Physics · AQA 7408 · Topic 3.3

Waves

Progressive and stationary waves, interference, diffraction, refraction and optical fibres.

  • 6 sub-topics
  • Taught in Year 12
  • 87 practice questions
  • 6 animated lessons
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What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.3.1 Progressive and stationary waves
  2. 3.3.1.1Progressive wavesY12
  3. 3.3.1.2Longitudinal and transverse wavesY12
  4. 3.3.1.3Principle of superposition of waves and formation of stationary wavesY12
  5. 3.3.2 Refraction, diffraction and interference
  6. 3.3.2.1InterferenceY12
  7. 3.3.2.2DiffractionY12
  8. 3.3.2.3Refraction at a plane surfaceY12

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 87 questions in this topic, free, no account needed.

Waves and the particle nature of light

Easy

The wave equation · Year 12 · 2 marks

A tuning fork produces a sound of frequency 440 Hz. The wavelength of the sound in air is 0.78 m.
  1. Calculate the speed of sound in air.
    (2)

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  1. Wave speed = frequency × wavelength.v=fλ=440×0.78v = f\lambda = 440 \times 0.78
  2. Round to match the data.v=343 m s−1v = 343\ \mathrm{m\,s^{-1}}

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Waves and the particle nature of light

Medium

Refraction · Year 12 · 4 marks

Figure 1 shows a ray of light passing from air into a material of refractive index 1.52 at an angle of incidence of 40°.
irairglass
Figure 1
  1. (a)
    Calculate the angle of refraction.
    (2)
  2. (b)
    Calculate the speed of light in the material.
    (1)
  3. (c)
    Explain why the ray bends towards the normal.
    (1)

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Part (a)2 marks

  1. sin⁡r\sin r=sin⁡40∘1.52{}= \frac{\sin 40^\circ}{1.52}
  2. r=25.0 ∘r = 25.0\ \mathrm{^\circ}

Part (b)1 mark

  1. vv=1.97×108 m s−1{}= 1.97 \times 10^{8}\ \mathrm{m\,s^{-1}}

Part (c)1 mark

  1. It slows down on entering the denser medium.

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Waves

Hard

Double-slit interference · Year 12 · 9 marks

Figure 1 shows a double-slit arrangement. Light of wavelength 520 nm passes through two slits a distance s = 0.25 mm apart and forms fringes on a screen D = 2.0 m away.
sDwlaser
Figure 1
  1. (a)
    Explain why the two slits must act as coherent sources.
    (2)
  2. (b)
    Calculate the fringe spacing w.
    (3)
  3. (c)
    Calculate the distance on the screen from the central bright fringe to the 4th bright fringe.
    (1)
  4. (d)
    State the path difference, in terms of λ, for light arriving at the first dark fringe.
    (1)
  5. (e)
    The light is replaced by light of wavelength 650 nm. Calculate the new fringe spacing and describe the change in the pattern.
    (2)

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Part (a)2 marks

  1. Say what coherent means.Coherent sources have a constant phase difference (and the same frequency).
  2. Link it to the pattern.Without this the positions of the bright and dark fringes would keep changing, so no steady pattern would be seen.

Part (b)3 marks

  1. The fringe spacing equation.w=λDsw = \frac{\lambda D}{s}
  2. Convert nm and mm to metres.ww=520×10−9×2.00.25×10−3{}= \frac{520 \times 10^{-9} \times 2.0}{0.25 \times 10^{-3}}
  3. w=0.00416 mw = 0.00416\ \mathrm{m}

Part (c)1 mark

  1. Bright fringes are evenly spaced.4w=0.0166 m4w = 0.0166\ \mathrm{m}

Part (d)1 mark

  1. Dark fringes are where the waves arrive in antiphase.Half a wavelength (λ/2).

Part (e)2 marks

  1. w is proportional to λ.w=0.00520 mw = 0.00520\ \mathrm{m}
  2. Compare the two spacings.The fringes are further apart (and a different colour).

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