Question
A source oscillates with frequency \(f\) to produce a progressive wave of wavelength \(\lambda\). The source takes time \(t\) to produce \(n\) complete oscillations.
(a)(i) State what is meant by a progressive wave. [1]
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(ii) State expressions, in terms of some or all of \(f\), \(\lambda\) and \(n\), for:
- the distance moved by a wavefront in time \(t\)
- time \(t\)
distance = ______________________________
time \(t\) = ______________________________ [2]
(iii) Use your answers in (ii) to determine an expression for the speed \(v\) of the wave in terms of \(f\) and \(\lambda\). [1]
\(v=\) ______________________________
(b) Two identical microwave sources X and Y emit waves in phase. The sources are separated by a distance of \(30\,\mathrm{cm}\), as shown in Fig. 4.1.
The intensity of the microwaves is to be investigated at points P and Q. Line PQ is parallel to line XY. Distance XP is equal to distance YP. Distance YQ is \(72\,\mathrm{cm}\) and angle \(XYQ\) is \(90^\circ\).
The wavelength of the microwaves is \(4.0\,\mathrm{cm}\).
(i) Calculate the frequency, in GHz, of the microwaves. [2]
frequency = ______________________________ \(\mathrm{GHz}\)
(ii) Show that the difference between the path lengths XQ and YQ is \(6\,\mathrm{cm}\). [1]
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(iii) State and explain what may be deduced about the intensity of the microwaves at point Q. [3]
_____________________________________
(iv) A microwave detector is positioned at P and connected to a cathode-ray oscilloscope (CRO). The controls of the CRO are adjusted so that a waveform is shown on the screen.
Describe the changes to the amplitude of the waveform as the detector is moved from P to Q. [2]
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 7.1 Progressive waves – wave motion, wavelength, frequency, speed, energy transfer and \(v=f\lambda\).
• 8.3 Interference – coherence, two-source interference, path difference and intensity maxima/minima.
▶️ Answer/Explanation
(a)(i) Progressive wave [1 mark]
A progressive wave is a wave that transfers or propagates energy.
(a)(ii) Expressions [2 marks]
In \(n\) complete oscillations, the wavefront travels through \(n\) wavelengths:
\(\mathrm{distance}=n\lambda\)
The frequency is the number of oscillations per unit time:
\(f=\dfrac{n}{t}\)
Therefore,
\(t=\dfrac{n}{f}\)
(a)(iii) Wave speed [1 mark]
Using \(v=\dfrac{\mathrm{distance}}{\mathrm{time}}\),
\(v=\dfrac{n\lambda}{n/f}\)
\(\boxed{v=f\lambda}\)
(b)(i) Frequency [2 marks]
\(v=f\lambda\)
\(f=\dfrac{v}{\lambda}\)
\(f=\dfrac{3.00\times10^8}{4.0\times10^{-2}}\)
\(f=7.5\times10^9\,\mathrm{Hz}\)
\(\boxed{f=7.5\,\mathrm{GHz}}\)
(b)(ii) Path difference [1 mark]
Triangle \(XYQ\) is right-angled at \(Y\).
\(XQ=\sqrt{72^2+30^2}\)
\(XQ=78\,\mathrm{cm}\)
Therefore,
\(\mathrm{path\ difference}=XQ-YQ=78-72\)
\(\boxed{6\,\mathrm{cm}}\)
(b)(iii) Intensity at Q [3 marks]
The path difference is \(6\,\mathrm{cm}\), while
\(\lambda=4\,\mathrm{cm}\)
Hence,
\(\dfrac{\mathrm{path\ difference}}{\lambda}=\dfrac{6}{4}=1.5\lambda\)
A path difference of \(1.5\lambda\) corresponds to a phase difference of \(540^\circ\), which is equivalent to \(180^\circ\).
The waves therefore arrive at Q in antiphase and undergo destructive interference.
\(\boxed{\text{The intensity at Q is a minimum.}}\)
(b)(iv) CRO amplitude [2 marks]
At P, the waves interfere constructively, so the amplitude is maximum.
As the detector moves from P to Q, the amplitude changes from maximum to minimum, then maximum, then minimum again at Q.
\(\boxed{\text{maximum} \rightarrow \text{minimum} \rightarrow \text{maximum} \rightarrow \text{minimum}}\)