without units, always between $0$ ⁠ and $1$ ⁠ here. To get the angle itself takes one more step: $\theta = \sin^{-1}(\ldots)$ ⁠, in degrees.
Feedback for a right answer
$\sin\theta_c = \dfrac{1}{1.33} \approx 0.752$ ⁠, so $\theta_c \approx 48.8^\circ$ ⁠. Raise the Angle of incidence slowly and watch the outgoing ray hug the surface and vanish at $49^\circ$ ⁠.

Hint
At the critical angle the refracted ray leaves along the surface, $\theta_2 = 90^\circ$ ⁠. Put that into Snell’s law and solve for $\theta_1$ ⁠.

Worked solution
1. At the critical angle $\theta_2 = 90^\circ$ ⁠, so $\sin\theta_2 = 1$ ⁠.
2. Snell’s law: $1.33 \sin\theta_c = 1 \cdot 1$ ⁠, that is $\sin\theta_c = \dfrac{1}{1.33} \approx 0.752$ ⁠.
3. $\theta_c = \sin^{-1}(0.752) \approx 48.8^\circ$ ⁠.
If you answered $0.752$ ⁠, you stopped at the sine and did not find the angle. And $41.2^\circ$ ⁠ is the angle from the surface. In the Scene, in the disc’s red light, the light stops getting out between $48.5^\circ$ ⁠ and $49^\circ$ ⁠.

Lesson 2 of 6 · About 35 minutes

Prisms and rainbows

White light is a mixture of colours, and each colour bends at a slightly different angle in glass and water: dispersion in a prism, a refractive index that depends on wavelength, and a raindrop that sends light back at up to 42° and makes a bow, red on the outside.

Open the lesson

The 9 steps and their answers

Step 1 of 9
Open this step
Colours that were there all along
Practice · Predict, then look
Question
White light enters the prism and leaves it as a fan of colours. Now we turn off White light, and a red beam alone enters at the same angle. What will leave the prism?

Correct answer
- Correct One red beam, bent about as much as before
Setup, choices and feedback
The Scene at this step
Angle of incidence
52 °
Shape
Glass prismfixed
White light
on
Show the angles
Off
Choices
- Correct One red beam, bent about as much as before
- Wrong A fan of colours again: the glass splits any light
  Misconception: Thinks the prism makes the colours, so it would split light of a single colour too.
  Feedback from the catalogue
  A prism does not make colours: it sorts them. White light is a mixture of every colour, and each colour bends at a slightly different angle, so they separate. Light of one colour bends, but stays one colour.
- Wrong A red beam that goes straight through, unbent
  Feedback
  Every colour bends going into the glass and coming out. What changes from colour to colour is only by how much.
Once the prediction is locked in, the Scene changes
White light
off
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The red beam bends, but stays one colour. The prism adds no colour: it only sorts the colours that were already in the white light. Newton showed this in the $1660$ ⁠s with a second prism.

Worked solution
1. White light is a mixture of all the colours.
2. Each colour bends at a slightly different angle in glass, so the colours separate.
3. Light of one colour has nothing to sort: it bends, and stays one colour.
If you expected a fan, you took the glass to make the colours. Newton sent a single colour through a second prism, and it did not split it.
Step 2 of 9
Open this step
Each colour bends its own way
Practice · Choose
Question
Look at the fan leaving the prism. Which colour bends the most, and why?

Correct answer
- Correct Violet: the glass has a higher refractive index for violet
Setup, choices and feedback
The Scene at this step
Angle of incidence
52 °
Shape
Glass prismfixed
White light
on
Show the angles
On
Choices
Shown in a random order.
- Correct Violet: the glass has a higher refractive index for violet
- Wrong Red: it is the strongest colour
  Misconception: Thinks red light bends the most in glass or water, or that glass has its highest refractive index for red.
  Feedback from the catalogue
  In glass and water the refractive index is higher for shorter wavelengths. Violet has the shortest, so it slows the most and bends the most. Red bends the least.
- Wrong They all bend equally, and only leave from different places
  Feedback
  All the colours enter at the same point and the same angle. If they bent equally, they would leave together, as one white beam.
Feedback for a right answer
Light slows down in glass, and violet slows more than red. So the glass’s refractive index is higher for violet, and violet bends more at every surface. This spreading into colours is called dispersion.

Worked solution
1. By Snell’s law, $\sin\theta_2 = \dfrac{\sin\theta_1}{n}$ ⁠: the larger $n$ ⁠, the more the ray bends.
2. In glass, $n$ ⁠ depends on the wavelength: it is higher for shorter wavelengths.
3. Violet has the shortest wavelength, so it bends the most, and red the least.
Light has no “strength” by colour. What sets the bending is the refractive index at that wavelength.
Step 3 of 9
Open this step
Index and wavelength
Practice · Calculate
Question
The flint glass of the prism has a refractive index of $1.62$ ⁠ in yellow light, 589 nm. It changes with wavelength by Cauchy’s formula, $n = A + \dfrac{B}{\lambda^2}$ ⁠, with B = 0.0089 µm². What is the refractive index for violet light, 410 nm?

Correct answer
1.647 ± 0.003
Setup, choices and feedback
The Scene at this step
Angle of incidence
52 °
Shape
Glass prismfixed
White light
offfixed
Show the angles
On
Other settings
Wavelength/nodes/bench/props/wavelength = 410
Accepted within
± 0.003
Computed from the Scene at this step’s setting
The element's refractive index at the beam's colour (Cauchy's formula; 589 nm for white light), the index the drawn ray is traced with/nodes/bench/outputs/index
Wrong answers it recognises
- 1.613 ± 0.002
  Misconception: Thinks red light bends the most in glass or water, or that glass has its highest refractive index for red.
  Feedback from the catalogue
  In glass and water the refractive index is higher for shorter wavelengths. Violet has the shortest, so it slows the most and bends the most. Red bends the least.
Feedback for a right answer
$n \approx 1.647$ ⁠ for violet, against $1.613$ ⁠ for deep red, 690 nm. The difference is small, under $3\%$ ⁠, but enough to spread the colours over a fan of a few degrees.

Hint
First find $A$ ⁠ from the value at 589 nm. Write the wavelengths in micrometres: $0.589$ ⁠ and $0.410$ ⁠.

Worked solution
1. At $0.589$ ⁠ micrometres: $A = 1.62 - \dfrac{0.0089}{0.589^2} = 1.62 - 0.0257 = 1.5943$ ⁠.
2. At $0.410$ ⁠ micrometres: $\dfrac{0.0089}{0.410^2} = \dfrac{0.0089}{0.1681} = 0.0529$ ⁠.
3. $n = 1.5943 + 0.0529 \approx 1.647$ ⁠.
If you got a number below $1.62$ ⁠, you swapped the colours: a shorter wavelength gives a higher refractive index.
Step 4 of 9
Open this step
Inside a raindrop
Practice · Reach the target
Question
In the raindrop the light bends going in, reflects off the back and bends coming out. The angle printed by the outgoing ray is the angle between it and the direction the sunlight came from. Change Angle of incidence until the light leaves at $42^\circ$ ⁠, and press Check.

Goal
- 42 ° ± 0.1 °
  At the start of the step: 28.12 °
Setup, choices and feedback
The Scene at this step
Angle of incidence
30 °
Shape
Raindropfixed
White light
on
Show the angles
On
Read from the Scene at
A drop: the angle between the light coming back out and the sunlight going in, 4r − 2i (the rainbow's angle from the point opposite the Sun); NaN for other elements/nodes/bench/outputs/returnAngle
One setting that reaches it
Angle of incidence
55.5 °
White light
off
Feedback for a right answer
Around $59^\circ$ ⁠ the light comes back at about $42^\circ$ ⁠. Notice how little the angle changes there: between $58^\circ$ ⁠ and $61^\circ$ ⁠ it moves less than a tenth of a degree. So light striking the drops at many heights comes back at the same angle, and piles up there into a bright bow.

Feedback for a wrong answer
Not yet. Read the angle by the outgoing ray. It never rises much above $42^\circ$ ⁠.

Hint
Try $59^\circ$ ⁠, then move a little either way.

Worked solution
1. The light bends going in to an angle $r$ ⁠, with $\sin r = \dfrac{\sin i}{1.333}$ ⁠.
2. Going in it turns by $i - r$ ⁠, at the reflection by $180^\circ - 2r$ ⁠, and coming out by $i - r$ ⁠ again. The angle between the returning light and the sunlight is $4r - 2i$ ⁠.
3. At $i = 59^\circ$ ⁠: $r \approx 40.0^\circ$ ⁠, and $4 \times 40.0 - 2 \times 59 \approx 42^\circ$ ⁠.
That is the maximum: at any other incidence the light comes back at less than $42^\circ$ ⁠.
Step 5 of 9
Open this step
A limit
Practice · Predict, then look
Question
At the rainbow ray, $59^\circ$ ⁠, the light comes back at about $42^\circ$ ⁠. What will happen to the return angle if we raise Angle of incidence to $75^\circ$ ⁠?

Correct answer
- Correct It will fall, below $42^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Angle of incidence
59 °
Shape
Raindropfixed
White light
on
Show the angles
On
Choices
- Correct It will fall, below $42^\circ$ ⁠
- Wrong It will grow, because the light strikes at a larger angle
  Misconception: Thinks the angle at which a raindrop sends light back keeps growing as the angle of incidence grows.
  Feedback from the catalogue
  The angle at which light returns from the drop is $4r - 2i$ ⁠. It rises with $i$ ⁠ up to about $59^\circ$ ⁠, reaches a maximum of about $42^\circ$ ⁠ there, and then falls. Because of that maximum the light piles up near $42^\circ$ ⁠, and that is where the bow is bright.
- Wrong It will stay $42^\circ$ ⁠: that is the rainbow’s angle
  Feedback
  $42^\circ$ ⁠ is the largest angle the drop sends light back at, not the only one. Far from the rainbow ray the light returns at smaller angles.
Once the prediction is locked in, the Scene changes
Angle of incidence
75 °
The Scene then shows: 35.75 ° · A drop: the angle between the light coming back out and the sunlight going in, 4r − 2i (the rainbow's angle from the point opposite the Sun); NaN for other elements/nodes/bench/outputs/returnAngle
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At $75^\circ$ ⁠ the light comes back at about $36^\circ$ ⁠. On both sides of the rainbow ray the return angle falls, so $42^\circ$ ⁠ is a limit: the drops send light at every angle up to it, and most of it right by the limit.

Worked solution
1. At $i = 75^\circ$ ⁠: $\sin r = \dfrac{\sin 75^\circ}{1.333} = 0.725$ ⁠, so $r \approx 46.5^\circ$ ⁠.
2. $4r - 2i = 186 - 150 \approx 36^\circ$ ⁠.
3. That is less than $42^\circ$ ⁠: the angle $4r - 2i$ ⁠ rises up to the rainbow ray and then falls.
If you expected it to grow, notice that $r$ ⁠ grows more slowly than $i$ ⁠ as the light strikes more steeply, so $2i$ ⁠ wins.
Step 6 of 9
Open this step
Why a bow
Practice · Choose
Question
For red the limit of a water drop is about $42.4^\circ$ ⁠, and for violet about $40.7^\circ$ ⁠. In a rainbow, which colour is on the outside, on the bigger arc?

Correct answer
- Correct Red
Setup, choices and feedback
The Scene at this step
Angle of incidence
59 °
Shape
Raindropfixed
White light
on
Show the angles
On
Choices
Shown in a random order.
- Correct Red
- Wrong Violet, because it bends more
  Misconception: Thinks violet is on the outside of a rainbow because it bends more.
  Feedback from the catalogue
  Light that bends more in the drop comes back at a smaller angle: the limit is about $40.7^\circ$ ⁠ for violet and $42.4^\circ$ ⁠ for red. The bow is a circle round the point opposite the Sun, so the colour with the larger angle, red, is on the outside.
- Wrong There is no fixed order: it depends on the size of the drops
  Feedback
  The limit depends only on water’s refractive index for each colour, not on the size of the drop. So the order is the same in every primary bow: red outside, violet inside.
Feedback for a right answer
Red comes back at a larger angle, so it reaches your eye from drops farther from the point opposite the Sun: from the outer arc. Violet bends more, its limit is smaller, and it is on the inner arc.

Hint
The bow is all the drops at a certain angle from the point opposite the Sun: the shadow of your head.

Worked solution
1. Every drop $42.4^\circ$ ⁠ from the point opposite the Sun sends red light to your eye. All those drops form a circle of radius $42.4^\circ$ ⁠.
2. The violet drops lie $40.7^\circ$ ⁠ from the same point: a smaller circle.
3. So red is outside and violet inside.
Violet does bend more, but that is exactly why its limit is smaller. Light that bends more in the drop comes back at a smaller angle.
Step 7 of 9
Open this step
How far the prism turns it
Practice · Calculate
Question
The prism’s two faces meet at $60^\circ$ ⁠. The yellow light strikes the first face at an Angle of incidence of $52^\circ$ ⁠ and leaves the second at $56.3^\circ$ ⁠ from its normal. By how many degrees has it turned from the direction it came in?

Correct answer
48.29 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Angle of incidence
52 °fixed
Shape
Glass prismfixed
White light
on
Show the angles
On
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
How far the beam's colour (589 nm for white light) is turned from its first direction: through a 60° prism i + e − 60° (NaN if it cannot get out of the second face); through a drop, in, one reflection and out, 180° + 2i − 4r; NaN for other elements/nodes/bench/outputs/deviation
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About $48.3^\circ$ ⁠. Violet light turns a few degrees more, so the colours spread into a fan.

Hint
At each face the ray turns by the difference between the angle outside and the angle inside. Add the two turns, and remember that inside, the two angles together equal the prism’s angle.

Worked solution
1. Inside the prism the two angles of refraction add up to the angle between the faces: $r_1 + r_2 = 60^\circ$ ⁠.
2. At the first face the ray turns by $i - r_1$ ⁠, and at the second by $e - r_2$ ⁠.
3. Together: $D = i + e - (r_1 + r_2) = 52 + 56.3 - 60 = 48.3^\circ$ ⁠.
Both faces turn the light the same way, towards the prism’s thick base, so the turns add up.
Step 8 of 9
Open this step
Back from the drop
Practice · Calculate
Question
The light strikes the raindrop at an Angle of incidence of $30^\circ$ ⁠ and bends inwards to $22.0^\circ$ ⁠. It bends going in, reflects once off the back and bends again coming out. The total turn is $D = 180^\circ + 2i - 4r$ ⁠. By how many degrees has it turned?

Correct answer
151.9 ° ± 0.5 °
Setup, choices and feedback
The Scene at this step
Angle of incidence
30 °fixed
Shape
Raindropfixed
White light
on
Show the angles
On
Accepted within
± 0.5 °
Computed from the Scene at this step’s setting
How far the beam's colour (589 nm for white light) is turned from its first direction: through a 60° prism i + e − 60° (NaN if it cannot get out of the second face); through a drop, in, one reflection and out, 180° + 2i − 4r; NaN for other elements/nodes/bench/outputs/deviation
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About $152^\circ$ ⁠: the light comes back at only $28^\circ$ ⁠ from the direction it came from.

Hint
Put in $i = 30^\circ$ ⁠ and $r = 22.0^\circ$ ⁠. What is missing up to $180^\circ$ ⁠ is the angle at which the light comes back to the watcher.

Worked solution
1. $2i = 60^\circ$ ⁠, and $4r = 88.0^\circ$ ⁠.
2. $D = 180 + 60 - 88.0 = 152^\circ$ ⁠.
3. The light has almost turned right back: it leaves $180 - 152 = 28^\circ$ ⁠ from the direction it came from.
At an angle of incidence of about $59^\circ$ ⁠ the turn is smallest, 138°, and the light comes back at $42^\circ$ ⁠: the rainbow ray.
Step 9 of 9
Open this step
Descartes’ sums
Practice · Calculate
Question
Descartes found that in a drop the turn is smallest when $\cos^2 i = \dfrac{n^2 - 1}{3}$ ⁠. For water $n = 1.333$ ⁠. Find $i$ ⁠, the angle of refraction $r$ ⁠, and the return angle $4r - 2i$ ⁠: the angle of the rainbow.

Correct answer
42.08 ° ± 0.4 °
Setup, choices and feedback
The Scene at this step
Angle of incidence
59 °
Shape
Raindropfixed
White light
on
Show the angles
On
Accepted within
± 0.4 °
Computed from the Scene at this step’s setting
A drop: the largest return angle any incidence gives at the beam's colour, where the returning light bunches into the bow (about 42° for water); NaN for other elements/nodes/bench/outputs/rainbowAngle
Any unit of the same quantity is accepted; it is converted before checking.
Feedback for a right answer
About $42^\circ$ ⁠, at an angle of incidence of about $59^\circ$ ⁠: the rainbow ray.

Hint
Work out $\cos i$ ⁠, then $i$ ⁠. Find $r$ ⁠ from Snell’s law: $\sin r = \dfrac{\sin i}{n}$ ⁠.

Worked solution
1. $\cos^2 i = \dfrac{1.777 - 1}{3} \approx 0.259$ ⁠, so $\cos i \approx 0.509$ ⁠ and $i \approx 59.4^\circ$ ⁠.
2. $\sin r = \dfrac{\sin 59.4^\circ}{1.333} \approx \dfrac{0.861}{1.333} \approx 0.646$ ⁠, so $r \approx 40.2^\circ$ ⁠.
3. The return angle: $4 \times 40.2 - 2 \times 59.4 \approx 160.9 - 118.8 = 42.1^\circ$ ⁠.
This is the largest angle at which a drop sends light back. The light bunches up around it, which is why the bow is bright there.

Lesson 3 of 6 · About 30 minutes

Lenses

A converging lens gathers light to a focus and a diverging lens spreads it; the thin-lens equation, magnification, real and virtual images, and what happens when half a lens is covered.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The focal point
Practice · Predict, then look
Question
Rays parallel to the axis meet at the focal point, 5 cm behind the converging lens. What will the same rays do in a diverging lens with the same focal length?

Correct answer
- Correct Spread apart, as if they came from a point 5 cm in front of the lens
Setup, choices and feedback
The Scene at this step
Object distance
12 cmdefaulthidden
Focal length
5 cm
Lens
Converging
Draw the rays
Off
Other settings
Source/nodes/bench/props/source = "beam"
Other settings
Incidence/nodes/bench/props/incidence = 0
Choices
- Correct Spread apart, as if they came from a point 5 cm in front of the lens
- Wrong Meet farther behind the lens
  Feedback
  A diverging lens is thinner in the middle than at the edges, so it turns the rays near the edges outwards, not inwards. They never meet.
- Wrong Pass straight through without changing direction
  Feedback
  Only the ray through the centre of the lens carries straight on. The others strike the curved surfaces at a slant, and are refracted.
Once the prediction is locked in, the Scene changes
Lens
Diverging
Draw the rays
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The rays leave spreading, and their dashed backward extensions meet at a focal point 5 cm in front of the lens. That is why a diverging lens’s focal length is written as a negative number, $f = -5\ \mathrm{cm}$ ⁠.

Worked solution
1. A converging lens is thicker in the middle: it turns every ray towards the axis, and the parallel rays meet at the focal point behind it, a distance $f = 5\ \mathrm{cm}$ ⁠ away.
2. A diverging lens is thinner in the middle: it turns every ray away from the axis, by the same amount.
3. So the rays leave spreading. Their backward extensions, the dashed lines drawn by Draw the rays, meet at a focal point 5 cm in front of the lens.
Such a focal point is called virtual: the light never actually passes through it, it only seems to come from it.
Step 2 of 6
Open this step
The thin-lens equation
Practice · Calculate
Question
A lit arrow stands 12 cm in front of a converging lens whose focal length is 5 cm. How far behind the lens does the image form?

Correct answer
8.571 cm ± 0.1 cm
Setup, choices and feedback
The Scene at this step
Object distance
12 cmfixed
Focal length
5 cmfixed
Lens
Convergingfixed
Draw the rays
Off
Other settings
Labels/nodes/bench/props/labels = false
Accepted within
± 0.1 cm
Computed from the Scene at this step’s setting
Lenses: the thin-lens image distance from 1/f = 1/dₒ + 1/dᵢ, a scene unit to the centimetre; negative for a virtual image on the object's side; NaN for other elements/nodes/bench/outputs/imageDistance
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3.53 cm ± 0.1 cm
  Misconception: Adds 1/f and 1/dₒ to find 1/dᵢ, instead of subtracting.
  Feedback from the catalogue
  In the thin-lens equation $\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$ ⁠ the sum is $\dfrac{1}{f}$ ⁠. To get $\dfrac{1}{d_i}$ ⁠ on its own, subtract: $\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o}$ ⁠. A quick check: for an object beyond the focal point, a converging lens forms its image beyond the focal point too, farther from the lens than $f$ ⁠.
- 0.117 cm ± 5%
  Misconception: Finds 1/dᵢ and gives it as dᵢ, forgetting to take the reciprocal.
  Feedback from the catalogue
  Working out $\dfrac{1}{f} - \dfrac{1}{d_o}$ ⁠ gives $\dfrac{1}{d_i}$ ⁠, in units of one over a centimetre. The image distance is its reciprocal: $d_i = \dfrac{1}{1/f - 1/d_o}$ ⁠.
Feedback for a right answer
$\dfrac{1}{d_i} = \dfrac{1}{5} - \dfrac{1}{12} = \dfrac{7}{60}$ ⁠, so $d_i \approx 8.57\ \mathrm{cm}$ ⁠. On the bench’s scale the upside-down image stands just past the 8 cm mark.

Hint
The thin-lens equation: $\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$ ⁠. Get $\dfrac{1}{d_i}$ ⁠ on its own, and remember to turn it over at the end.

Worked solution
1. Given: $f = 5\ \mathrm{cm}$ ⁠, $d_o = 12\ \mathrm{cm}$ ⁠.
2. Rearrange: $\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o} = \dfrac{1}{5} - \dfrac{1}{12}$ ⁠.
3. Over a common denominator: $\dfrac{12 - 5}{60} = \dfrac{7}{60}$ ⁠.
4. Turn it over: $d_i = \dfrac{60}{7} \approx 8.57\ \mathrm{cm}$ ⁠.
If you got about 3.53 cm, you added $\dfrac{1}{5}$ ⁠ and $\dfrac{1}{12}$ ⁠ instead of subtracting. And $0.117$ ⁠ is $\dfrac{1}{d_i}$ ⁠, before turning it over.
Step 3 of 6
Open this step
Magnification
Practice · Calculate
Question
In the same set-up, the object is 12 cm from the lens and the image about 8.57 cm behind it. What is the magnification $\dfrac{d_i}{d_o}$ ⁠?

Correct answer
0.7143 ± 0.02
Setup, choices and feedback
The Scene at this step
Object distance
12 cmfixed
Focal length
5 cmfixed
Lens
Convergingfixed
Draw the rays
Off
Other settings
Labels/nodes/bench/props/labels = false
Accepted within
± 0.02
Computed from the Scene at this step’s setting
Lenses: −dᵢ/dₒ, negative for an upside-down (real) image, its size beyond ±1 when enlarged; NaN for other elements/nodes/bench/outputs/magnification
Wrong answers it recognises
- 1.4 ± 0.02
  Misconception: Takes the magnification as dₒ/dᵢ instead of dᵢ/dₒ.
  Feedback from the catalogue
  The magnification is the image’s height over the object’s height, and that equals $\dfrac{d_i}{d_o}$ ⁠: the image distance on top. An image closer to the lens than the object is smaller than it, so its magnification is less than $1$ ⁠.
Feedback for a right answer
$\dfrac{8.57}{12} \approx 0.71$ ⁠: the image is smaller than the object, about $71\%$ ⁠ of its height, and upside down. This is how a camera works: a distant object, a small upside-down image on the sensor.

Hint
The magnification is the image’s height over the object’s height, and that equals $\dfrac{d_i}{d_o}$ ⁠.

Worked solution
1. The magnification is the ratio of the image’s height to the object’s, and equals $\dfrac{d_i}{d_o}$ ⁠.
2. Substitute: $\dfrac{8.57}{12} \approx 0.714$ ⁠.
3. The arrow in the Scene is 2.4 cm tall, so the image is $0.714 \cdot 2.4 \approx 1.71\ \mathrm{cm}$ ⁠ tall, and upside down.
If you got $1.4$ ⁠, you turned the ratio over. The image is closer to the lens than the object, so it is smaller than it and the magnification is less than $1$ ⁠.
Step 4 of 6
Open this step
The same size
Practice · Reach the target
Question
Change Object distance and Focal length until the image is exactly the same size as the object (upside down), then press Check.

Goal
- -1 ± 0.01
  At the start of the step: -0.5
Setup, choices and feedback
The Scene at this step
Object distance
15 cm
Focal length
5 cm
Lens
Convergingfixed
Draw the rays
Off
Read from the Scene at
Lenses: −dᵢ/dₒ, negative for an upside-down (real) image, its size beyond ±1 when enlarged; NaN for other elements/nodes/bench/outputs/magnification
One setting that reaches it
Object distance
6 cm
Focal length
3 cm
Feedback for a right answer
The object at $2f$ ⁠, and the image at $2f$ ⁠ on the other side: upside down and exactly the same size. Any closer and the image is bigger than the object, which is the arrangement in a projector.

Feedback for a wrong answer
Not yet. If the image is smaller than the object, move the arrow closer to the lens; if it is bigger, move it away. Aim for $d_o = 2f$ ⁠.

Hint
The image is the object’s size when $d_i = d_o$ ⁠. Put that into the thin-lens equation and see what object distance comes out.

Worked solution
1. An image the object’s size means a magnification of $1$ ⁠, that is $d_i = d_o$ ⁠.
2. In the thin-lens equation: $\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_o} = \dfrac{2}{d_o}$ ⁠.
3. So $d_o = 2f$ ⁠: the object at twice the focal length.
With $f = 5\ \mathrm{cm}$ ⁠ put the arrow at 10 cm, and the image forms 10 cm behind the lens, upside down and the same height. Any other pair with $d_o = 2f$ ⁠ works too, such as 12 cm with 6 cm.
Step 5 of 6
Open this step
A magnifying glass
Practice · Predict, then look
Question
The arrow stands 7.5 cm from the lens ( $f = 5\ \mathrm{cm}$ ⁠), and the image is real, upside down and enlarged. What will happen when it is brought to 3 cm, closer than the focal point?

Correct answer
- Correct A virtual image will form: upright and enlarged, on the same side as the object
Setup, choices and feedback
The Scene at this step
Object distance
7.5 cm
Focal length
5 cmfixed
Lens
Convergingfixed
Draw the rays
Off
Choices
- Correct A virtual image will form: upright and enlarged, on the same side as the object
- Wrong The real image will grow even more and move farther away
  Feedback
  When the object is closer than the focal point, the lens cannot bend the rays enough for them to meet behind it, so there is no real image there.
- Wrong There will be no image at all
  Misconception: Thinks a converging lens forms no image at all when the object is closer than the focal point.
  Feedback from the catalogue
  When the object is closer than the focal point, the rays leave the lens still spreading, so they never meet on any screen. But an eye looking through the lens sees them as if they came from a point further back: that is a virtual image, upright and enlarged. This is how a magnifying glass works.
Once the prediction is locked in, the Scene changes
Object distance
3 cm
Draw the rays
On
The Scene then shows: 2.5 · Lenses: −dᵢ/dₒ, negative for an upside-down (real) image, its size beyond ±1 when enlarged; NaN for other elements/nodes/bench/outputs/magnification
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The rays leave the lens spreading, and their backward extensions meet 7.5 cm in front of the lens: a virtual image, upright, 2.5 times as big.

Worked solution
1. The thin-lens equation with $f = 5\ \mathrm{cm}$ ⁠ and $d_o = 3\ \mathrm{cm}$ ⁠: $\dfrac{1}{d_i} = \dfrac{1}{5} - \dfrac{1}{3} = -\dfrac{2}{15}$ ⁠.
2. So $d_i = -7.5\ \mathrm{cm}$ ⁠. The minus sign says the image is on the same side as the object, in front of the lens: a virtual image.
3. The magnification: $\dfrac{7.5}{3} = 2.5$ ⁠, and the image is upright.
The rays never meet behind the lens, so no screen will catch the image. But an eye looking through the lens sees an upright arrow 2.5 times as big: a magnifying glass.
Step 6 of 6
Open this step
Half a lens
Practice · Choose
Question
The lens forms a sharp image of the arrow on a screen. The top half of the lens is covered with card. What will happen to the image on the screen?

Correct answer
- Correct The image will stay whole, but dimmer
Setup, choices and feedback
The Scene at this step
Object distance
12 cm
Focal length
5 cm
Lens
Converging
Draw the rays
Off
Choices
Shown in a random order.
- Correct The image will stay whole, but dimmer
- Wrong Half of the image will disappear
  Misconception: Thinks covering half a lens cuts off half of the image.
  Feedback from the catalogue
  Every point of the object sends light to every part of the lens, and every part of the lens sends that light to the same point of the image. Cover half the lens and each point of the image gets half the light, but it still gets it. The image stays whole, only dimmer.
- Wrong The image will move and blur
  Feedback
  The bottom half of the lens still sends the rays from each point of the arrow to the same point of the image, so the image stays sharp and in the same place.
- Wrong Nothing will change
  Feedback
  The card stops half the light reaching the lens, so something does change: the brightness.
Feedback for a right answer
Every part of the lens forms the whole image, because light from every point of the object reaches the whole lens. Half a lens means half the light, not half the image.

Worked solution
1. Every point of the arrow sends light to every part of the lens. The Scene shows it at the arrow’s tip: several rays leave it for different heights of the lens.
2. The lens sends all of those rays, from every height, to the same point of the image.
3. With the top half covered, only the rays through the bottom half are left, but they still reach the same points of the image.
So every point of the image gets about half the light: the image is whole, in the same place and the same size, and dimmer.

Lesson 4 of 6 · About 20 minutes

Diffraction and two slits

A wave passing through an opening spreads more the narrower the opening is compared with its wavelength; why you hear round a corner but cannot see, and the fringes of two slits, in water and in light.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
A narrow gap
Practice · Predict, then look
Question
The opening is 10 cm wide, four wavelengths, and the wave passes through it as a straight beam. What will happen when the opening is narrowed to 1.5 cm, less than one wavelength?

Correct answer
- Correct The wave will spread in half-circles through the whole tank behind the wall
Setup, choices and feedback
The Scene at this step
Opening width
10 cm
Barrier
One gapfixed
Show the calm water
Off
Choices
- Correct The wave will spread in half-circles through the whole tank behind the wall
- Wrong A narrower straight beam will pass through
  Misconception: Thinks a narrower gap lets through a narrower straight beam, as if the wave were a stream of particles.
  Feedback from the catalogue
  A stream of particles would behave like that, but a wave does the opposite. The narrower the opening compared with the wavelength, the more it acts like a single point source, and the wave spreads from it in half-circles in every direction. A straight beam with sharp shadows comes from an opening much wider than the wavelength.
- Wrong The wave will not pass at all
  Feedback
  A wave passes even through an opening narrower than its wavelength. What gets through is weaker, but it spreads in every direction.
Once the prediction is locked in, the Scene changes
Opening width
1.5 cm
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
An opening narrower than the wavelength becomes a new source, and the wave spreads from it in half-circles through the whole tank. The narrower the opening compared with the wavelength, the stronger the diffraction.

Worked solution
1. The wavelength in the tank is $\lambda = \dfrac{v}{f} = \dfrac{25}{10} = 2.5\ \mathrm{cm}$ ⁠.
2. The wide opening, 10 cm, is $4$ ⁠ wavelengths: ripples from different parts of it cancel each other to the sides, leaving a straight beam.
3. The narrow opening, 1.5 cm, is less than one wavelength: it acts like a single point source, and the wave spreads from it in half-circles.
This is diffraction: the narrower the opening compared with the wavelength, the more the wave spreads.
Step 2 of 4
Open this step
Sound and light
Practice · Choose
Question
Through an open door you hear a friend standing round the corner in the next room, but you cannot see them. Why?

Correct answer
- Correct Sound’s wavelength is about the width of the door, and light’s is about a million times smaller
Setup, choices and feedback
The Scene at this step
Opening width
2.5 cm
Barrier
One gap
Show the calm water
Off
Choices
Shown in a random order.
- Correct Sound’s wavelength is about the width of the door, and light’s is about a million times smaller
- Wrong Light is not a wave, so it does not diffract
  Misconception: Thinks light does not diffract (only sound or water waves do), so it always casts sharp shadows.
  Feedback from the catalogue
  Light diffracts too, like every wave. But its wavelength is less than a thousandth of a millimetre, far smaller than a door or a window, so through everyday openings the diffraction is negligible and light travels in straight lines. Through a narrow enough opening, such as a slit tenths of a millimetre wide, you see light spread out into bright and dark fringes.
- Wrong Sound passes through the walls, and light does not
  Feedback
  A little sound does pass through walls, but through an open door it mostly goes through the opening, and spreads round the corner.
- Wrong Light is too fast to bend
  Feedback
  Speed does not decide diffraction. What decides it is the ratio of the wavelength to the size of the opening.
Feedback for a right answer
Sound has a wavelength of about a metre, the size of a door, so it spreads round the corner. Light’s wavelength is less than a thousandth of a millimetre, and the door is a million times wider: it goes through almost in a straight line.

Worked solution
1. How much a wave spreads beyond an opening depends on the ratio of its wavelength to the opening’s width.
2. The sound of speech, at a few hundred hertz, has a wavelength of about a metre: $\lambda = \dfrac{v}{f} \approx \dfrac{340}{340} = 1\ \mathrm{m}$ ⁠, about the width of a door. So it diffracts strongly and reaches round the corner.
3. Visible light has a wavelength of about 500 nm, $5 \times 10^{-7}$ ⁠ metres, about a million times smaller than the door. So it passes through almost in a straight line and leaves a sharp shadow.
Light diffracts too, but only through very narrow openings, tenths of a millimetre wide or less.
Step 3 of 4
Open this step
Two slits
Practice · Calculate
Question
The two slits are 10 cm apart, centre to centre, and the waves are 2.5 cm long. Far from the slits, at what angle from straight ahead is the first calm line?

Correct answer
7.18 ° ± 0.3 °
Setup, choices and feedback
The Scene at this step
Opening width
2.5 cm
Barrier
Two slitsfixed
Show the calm water
On
Accepted within
± 0.3 °
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 14.48 ° ± 0.4 °
  Misconception: Uses d·sin θ = λ (the first bright line beside the middle) for the first calm or dark line.
  Feedback from the catalogue
  When $d \sin\theta = \lambda$ ⁠ the waves from the two slits arrive a whole wavelength apart, crest meets crest, and that is a bright fringe. A calm line lies where they arrive half a wavelength apart, crest meeting trough: the first one at $d \sin\theta = \dfrac{\lambda}{2}$ ⁠.
- 0.125 ° ± 5%
  Misconception: Stops at the sine (a ratio between −1 and 1) and gives it as the angle.
  Feedback from the catalogue
  The number that comes out of Snell’s law or of $\sin\theta_c = \dfrac{1}{n}$ ⁠ is the sine of the angle, a ratio without units, always between $0$ ⁠ and $1$ ⁠ here. To get the angle itself takes one more step: $\theta = \sin^{-1}(\ldots)$ ⁠, in degrees.
Feedback for a right answer
$\sin\theta = \dfrac{\lambda / 2}{d} = \dfrac{1.25}{10} = 0.125$ ⁠, so $\theta \approx 7.2^\circ$ ⁠. With Show the calm water on you can see two such calm lines, one on each side of the middle.

Hint
In a direction $\theta$ ⁠ the path from one slit is longer than from the other by $d \sin\theta$ ⁠. A calm line lies where that difference is half a wavelength.

Worked solution
1. Given: $d = 10\ \mathrm{cm}$ ⁠, $\lambda = 2.5\ \mathrm{cm}$ ⁠.
2. On the first calm line the waves from the two slits arrive half a wavelength apart, and crest meets trough: $d \sin\theta = \dfrac{\lambda}{2}$ ⁠.
3. Substitute: $\sin\theta = \dfrac{1.25}{10} = 0.125$ ⁠.
4. $\theta = \sin^{-1}(0.125) \approx 7.2^\circ$ ⁠.
If you got about $14.5^\circ$ ⁠, you used $d \sin\theta = \lambda$ ⁠: there the difference is a whole wavelength, crest meets crest, and that is the first bright fringe beside the middle.
Step 4 of 4
Open this step
Young's experiment
Practice · Calculate
Question
Red laser light, of wavelength 650 nm, passes through two slits 0.25 mm apart and lights a wall 2 m away. How far apart are two neighbouring bright fringes on the wall?

Correct answer
5.2 mm ± 3%
Setup, choices and feedback
The Scene at this step
Opening width
2.5 cm
Barrier
Two slits
Show the calm water
On
Accepted within
± 3%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.0052 mm ± 5%
  Misconception: Leaves nanometres or millimetres unconverted, so the answer is off by a power of ten.
  Feedback from the catalogue
  Before substituting into a formula, turn every length into metres: 1 nm = 10⁻⁹ m and 1 mm = 10⁻³ m. The speed of light, 3 × 10⁸ m/s, and Planck’s constant are written in SI units, so the wavelength has to be in metres for the units to cancel.
Feedback for a right answer
$\Delta y = \dfrac{\lambda L}{d} = \dfrac{650 \times 10^{-9} \cdot 2}{0.25 \times 10^{-3}} = 5.2 \times 10^{-3}\ \mathrm{m}$ ⁠: a fringe about every half centimetre. This is how Thomas Young showed that light is a wave.

Hint
At small angles $\sin\theta \approx \tan\theta = \dfrac{y}{L}$ ⁠, so the bright fringes lie at $y = \dfrac{m \lambda L}{d}$ ⁠. Turn every length into metres first.

Worked solution
1. Given, in metres: $\lambda = 650 \times 10^{-9}\ \mathrm{m}$ ⁠, $d = 0.25 \times 10^{-3}\ \mathrm{m}$ ⁠, $L = 2\ \mathrm{m}$ ⁠.
2. Bright fringe number $m$ ⁠ lies at the angle where $d \sin\theta = m\lambda$ ⁠. The angles are small, so $\sin\theta \approx \tan\theta = \dfrac{y}{L}$ ⁠, and $y_m = \dfrac{m \lambda L}{d}$ ⁠.
3. The distance between neighbouring fringes: $\Delta y = \dfrac{\lambda L}{d} = \dfrac{650 \times 10^{-9} \cdot 2}{0.25 \times 10^{-3}} = 5.2 \times 10^{-3}\ \mathrm{m}$ ⁠.
4. That is $5.2\ \mathrm{mm}$ ⁠.
If you got $0.0052$ ⁠ millimetres, you put $d$ ⁠ in millimetres and $\lambda$ ⁠ in metres. The slits have to be very close together because light’s wavelength is so small: with slits a centimetre apart the fringes would be too crowded to see.

Lesson 5 of 6 · About 25 minutes

Polarisation

Light is a transverse wave, and a polarising filter picks the direction it shakes in: half of unpolarised light, Malus's law, a third filter that brings light back, and polarising sunglasses.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
The first filter
Practice · Choose
Question
The lamp’s light is unpolarised. What share of it gets through the first polarising filter?

Correct answer
- Correct Half, whichever way the filter is turned
Setup, choices and feedback
The Scene at this step
Angle between filters
0 °
Add a third filter
Off
Choices
Shown in a random order.
- Correct Half, whichever way the filter is turned
- Wrong All of it, because the filter is not crossed with anything yet
  Feedback
  A polarising filter passes only the part of the shaking along its axis. Unpolarised light shakes in every direction, so some of it is absorbed already in the first filter.
- Wrong It depends on which way the filter is turned
  Misconception: Thinks how much unpolarised light the first filter passes depends on which way the filter is turned.
  Feedback from the catalogue
  In unpolarised light the electric field shakes in every direction across the beam equally, switching direction very fast. So there is no preferred direction, and a polarising filter passes half of it on average, whichever way it is turned. What the filter’s direction does decide is the direction the outgoing light is polarised in.
- Wrong None, it takes two filters for light to pass
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
Feedback for a right answer
Unpolarised light has no preferred direction, so any filter passes half of it on average, whichever way it faces. What the filter does decide is the polarisation of the light that leaves.

Worked solution
1. In unpolarised light the electric field shakes in every direction across the beam, switching direction at random.
2. From a shaking at an angle $\theta$ ⁠ to the filter’s axis, the share $\cos^2\theta$ ⁠ of the intensity passes.
3. The average of $\cos^2\theta$ ⁠ over every direction is $\tfrac12$ ⁠, because $\cos^2\theta + \sin^2\theta = 1$ ⁠ and the directions where $\cos^2\theta$ ⁠ is large are balanced exactly by those where it is small.
So the first filter passes half the light, and the light leaving it is polarised along its axis.
Step 2 of 5
Open this step
Malus's law
Practice · Calculate
Question
The last filter is turned $30^\circ$ ⁠ from the first. What percentage of the lamp’s light reaches the screen?

Correct answer
37.5 ± 0.5
Setup, choices and feedback
The Scene at this step
Angle between filters
30 °fixed
Add a third filter
Offfixed
Accepted within
± 0.5
Computed from the Scene at this step’s setting
Share of the lamp's light that reaches the screen, ½ × transmitted/nodes/bench/outputs/screen
Wrong answers it recognises
- 75 ± 0.5
  Misconception: Applies Malus's law to the lamp's light and forgets that the first filter already passes only half.
  Feedback from the catalogue
  Malus’s law, $I = I_1 \cos^2\theta$ ⁠, starts from $I_1$ ⁠, the polarised light that left the first filter. The lamp’s light is unpolarised, and the first filter passes only half of it. So from the lamp to the screen: $\tfrac12 \cos^2\theta$ ⁠.
- 43.3 ± 0.5
  Misconception: Uses cos θ instead of cos²θ for the share of polarised light a filter passes.
  Feedback from the catalogue
  The filter passes the part of the field along its axis, which is the field times $\cos\theta$ ⁠. But brightness goes as the square of the field, so the intensity that passes is $I_1 \cos^2\theta$ ⁠. At $60^\circ$ ⁠, for instance, half the field gets through and only a quarter of the light.
Feedback for a right answer
$\tfrac12 \cos^2 30^\circ = \tfrac12 \cdot 0.75 = 0.375$ ⁠, that is $37.5\%$ ⁠. The chart shows $75\%$ ⁠ because it measures from the polarised light, after the first filter.

Hint
Two stages: the first filter passes half of the unpolarised light, and the second passes the share $\cos^2\theta$ ⁠ of the polarised light.

Worked solution
1. The first filter passes half of the unpolarised light: $I_1 = \tfrac12 I_0$ ⁠.
2. Malus’s law for the second filter: $I = I_1 \cos^2 30^\circ = I_1 \cdot \left(\tfrac{\sqrt3}{2}\right)^2 = 0.75\, I_1$ ⁠.
3. Together: $I = \tfrac12 \cdot 0.75\, I_0 = 0.375\, I_0$ ⁠, that is $37.5\%$ ⁠.
The chart beside the Scene shows $75\%$ ⁠: that is the share of the polarised light, after the first filter. If you answered $75$ ⁠, you forgot the half the first filter stops. And $43.3$ ⁠ comes from using $\cos\theta$ ⁠ instead of $\cos^2\theta$ ⁠.
Step 3 of 5
Open this step
A third filter
Practice · Predict, then look
Question
The filters are crossed, at $90^\circ$ ⁠, and the screen is dark. A third filter, turned to $45^\circ$ ⁠, slides in between them. What will happen to the screen?

Correct answer
- Correct It will brighten: some light will reach it
Setup, choices and feedback
The Scene at this step
Angle between filters
90 °fixed
Add a third filter
Off
Choices
- Wrong It will stay dark: another filter can only stop more light
  Misconception: Thinks adding a filter can only take light away, so a third filter cannot brighten crossed filters.
  Feedback from the catalogue
  A polarising filter does not only block: it also turns the polarisation of the light it passes to the direction of its axis. A middle filter at $45^\circ$ ⁠ passes half the light and polarises it at $45^\circ$ ⁠, and now the last filter is only $45^\circ$ ⁠ away from it rather than $90^\circ$ ⁠, so it passes half too. A quarter of the light reaches the screen instead of none.
- Correct It will brighten: some light will reach it
- Wrong It will be as bright as with aligned filters
  Feedback
  Each of the two steps, $45^\circ$ ⁠ and another $45^\circ$ ⁠, passes only half the light. Aligned filters stop none of the polarised light.
Once the prediction is locked in, the Scene changes
Add a third filter
On
The Scene then shows: 0.25 · Share of the polarised light (past the first filter) that reaches the screen: cos²θ (Malus's law), cos⁴(θ/2) with the middle filter in the beam; the chart's percentage/nodes/bench/outputs/transmitted
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A quarter of the polarised light reaches the screen: the middle filter passes half and turns the polarisation to $45^\circ$ ⁠, and the last filter passes another half. On the chart, the dot jumps to the orange curve.

Worked solution
1. After the first filter the light is polarised upright.
2. The middle filter at $45^\circ$ ⁠ passes $\cos^2 45^\circ = \tfrac12$ ⁠ of the light, and polarises it afresh at $45^\circ$ ⁠.
3. The last filter, at $90^\circ$ ⁠, is now only $45^\circ$ ⁠ from the direction of polarisation, so it passes $\tfrac12$ ⁠ too.
4. Together: $\tfrac12 \cdot \tfrac12 = \tfrac14$ ⁠ of the polarised light, that is $0.25$ ⁠.
Without the middle filter, $\cos^2 90^\circ = 0$ ⁠. The extra filter adds no light; it turns the polarisation in two smaller steps, and each step loses less.
Step 4 of 5
Open this step
Three quarters
Practice · Reach the target
Question
Turn Angle between filters so that exactly three quarters of the polarised light reaches the screen, then press Check. Try to work out the angle before you move anything.

Goal
- 0.75 ± 0.01
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Angle between filters
0 °
Add a third filter
Offfixed
Read from the Scene at
Share of the polarised light (past the first filter) that reaches the screen: cos²θ (Malus's law), cos⁴(θ/2) with the middle filter in the beam; the chart's percentage/nodes/bench/outputs/transmitted
One setting that reaches it
Angle between filters
30 °
Feedback for a right answer
$\cos^2 30^\circ = \left(\tfrac{\sqrt3}{2}\right)^2 = \tfrac34$ ⁠. On the chart, the dot sits on the blue curve at $30^\circ$ ⁠ and $75\%$ ⁠.

Feedback for a wrong answer
Not yet. If more than $75\%$ ⁠ gets through, increase the angle; if less, decrease it.

Hint
Solve $\cos^2\theta = \tfrac34$ ⁠: what is $\cos\theta$ ⁠, and at what angle?

Worked solution
1. Malus’s law: the share that passes is $\cos^2\theta$ ⁠, and you need $\cos^2\theta = 0.75$ ⁠.
2. $\cos\theta = \sqrt{0.75} = \dfrac{\sqrt3}{2} \approx 0.866$ ⁠.
3. $\theta = \cos^{-1}(0.866) = 30^\circ$ ⁠.
At $29^\circ$ ⁠ already $76.5\%$ ⁠ passes, and at $31^\circ$ ⁠ only $73.5\%$ ⁠: Malus’s law is sensitive to the angle in the middle of its range.
Step 5 of 5
Open this step
Glare
Practice · Choose
Question
Polarising sunglasses cut the glare from a wet road, and colours look the same through them. What do they do?

Correct answer
- Correct Their axis is upright, so they stop most of the reflected light, which is mostly polarised sideways
Setup, choices and feedback
The Scene at this step
Angle between filters
90 °
Add a third filter
Off
Choices
Shown in a random order.
- Correct Their axis is upright, so they stop most of the reflected light, which is mostly polarised sideways
- Wrong They filter out the brightest colours
  Misconception: Thinks a polarising filter works by removing colours or dimming some colours more than others.
  Feedback from the catalogue
  A polarising filter picks a direction of shaking, not a colour: it passes the part of the field along its axis at almost every wavelength alike. So polarising sunglasses do not change colours; they mainly weaken glare, light reflected from water or a road whose shaking is mostly sideways.
- Wrong They block light coming from below, like a blind
  Feedback
  A polarising filter picks the direction the electric field shakes in, not the direction the light comes from.
- Wrong They simply darken all the light equally
  Feedback
  Ordinary dark glasses do that. Polarising glasses weaken the reflected light far more than the rest, so the glare goes and the view stays.
Feedback for a right answer
Glare is mostly polarised sideways, and the glasses’ axis is upright, like crossed filters. A polarising filter picks a direction of shaking, not a colour, so the colours do not change.

Worked solution
1. Light reflected from a level surface, like water or a wet road, is mostly polarised sideways.
2. The sunglasses’ axis is upright, at $90^\circ$ ⁠ to the glare’s polarisation. By Malus’s law, $\cos^2 90^\circ = 0$ ⁠, so most of the glare is stopped.
3. The rest of the light is unpolarised, and the glasses pass half of it, at almost every colour alike.
Tilt your head by $90^\circ$ ⁠ wearing such glasses and the glare comes back: their axis is now sideways.

Lesson 6 of 6 · About 25 minutes

The electromagnetic spectrum

Radio, light and X-rays are the same wave at different lengths: one speed in a vacuum, frequency from wavelength, brightness against colour, and the energy of a single photon.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
One family
Practice · Choose
Question
A radio wave, visible light and X-rays set out together from a distant star. Which reaches Earth first, through empty space?

Correct answer
- Correct They all arrive together: in a vacuum they all travel at the speed of light
Setup, choices and feedback
The Scene at this step
Wavelength (powers of ten)
1 log m
Brightness
60 %
Add the magnetic field
Off
Choices
Shown in a random order.
- Correct They all arrive together: in a vacuum they all travel at the speed of light
- Wrong The X-rays, because they shake fastest
  Misconception: Thinks electromagnetic waves of different wavelengths travel at different speeds in a vacuum.
  Feedback from the catalogue
  In a vacuum all electromagnetic waves travel at the same speed, the speed of light $c \approx 3 \times 10^8\ \mathrm{m/s}$ ⁠, from radio waves to gamma rays. What tells them apart is the wavelength, and so the frequency: $f = \dfrac{c}{\lambda}$ ⁠. A shorter wave shakes faster, but does not move faster.
- Wrong The radio wave, because each of its strides is long
  Misconception: Thinks electromagnetic waves of different wavelengths travel at different speeds in a vacuum.
  Feedback from the catalogue
  In a vacuum all electromagnetic waves travel at the same speed, the speed of light $c \approx 3 \times 10^8\ \mathrm{m/s}$ ⁠, from radio waves to gamma rays. What tells them apart is the wavelength, and so the frequency: $f = \dfrac{c}{\lambda}$ ⁠. A shorter wave shakes faster, but does not move faster.
- Wrong The radio wave will never arrive, because space has no air to carry it
  Misconception: Thinks radio waves are sound waves.
  Feedback from the catalogue
  Radio waves are electromagnetic waves, like light: electric and magnetic fields that need no medium, so they cross empty space too. Sound is a wave of pressure in the air, about a million times slower. A radio receiver picks up the electromagnetic wave, and only its loudspeaker turns the information in it into sound.
Feedback for a right answer
The whole electromagnetic spectrum travels through a vacuum at $c$ ⁠. What changes along it is the wavelength, and with it the frequency: $f = \dfrac{c}{\lambda}$ ⁠.

Worked solution
1. All these waves are electromagnetic waves: electric and magnetic fields that need no medium.
2. In a vacuum they all travel at the same speed, $c \approx 3 \times 10^8\ \mathrm{m/s}$ ⁠.
3. They differ only in wavelength, and so in frequency: $f = \dfrac{c}{\lambda}$ ⁠. A short wave has a high frequency, but not a higher speed.
In the Scene, going from Radio to X-ray, the wave gets shorter and shakes faster, but keeps moving forward at the same pace.
Step 2 of 5
Open this step
Frequency
Practice · Calculate
Question
The wavelength in the Scene is $10^{-6.3}$ ⁠ metres, about 501 nm: green light. What is its frequency? Use $c = 3.00 \times 10^8\ \mathrm{m/s}$ ⁠.

Correct answer
598.2 THz ± 2%
Setup, choices and feedback
The Scene at this step
Wavelength (powers of ten)
-6.3 log mfixed
Brightness
60 %
Add the magnetic field
Offdefault
Accepted within
± 2%
Computed from the Scene at this step’s setting
The frequency in a vacuum, f = c/λ/nodes/wave/outputs/frequency
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 5.99 × 10⁻⁷ THz ± 5%
  Misconception: Leaves nanometres or millimetres unconverted, so the answer is off by a power of ten.
  Feedback from the catalogue
  Before substituting into a formula, turn every length into metres: 1 nm = 10⁻⁹ m and 1 mm = 10⁻³ m. The speed of light, 3 × 10⁸ m/s, and Planck’s constant are written in SI units, so the wavelength has to be in metres for the units to cancel.
Feedback for a right answer
$f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8}{5.01 \times 10^{-7}} \approx 5.99 \times 10^{14}\ \mathrm{Hz}$ ⁠, about 600 THz.

Hint
Every wave obeys $v = f \lambda$ ⁠, and in a vacuum $v = c$ ⁠. Turn the wavelength into metres first. A terahertz, THz, is $10^{12}$ ⁠ hertz.

Worked solution
1. Given: $c = 3.00 \times 10^8\ \mathrm{m/s}$ ⁠, $\lambda = 501 \times 10^{-9}\ \mathrm{m} = 5.01 \times 10^{-7}\ \mathrm{m}$ ⁠.
2. From $c = f\lambda$ ⁠: $f = \dfrac{c}{\lambda}$ ⁠.
3. Substitute: $f = \dfrac{3.00 \times 10^8}{5.01 \times 10^{-7}} \approx 5.99 \times 10^{14}\ \mathrm{Hz}$ ⁠.
4. That is about 599 THz: nearly six hundred trillion shakes a second.
If you got about $6 \times 10^5$ ⁠ hertz, you put in $501$ ⁠ instead of $5.01 \times 10^{-7}$ ⁠: nanometres instead of metres.
Step 3 of 5
Open this step
Brightness and colour
Practice · Predict, then look
Question
The screen shows red light, at a Brightness of $30\%$ ⁠. What will change in the wave when the brightness is turned up to $100\%$ ⁠?

Correct answer
- Correct Only its height: the wavelength, the frequency and the colour will stay
Setup, choices and feedback
The Scene at this step
Wavelength (powers of ten)
-6.2 log mfixed
Brightness
30 %
Add the magnetic field
Offdefault
Choices
- Correct Only its height: the wavelength, the frequency and the colour will stay
- Wrong The wavelength will shorten and the light will turn bluer
  Misconception: Thinks a brighter beam has a higher frequency or more energy per photon (or a different colour).
  Feedback from the catalogue
  Brightness is how many photons arrive each second, and the height of the wave. The frequency, the wavelength and the colour do not change when the light is turned up, and neither does the energy of each photon: $E = hf$ ⁠ depends only on the frequency. Bright red light is many red photons, each one weaker than a single blue photon.
- Wrong The wave will move forward faster
  Misconception: Thinks electromagnetic waves of different wavelengths travel at different speeds in a vacuum.
  Feedback from the catalogue
  In a vacuum all electromagnetic waves travel at the same speed, the speed of light $c \approx 3 \times 10^8\ \mathrm{m/s}$ ⁠, from radio waves to gamma rays. What tells them apart is the wavelength, and so the frequency: $f = \dfrac{c}{\lambda}$ ⁠. A shorter wave shakes faster, but does not move faster.
Once the prediction is locked in, the Scene changes
Brightness
100 %
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The wave gets taller, and nothing else changes: the wavelength, the frequency, the colour and the speed. Brighter light is more photons, not stronger photons.

Worked solution
1. Brightness sets the wave’s amplitude, the height of the electric field, and with it how much energy arrives each second.
2. The wavelength ( $10^{-6.2}$ ⁠ metres, about 631 nm) does not depend on brightness, so the frequency $f = \dfrac{c}{\lambda} \approx 4.75 \times 10^{14}\ \mathrm{Hz}$ ⁠ and the colour stay too.
3. The speed is $c$ ⁠, always.
In terms of photons: brighter light is more photons each second, and each one carries the same energy $E = hf$ ⁠ as before.
Step 4 of 5
Open this step
The energy of a photon
Practice · Calculate
Question
Violet light, of wavelength $10^{-6.4}$ ⁠ metres, about 398 nm. How much energy does one of its photons carry? Use $h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s}$ ⁠ and $c = 3.00 \times 10^8\ \mathrm{m/s}$ ⁠, and answer in joules or electronvolts.

Correct answer
3.114 eV ± 2%
Setup, choices and feedback
The Scene at this step
Wavelength (powers of ten)
-6.4 log mfixed
Brightness
60 %
Add the magnetic field
Offdefault
Accepted within
± 2%
Computed from the Scene at this step’s setting
The energy of one photon, E = h·f (Planck's constant h)/nodes/wave/outputs/photonEnergy
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3.12 × 10⁻⁹ eV ± 5%
  Misconception: Leaves nanometres or millimetres unconverted, so the answer is off by a power of ten.
  Feedback from the catalogue
  Before substituting into a formula, turn every length into metres: 1 nm = 10⁻⁹ m and 1 mm = 10⁻³ m. The speed of light, 3 × 10⁸ m/s, and Planck’s constant are written in SI units, so the wavelength has to be in metres for the units to cancel.
Feedback for a right answer
$E = \dfrac{hc}{\lambda} = \dfrac{6.63 \times 10^{-34} \cdot 3.00 \times 10^8}{3.98 \times 10^{-7}} \approx 5.0 \times 10^{-19}\ \mathrm{J}$ ⁠, about 3.1 eV. That is already close to the energy of a chemical bond.

Hint
$E = hf$ ⁠, and $f = \dfrac{c}{\lambda}$ ⁠, so $E = \dfrac{hc}{\lambda}$ ⁠, with $\lambda$ ⁠ in metres. One electronvolt is $1.60 \times 10^{-19}$ ⁠ joules.

Worked solution
1. Given: $\lambda = 3.98 \times 10^{-7}\ \mathrm{m}$ ⁠, $h = 6.63 \times 10^{-34}\ \mathrm{J \cdot s}$ ⁠, $c = 3.00 \times 10^8\ \mathrm{m/s}$ ⁠.
2. The frequency: $f = \dfrac{c}{\lambda} = \dfrac{3.00 \times 10^8}{3.98 \times 10^{-7}} \approx 7.54 \times 10^{14}\ \mathrm{Hz}$ ⁠.
3. The energy: $E = hf \approx 6.63 \times 10^{-34} \cdot 7.54 \times 10^{14} \approx 5.00 \times 10^{-19}\ \mathrm{J}$ ⁠.
4. In electronvolts: $\dfrac{5.00 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 3.1\ \mathrm{eV}$ ⁠.
If you got a number a billion times too small, you put in $\lambda = 398$ ⁠, in nanometres. A red photon, about 700 nm, carries only about 1.8 eV.
Step 5 of 5
Open this step
Energy per photon
Practice · Choose
Question
A radio mast transmits thousands of watts, far more energy each second than a dentist’s X-ray machine. Yet X-rays can break molecules in cells, and radio waves cannot. Why?

Correct answer
- Correct Each X-ray photon carries about ten billion times the energy of a radio photon, and a molecule breaks only for a photon with enough energy
Setup, choices and feedback
The Scene at this step
Wavelength (powers of ten)
-10 log m
Brightness
60 %
Add the magnetic field
Offdefault
Choices
Shown in a random order.
- Correct Each X-ray photon carries about ten billion times the energy of a radio photon, and a molecule breaks only for a photon with enough energy
- Wrong X-rays travel faster, so they hit harder
  Misconception: Thinks electromagnetic waves of different wavelengths travel at different speeds in a vacuum.
  Feedback from the catalogue
  In a vacuum all electromagnetic waves travel at the same speed, the speed of light $c \approx 3 \times 10^8\ \mathrm{m/s}$ ⁠, from radio waves to gamma rays. What tells them apart is the wavelength, and so the frequency: $f = \dfrac{c}{\lambda}$ ⁠. A shorter wave shakes faster, but does not move faster.
- Wrong Radio waves are sound waves, so they do not affect atoms
  Misconception: Thinks radio waves are sound waves.
  Feedback from the catalogue
  Radio waves are electromagnetic waves, like light: electric and magnetic fields that need no medium, so they cross empty space too. Sound is a wave of pressure in the air, about a million times slower. A radio receiver picks up the electromagnetic wave, and only its loudspeaker turns the information in it into sound.
- Wrong Radio would break molecules too, if only it were strong enough
  Misconception: Thinks a brighter beam has a higher frequency or more energy per photon (or a different colour).
  Feedback from the catalogue
  Brightness is how many photons arrive each second, and the height of the wave. The frequency, the wavelength and the colour do not change when the light is turned up, and neither does the energy of each photon: $E = hf$ ⁠ depends only on the frequency. Bright red light is many red photons, each one weaker than a single blue photon.
Feedback for a right answer
What decides whether a photon breaks a molecule is the energy of that one photon, $E = hf$ ⁠, not how many photons arrive. An X-ray photon carries about $10^{10}$ ⁠ times as much as a radio photon.

Worked solution
1. Photon energy: $E = hf = \dfrac{hc}{\lambda}$ ⁠, inversely proportional to the wavelength.
2. A radio photon with a wavelength of a metre: $E \approx \dfrac{6.63 \times 10^{-34} \cdot 3 \times 10^8}{1} \approx 2 \times 10^{-25}\ \mathrm{J}$ ⁠.
3. An X-ray photon with a wavelength of $10^{-10}$ ⁠ metres: $E \approx 2 \times 10^{-15}\ \mathrm{J}$ ⁠, $10^{10}$ ⁠ times larger.
Breaking a chemical bond takes a few electronvolts, about $10^{-19}$ ⁠ joules, in one photon. A strong mast sends out a great many photons, but each one is far too weak, and they do not team up to break a bond.

Across the unit · 26 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

Reflection and refraction

An angle from the normal

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Towards the normal

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Snell's law

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Out of the glass

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The critical angle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Prisms and rainbows

Colours that were there all along

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Each colour bends its own way

Question

Correct answer

The Scene at this step

Choices

Worked solution

Index and wavelength

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Inside a raindrop

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

A limit

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Why a bow

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

How far the prism turns it

Question