Answers · Waves in a ripple tank

Reflection, refraction and diffraction — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. VQ15 · Practice

Options as numbered on the pupil sheet: 1 — Diffraction; 2 — Reflection; 3 — Refraction.

Model answer. At the concrete edge the ripples stay in the same water but travel away from the edge: they are reflected, and the ripples moving away from the harbour wall have been reflected straight back. Over the sandbar the water is shallower, so the waves slow down; their wavelength gets shorter and they turn: refraction. Curved wavefronts beyond the opening, and moving water behind the end of the barrier, out of line with the waves, show waves spreading out after a gap and bending round an edge: diffraction.

A full answer includes

I matched the concrete edge to reflection.
I matched the sandbar to refraction.
I matched the opening between the rocks to diffraction.
I matched the harbour wall met head-on to reflection.
I matched the water behind the rock barrier to diffraction.
I did not match the sandbar to reflection.

Also accept

concrete edge reflection; sandbar refraction; opening between rocks diffraction; harbour wall reflection; behind the rock barrier diffraction

Watch for

“The curved wavefronts beyond the opening show refraction.” — Refraction is a change of direction caused by a change of speed, where the depth changes. Waves spreading out beyond a gap are diffracted.
“The waves over the sandbar are reflected by it.” — They cross the sandbar and turn because they slow down in the shallow water over it: that is refraction.
“The water behind the barrier moves because the waves are reflected round its end.” — Reflected waves travel back from the barrier. Waves that bend round its end into the water behind it are diffracted.

2. VQ16 · Practice

Model answer. The reflected waves are still in the same water, so they travel at the same speed. The dipper sets the frequency, so it is unchanged too, and so is the wavelength. Only the direction of the waves changes.

A full answer includes

I chose their direction.
I did not choose their wavelength.

Also accept

their direction

Watch for

“Their wavelength” — They travel in the same water at the same frequency, so their wavelength is unchanged.
“Their speed” — The speed of water waves depends on the water. The reflected waves are in the same water, so their speed is unchanged.
“Their frequency” — The dipper sets the frequency, and every wave that arrives is reflected, so the frequency is unchanged.

3. VQ17 · Practice

Model answer. (a) The dipper still vibrates 6 times each second, and every wave that reaches the sheet carries on across it, so the frequency over the sheet is still 6 Hz. (b) λ = v ÷ f = 0.33 m/s ÷ 6 Hz = 0.055 m = 5.5 cm. (c) λ = 0.27 m/s ÷ 6 Hz = 0.045 m = 4.5 cm. Over the sheet the waves are slower at the same frequency, so their wavelength is shorter. This change of speed is what makes waves that cross the edge of the sheet at an angle change direction: refraction.

Checked automatically

(a) Frequency over the sheet: 6 Hz (within ±0.1 Hz)
- Common wrong answer 4.9 : You lowered the frequency with the speed. The dipper sets the frequency, and every wave that arrives crosses the sheet, so it stays the same.
(b) Wavelength in the deep water: 5.5 cm (within ±0.05 cm)
- Common wrong answer 0.055 : That is in metres. Give the wavelength in centimetres.
- Common wrong answer 18 : That is f ÷ v. The wavelength is λ = v ÷ f.
(c) Wavelength over the sheet: 4.5 cm (within ±0.05 cm)
- Common wrong answer 0.045 : That is in metres. Give the wavelength in centimetres.
- Common wrong answer 22 : That is f ÷ v. The wavelength is λ = v ÷ f.
- Common wrong answer 5.5 : That is the wavelength in the deep water. Over the sheet the waves are slower at the same frequency, so their wavelength is shorter.

A full answer includes

(a) I wrote 6 Hz.
(a) I did not write a lower frequency over the sheet.
(b) I wrote λ = v ÷ f = 0.33 m/s ÷ 6 Hz = 0.055 m.
(b) I wrote 5.5 cm.
(c) I wrote λ = 0.27 m/s ÷ 6 Hz = 0.045 m.
(c) I wrote 4.5 cm.
(c) I did not give the same wavelength as in the deep water.

Also accept

(a) 6 Hz (b) 5.5 cm (c) 4.5 cm

Watch for

“The frequency falls over the sheet, because the waves slow down.” — The dipper sets the frequency, and every wave that arrives carries on across: it stays 6 Hz. The speed and the wavelength fall.
“λ = f ÷ v.” — That is upside down. Dividing both sides of v = fλ by f leaves λ = v ÷ f.

4. VQ18 · Practice

Model answer. (a) In the deep water λ = 1.1 cm = 0.011 m, so v = fλ = 20 Hz × 0.011 m = 0.22 m/s. (b) Every wave that reaches the tile carries on across it, so the frequency over the tile is still 20 Hz. λ = 0.70 cm = 0.0070 m, so v = 20 Hz × 0.0070 m = 0.14 m/s: the waves are slower over the tile, where the water is shallower.

Checked automatically

(a) Speed in the deep water: 0.22 m/s (within ±0.005 m/s)
- Common wrong answer 22 : That is in cm/s: change 1.1 cm to metres first.
(b) Speed over the tile: 0.14 m/s (within ±0.005 m/s)
- Common wrong answer 14 : That is in cm/s: change 0.70 cm to metres first.
- Common wrong answer 0.22 : Over the tile the wavelength is shorter, so at the same frequency the speed is lower.

A full answer includes

(a) I changed 1.1 cm to 0.011 m.
(a) I wrote v = fλ = 20 Hz × 0.011 m.
(a) I wrote 0.22 m/s.
(b) I used the same frequency, 20 Hz, over the tile.
(b) I wrote v = 20 Hz × 0.0070 m.
(b) I wrote 0.14 m/s.
I did not give my answers in cm/s.

Also accept

(a) 0.22 m/s (b) 0.14 m/s

Watch for

“The wavelength is shorter over the tile, so the frequency is lower there too.” — The frequency is set by the dipper and does not change; the speed falls, and so does the wavelength.

5. VQ19 · Practice

Model answer. (a) Region B. Its wavefronts are closer together, so the wavelength is shorter. The frequency is the same in both regions, so the waves must be slower in B (v = fλ), and water waves are slower in shallower water. (b) The wavefronts reach the boundary at an angle. One end of each wavefront crosses into B and slows down first, while the other end is still moving faster in A, so the wavefront swings round and the waves travel in a new direction. This is refraction. (c) In region A the wavefronts are 3.0 cm = 0.030 m apart, so f = v ÷ λ = 0.18 m/s ÷ 0.030 m = 6.0 Hz.

A full answer includes

(a) I identified region B.
(a) I wrote that its wavefronts are closer together, so its wavelength is shorter.
(a) I wrote that at the same frequency a shorter wavelength means the waves are slower.
(a) I wrote that water waves are slower in shallower water.
(a) I did not write that the waves bunch up in B because the water is deeper.
(b) I wrote that each wavefront meets the boundary at an angle.
(b) I wrote that one end of each wavefront enters B and slows down first.
(b) I wrote that the wavefront swings round: refraction.
(c) I read the wavelength in A from the arrow as 3.0 cm.
(c) I changed it to 0.030 m.
(c) I wrote f = v ÷ λ = 0.18 m/s ÷ 0.030 m.
(c) I wrote 6.0 Hz.
(c) I did not write 0.060 Hz.

Also accept

(a) B: shorter wavelength, so slower (b) one end slows first, so the wavefront swings round (c) 6.0 Hz

Watch for

“Region B is deeper, because the waves bunch up where there is more water.” — The waves bunch up because they slow down, and they slow down in shallow water.
“The waves change direction because the boundary reflects them.” — They cross the boundary and turn because their speed changes: that is refraction.
“f = 0.18 ÷ 3.0 = 0.060 Hz.” — That uses the wavelength in centimetres. Change 3.0 cm to metres first.

6. VQ20 · Practice

Model answer. Over the rock the water is shallower, and water waves travel more slowly in shallower water. The wavefronts arrive at an angle, so one end of each slows down first and the wavefront swings round: the waves change direction. This is refraction; the frequency stays the same.

A full answer includes

I chose: the water is shallower there, so the waves slow down.
I did not choose that their frequency falls.

Also accept

the water is shallower, so the waves slow down

Watch for

“The rock reflects them, so they turn back the way they came.” — The waves carry on across the rock; they are not sent back. They turn because their speed changes.
“They bend round the edge of the rock, so they spread out.” — Bending round an edge is diffraction, at a barrier that stands up out of the water. Here the waves pass over the rock and change speed.
“The water is shallower there, so their frequency falls.” — The frequency is set by the source and does not change. It is the speed, and so the wavelength, that falls.

7. VQ21 · Practice

Options as numbered on the pupil sheet: 1 — Diffraction by an edge; 2 — Reflection at a plane surface; 3 — Refraction; 4 — Diffraction by a gap.

Model answer. A flat face standing in the water, such as a brick or a long block, is a plane surface: it reflects the waves, straight back the way they came if it faces them head-on. A sheet lying on the bottom makes the water shallower, so the waves slow down over it and are refracted. A narrow gap between two blocks makes the waves spread out beyond it: diffraction by a gap. Waves passing the end of a single ruler bend round it into the water behind: diffraction by an edge.

A full answer includes

I matched the brick at an angle to reflection at a plane surface.
I matched the plastic sheet to refraction.
I matched the two blocks with a gap to diffraction by a gap.
I matched the ruler on its edge to diffraction by an edge.
I matched the long block facing the waves to reflection at a plane surface.
I did not match the plastic sheet to reflection.

Also accept

brick reflection; plastic sheet refraction; two blocks diffraction by a gap; ruler diffraction by an edge; long block reflection

Watch for

“The plastic sheet reflects the waves, because they meet its surface.” — The sheet lies under the water and makes it shallower, so the waves slow down over it and are refracted.
“The ruler on its edge reflects the waves, because it is a flat surface.” — Its flat face reflects waves that hit it, but waves passing its end bend round into the water behind it: diffraction by an edge.
“The two blocks show diffraction by an edge, and the ruler shows diffraction by a gap.” — The waves pass through the opening between the two blocks: diffraction by a gap. At the single ruler they pass its end, with open water beyond: diffraction by an edge.

8. VQ22 · Practice

Model answer. (a) Lay a flat sheet, such as a glass plate, on the bottom of the tank, so that the water over it is shallower. (b) Turn the sheet so that its edge is at an angle to the straight wavefronts from the dipper. (c) On the screen, the wavefronts over the sheet are closer together, because the waves are slower there, and they have turned to travel in a new direction. The frequency is the same on both sides of the edge.

A full answer includes

(a) I wrote a flat sheet, such as a glass plate, on the bottom of the tank.
(a) I wrote that it makes the water shallower over part of the tank.
(b) I wrote that the edge of the sheet is at an angle to the wavefronts.
(b) I did not set the edge parallel to the wavefronts.
(c) I wrote that over the sheet the wavefronts are closer together.
(c) I wrote that over the sheet the waves travel in a new direction.

Also accept

(a) a flat sheet on the bottom (b) edge at an angle to the wavefronts (c) closer wavefronts, new direction

Watch for

“The sheet should be parallel to the wavefronts, so that the waves meet it fairly.” — Waves that meet the edge head-on slow down but carry straight on. To see a change of direction, the edge must be at an angle.

9. VQ23 · Practice

Model answer. At a plane surface, waves are reflected at the same angle as they arrive: the reflected wavefronts also make 50° with the barrier, turned the other way. The reflected waves are in the same water and come from the same dipper, so their speed, frequency and wavelength are unchanged. Row C.

A full answer includes

I chose row C.
I did not choose row B, where the wavelength is shorter.

Also accept

Row C

Watch for

“Row A.” — 40° is the angle between the barrier and the direction the waves travel. The reflected wavefronts make the same angle with the barrier as the arriving ones.
“Row B.” — The reflected waves are in the same water at the same frequency, so their speed and their wavelength are unchanged.
“Row D.” — The reflected wavefronts make the same angle with the barrier as the arriving ones, and in the same water at the same frequency the wavelength does not change.

10. VQ24 · Practice

Model answer. (a) As the waves pass the open end of the barrier, they bend round it into the water behind it, and their wavefronts there become curved. (b) Yes. The waves that bend round the edge reach the water behind the barrier, so the ball bobs up and down, even though it is out of line with the waves arriving. This is diffraction by an edge. (c) The water behind the barrier is as deep as the rest and the dipper is the same, so the wavelength, the frequency and the speed of the waves are all the same as before; only their direction changes.

A full answer includes

(a) I wrote that the waves bend round the end of the barrier into the water behind it.
(a) I wrote that their wavefronts there are curved.
(b) I wrote that the ball bobs up and down.
(b) I explained that the waves bending round the end reach it.
(b) I did not write that the water behind the barrier stays still.
(c) I gave one property that stays the same: the wavelength, the frequency or the speed.
(c) I did not write that the waves behind the barrier are longer.

Also accept

(a) bend round the end (b) yes, it bobs (c) the wavelength (or the frequency, or the speed)

Watch for

“The water behind the barrier stays still, because waves travel only in straight lines.” — Past an edge, waves bend round into the water behind it: this is diffraction.
“The waves that bend round the end get longer as they spread out.” — They are in the same water at the same frequency, so their wavelength is unchanged; only their direction changes.

11. VQ25 · Practice

Model answer. Waves spread out most when the gap is about the same size as their wavelength. The 4 cm gap is closest to one wavelength, about 1.3 wavelengths wide, so beyond it the waves spread out the most, in curves. The 10 cm and 20 cm gaps are about 3.3 and 6.7 wavelengths wide, so the waves go mostly straight on and curve only at the sides.

A full answer includes

I chose the gap 4 cm wide.
I did not choose the widest gap.

Also accept

the 4 cm gap

Watch for

“A gap 10 cm wide” — A wider gap lets more of the wave through, but the waves spread out less: 10 cm is more than three wavelengths wide.
“A gap 20 cm wide” — The widest gap lets most through, but the waves pass mostly straight on: 20 cm is more than six wavelengths wide.
“All gaps equally” — How much the waves spread depends on the gap compared with the wavelength, so different gaps give different spreading.

12. VQ26 · Practice

Model answer. (a) The gap is five wavelengths wide, so the waves pass through mostly straight on, in a broad band, and curve only a little at its sides. (b) A 1.0 cm gap is about one wavelength wide, so the waves spread out much more, in semicircles, into the water on both sides beyond it. (c) λ = v ÷ f. The speed is the same and the frequency is one third as big, so the wavelength is three times as long: 3 × 1.0 cm = 3.0 cm. The 5.0 cm gap is now less than two wavelengths wide, so the waves spread out more than in (a).

A full answer includes

(a) I wrote that the waves go mostly straight on.
(a) I wrote that they curve only a little at the sides.
(b) I wrote that they spread out much more, in curves (semicircles).
(b) I explained that the gap is now about one wavelength wide.
(c) I wrote that one third of the frequency at the same speed gives three times the wavelength.
(c) I wrote 3.0 cm.
(c) I did not divide the wavelength by 3.
(c) I wrote that the waves spread out more.

Also accept

(a) mostly straight on, curving a little at the sides (b) spread out much more: the gap is about one wavelength (c) 3.0 cm; they spread out more

Watch for

“A lower frequency gives a shorter wavelength.” — At the same speed, λ = v ÷ f: fewer waves each second means each wave is longer.
“Narrowing the gap makes the waves spread out less, because less of the wave gets through.” — Less of the wave gets through, but it spreads out more: most when the gap is about one wavelength wide.

13. VQ27 · Practice

Model answer. The depth is the same, so the speed is the same. From λ = v ÷ f, the lower frequency gives the longer wavelength: the 4 Hz waves are three times as long as the 12 Hz waves. The longer the wavelength, the further waves bend round an edge, so the 4 Hz waves bend further. Row D.

A full answer includes

I chose row D.
I did not choose row B, where the longer waves bend less.

Also accept

Row D

Watch for

“Row A.” — At the same speed, more waves each second means each wave is shorter: the 12 Hz waves have the shorter wavelength, and shorter waves bend less round an edge.
“Row B.” — The 4 Hz waves do have the longer wavelength, but it is the longer waves that bend further round an edge.
“Row C.” — At the same speed, λ = v ÷ f, so the 4 Hz waves are the longer ones. Your two columns also disagree: the waves that bend further are the ones with the longer wavelength.

14. VT2 · Apply

Model answer. (a) The opening is a gap in the walls, so the waves diffract: beyond it the wavefronts are curved and spread out into the water behind the walls. They are in the same water and come from the same machine, so their wavelength is still 2.4 m. (b) As the water gets shallower, the waves slow down. The frequency stays 0.80 Hz, so the wavelength gets shorter (λ = v ÷ f). (c) Waves spread out most through a gap about the same size as their wavelength. The 2.4 m waves are about the width of the 2.5 m opening, so they spread out a lot and reach the swimmers behind the walls. The 0.20 m waves are more than ten times shorter than the opening is wide, so they pass through mostly straight on and hardly reach the water behind the walls.

A full answer includes

(a) I wrote that the wavefronts beyond the opening are curved and spread out.
(a) I wrote that the wavelength is still 2.4 m.
(a) I did not write that the wavelength changes as the waves spread out.
(b) I wrote that the speed decreases.
(b) I wrote that the wavelength decreases.
(b) I wrote that the frequency stays the same, 0.80 Hz.
(c) Supplement: I wrote that the 0.20 m waves are far shorter than the 2.5 m opening is wide.
(c) Supplement: I wrote that these waves hardly spread out beyond the opening.
(c) Supplement: I wrote that they pass mostly straight on, so little of them reaches the water behind the wall.
(c) Supplement: I did not explain it by how tall the waves are.

Also accept

(a) curved, spreading out; still 2.4 m (b) slower, shorter wavelength; frequency the same (c) 0.20 m is far smaller than the opening, so they hardly spread behind the wall

Watch for

“The walls reflect the waves round the corner to the swimmers.” — The walls reflect waves back towards the machine; it is diffraction at the opening that spreads them behind the walls.
“The waves get longer as they spread out beyond the opening.” — They are in the same water at the same frequency, so their speed and wavelength are unchanged; only their direction changes.
“Bigger waves always spread out more.” — How much waves spread through a gap depends on the wavelength compared with the gap, not on how tall the waves are.