A-Level Waves and interferences exam-style questions
Use Clevolab for A-Level exam-style practice in waves and interferences. This page shows what the topic covers, what skills the current set targets, and a few real examples from the reviewed question bank.
About this topic
Clevolab treats waves and interferences as repeated practice with explanation, not just answer checking. This page is designed to make the topic legible before you open the app.
The current A-Level set covers superposition, diffraction, standing waves, and interference patterns. 32 reviewed questions currently published for this page.
What you can practise
- Wave behaviour
- Superposition
- Interference patterns
- Standing waves
- Diffraction reasoning
Real sample questions from the current set
These examples come from the reviewed questions currently stored for this topic. They are here so the page shows the actual flavour of Clevolab, not just a summary.
Sample question
A stationary source emits $1000\,\mathrm{Hz}$. An observer moves toward it at $10\,\mathrm{m\,s^{-1}}$; sound speed is $340\,\mathrm{m\,s^{-1}}$. What frequency is heard?
- A1029 HzCorrect answer
- B971 HzAnswer option
- C1000 HzAnswer option
- D1039 HzAnswer option
Why this answer is right
For a moving observer, $f'=f\,\dfrac{v+v_o}{v}$. Here $f'=1000\times\dfrac{340+10}{340}\approx1029\,\mathrm{Hz}$.
When the observer moves, the effective sampling rate of wavefronts changes: $$f'=f\,\frac{v+v_o}{v}.$$ With $f=1000\,\mathrm{Hz}$, $v_o=10\,\mathrm{m\,s^{-1}}$, $v=340\,\mathrm{m\,s^{-1}}$: $$f'=1000\times\frac{350}{340}\approx1029\,\mathrm{Hz}.$$ Moving toward the source increases the received frequency.
Sample question
A string of length $0.80\,\mathrm{m}$ is fixed at both ends. A transverse wave speed of $120\,\mathrm{m\,s^{-1}}$ is measured. What is the fundamental frequency?
- A60 HzAnswer option
- B120 HzAnswer option
- C150 HzAnswer option
- D75 HzCorrect answer
Why this answer is right
For a string with both ends fixed, $f_1=\tfrac{v}{2L}$. Here $f_1=\tfrac{120}{2\times0.80}=75\,\mathrm{Hz}$.
Standing waves on a string fixed at both ends have harmonics $$f_n=\frac{nv}{2L}.$$ For the fundamental $n=1$ with $L=0.80\,\mathrm{m}$ and $v=120\,\mathrm{m\,s^{-1}}$: $$f_1=\frac{120}{2\times0.80}=75\,\mathrm{Hz}.$$ This corresponds to a half-wavelength fitting the string length.
Sample question
Two in-phase coherent sources emit waves of wavelength $0.50\,\mathrm{m}$. At point $P$, the path difference is $0.75\,\mathrm{m}$. What is observed at $P$?
- AA bright maximumAnswer option
- BPartially bright, about half-intensityAnswer option
- CIn-phase additionAnswer option
- DA dark minimumCorrect answer
Why this answer is right
Destructive interference occurs when path difference is $(m+\tfrac{1}{2})\lambda$. Here $0.75=1.5\lambda$, so it is fully destructive.
Interference depends on phase difference, which is set by path difference $\Delta$. Destructive minima occur for $$\Delta=(m+\tfrac{1}{2})\lambda.$$ With $\lambda=0.50\,\mathrm{m}$ and $\Delta=0.75\,\mathrm{m}$, we get $\Delta=1.5\lambda$, matching $m=1$ plus a half. Hence $P$ is a dark minimum.
How this page fits into Clevolab
Clevolab is broader than any one exam mode. GCSE and A-level pages are useful entry points, while the wider project is about sharpening understanding through repeated topic practice.