Question 1 Report
A student sets up a ripple tank experiment to compare the diffraction of waves through gaps of different widths. She uses a motor to generate plane waves and records her observations.
| Gap width / cm | Wavelength / cm | Ratio gap/wavelength | Observed pattern after gap |
|---|---|---|---|
| 1.0 | 2.0 | 0.5 | almost fully circular wavefronts |
| 2.0 | 2.0 | 1.0 | circular wavefronts with strong spreading |
| 4.0 | 2.0 | 2.0 | moderate spreading at edges, mostly plane in centre |
| 8.0 | 2.0 | 4.0 | slight edge spreading only |
| 16.0 | 2.0 | 8.0 | very little spreading, nearly all plane |
(a) State the conclusion the student can draw about the relationship between the gap/wavelength ratio and the amount of diffraction. [2]
(b) State the gap/wavelength ratio that produces the most effective diffraction (circular wavefronts). [1]
(c) The student wants to increase the wavelength without changing the motor. Describe how she could do this. [1]
(d) Calculate the wave speed if the motor vibrates at 5.0 Hz and the wavelength is 2.0 cm. [1]
(e) The student notices that at the widest gap (16.0 cm), the wavefronts are almost plane. Explain why this is similar to light passing through a window. [2]
(f) Suggest one improvement to make the experiment more quantitative. [1]
(g) State one safety precaution when using the ripple tank. [1]
(h) Explain why the student should use sloped edges (beaches) at the sides of the tank. [1]
(a)
The table shows this trend clearly: at a ratio of 0.5, the wavefronts are almost fully circular (maximum spreading), while at a ratio of 8.0, there is very little spreading. Diffraction is strongest when the gap is comparable to the wavelength.
(b) The most effective diffraction (fully circular wavefronts) occurs at a gap/wavelength ratio of 0.5 (or 1.0). [1]
At ratio 0.5 the gap is smaller than the wavelength, forcing the waves to spread in all directions as if the gap were a point source.
(c) She could increase the depth of the water in the tank. [1]
Water waves travel faster in deeper water. Since the motor keeps the frequency constant, \(v = f\lambda\) means a higher speed produces a longer wavelength.
(d)
\(v = f \times \lambda = 5.0 \times 2.0 = 10\text{ cm/s}\) [1]
(e)
This is exactly the same physics as the 16.0 cm gap in the experiment: when the gap is many times larger than the wavelength, the waves pass through with almost no spreading.
(f) Measure the angle of spread of the diffracted wavefronts (e.g. using a protractor or by photographing the pattern and measuring digitally). [1]
Quantifying the angle allows a numerical comparison between different gap/wavelength ratios, rather than relying on qualitative descriptions.
(g) Mop up any water spills immediately to prevent slipping, or keep electrical connections away from the water. [1]
(h) The sloped edges (beaches) absorb the waves and prevent reflections from the tank walls. [1]
Without beaches, reflected waves would overlap with the incident waves and create a confused interference pattern, making it impossible to observe the diffraction clearly.
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