Describe ripple tank refraction across a depth change
Use a shallow region to compare transmitted spacing and direction while retaining the source cycle rate.
Cross-referenced from the related Cambridge IGCSE Chemistry 0620 or Physics 0625 teaching to the 0652 syllabus. Exact 0652 Core/Supplement objective coverage and this final-series course’s examiner fit still need specialist review. Written practice is self-checked and does not predict a Cambridge grade or demonstrate supervised laboratory competence.
Before you begin
- Explain refraction and normal incidence.
- Use v = fλ and consistent front spacing.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the intended role of the plate in this experiment?
What you will learn
- Describe a ripple-tank refraction experiment using a change of depth.
- Interpret speed, frequency and wavelength under the stated stationary conditions.
Use a raised region to change water depth
A teacher-supervised ripple tank can contain a suitable transparent plate that makes water shallower over part of the tank. Use a straight vibrating bar to send fronts toward the depth boundary, with suitable source amplitude, illumination and viewing screen. Under the selected wave conditions, the shallower region has lower propagation speed. Observe spacing on both sides and direction for an oblique boundary.
Keep the source frequency steady and the boundary stationary. Fronts closer together over the plate show shorter wavelength; for unchanged frequency, v = fλ gives lower speed. With oblique entry, the direction is closer to the normal in the slower region. For normal entry, spacing can change without a change of propagation direction.
Compare consistent repeated features
Measure adjacent corresponding fronts along their propagation direction using the observation scale, not an arbitrary distance along the boundary. Several complete wavelengths can improve the spacing estimate when a clear pattern permits it. A supplied 4.0 Hz source with 0.15 m and 0.10 m spacings gives 0.60 and 0.40 m/s; these are teaching values, not real tank measurements.
Disturbed flow, tank-edge reflections or an uncontrolled source change can complicate the comparison. State the actual depth/source conditions and observation method. The plate must create a water-depth boundary rather than a dry solid wall blocking the whole path. A description and supplied calculations do not establish apparatus competence or a completed practical assessment.
Sources for this lesson
Pause and explain
At 4.0 Hz, a measured-model spacing changes from 0.15 to 0.10 m. Which speed comparison follows?
Worked example
Supplied fixed-frequency fronts have spacings 0.15 m in the deeper region and 0.10 m over a shallow plate, at 4.0 Hz. Find speeds.
Show the worked solution
- Use v = fλ for each region.
- Deeper-region speed = 4.0 × 0.15 = 0.60 m/s.
- Shallower-region speed = 4.0 × 0.10 = 0.40 m/s.
Answer 0.60 m/s then 0.40 m/s; source frequency remains 4.0 Hz.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe apparatus and observations for refraction over a shallow region in a ripple tank.
Give me a hint
Use transmission through a depth change with a steady source.
Compare my reasoning
- Use a suitable plate below the water to create two depths.
- Send approximately straight waves at a stationary boundary using a steady source.
- Compare front spacing and, for oblique incidence, normal-referenced direction.
A steady-source depth-boundary comparison shows changed spacing and possible oblique bending.
Look for these in your work
- Describe the plate's purpose.
- State the measured pattern features.
For supplied f = 5.0 Hz, deeper spacing 0.060 m and shallower spacing 0.040 m, calculate speeds.
Give me a hint
Apply the same frequency separately to both repeat lengths.
Compare my reasoning
- Deeper speed = 5.0 × 0.060 = 0.30 m/s.
- Shallower speed = 5.0 × 0.040 = 0.20 m/s.
- The smaller spacing accompanies the lower speed.
0.30 m/s and 0.20 m/s.
Look for these in your work
- Use consistent units.
- Retain fixed frequency.
The source rate changes between the two depth observations. Can spacing alone isolate the depth effect on speed?
Give me a hint
Both frequency and wavelength now enter the comparison.
Compare my reasoning
- v = fλ uses each observation's actual frequency.
- The changed source rate affects spacing independently of a depth change.
- Use a matched steady source or record both frequencies before attributing the difference.
No depth-only conclusion follows from unmatched source frequencies and spacing alone.
Look for these in your work
- Identify the changed variable.
- Specify a fairer comparison.
Common mistakes
- The plate changes depth rather than being a full dry barrier.
- Measure spacing along propagation, not along the boundary.
- No bending is required at normal incidence.
What can you explain now?
Do closer transmitted fronts at unchanged frequency imply a higher speed?
Compare with the explanation
No; lower speed.
The smaller wavelength gives smaller fλ when frequency stays fixed.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Describe a controlled depth-change tank method and use matched frequency to interpret transmitted spacing.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.