A-level Physics
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    6.2 Diffraction and polarisation — Practise with help

    Work each problem first. Open a hint only if you are stuck, and the full working only to check. Angles from a grating are measured from the straight-through direction.

    Warm-up: remember

    1. A wave has a wavelength of 0.60 m and a frequency of 550 Hz. Find its speed.
    2. Two points on a wave are 90∘90^\circ apart in phase. What fraction of a wavelength is the distance between them?
    3. Write 6.33×10−7 m6.33 \times 10^{-7}\,\text{m} in nanometres.
    Answers
    1. v=fλ=550×0.60=330 m s−1v = f\lambda = 550 \times 0.60 = 330\,\text{m}\,\text{s}^{-1}.
    2. 90∘90^\circ is one quarter of 360∘360^\circ, so the distance is λ4\dfrac{\lambda}{4}.
    3. 6.33×10−7 m=633 nm6.33 \times 10^{-7}\,\text{m} = 633\,\text{nm}.

    If question 1 goes wrong, use v=fλv = f\lambda. If question 2 goes wrong, remember that a full cycle is 360∘360^\circ. If question 3 goes wrong, one nanometre is 10−9 m10^{-9}\,\text{m}.

    Practise with help

    1. A diffraction grating has 500 lines per millimetre. Find the spacing dd between neighbouring slits, in metres.

    Hint

    The spacing is one divided by the number of lines per metre. Change the lines per millimetre into lines per metre first.

    Next step

    500 lines mm−1=5.00×105 lines m−1500\,\text{lines mm}^{-1} = 5.00 \times 10^5\,\text{lines m}^{-1}. Then use d=1Nd = \dfrac{1}{N}.

    Full working N=5.00×105 m−1d=1N=15.00×105=2.00×10−6 m\begin{aligned} N &= 5.00 \times 10^{5}\,\text{m}^{-1} \\ d &= \frac{1}{N} = \frac{1}{5.00 \times 10^{5}} = 2.00 \times 10^{-6}\,\text{m} \end{aligned}

    The slits are 2.00×10−6 m2.00 \times 10^{-6}\,\text{m} apart.

    2. Light of wavelength 633 nm falls on a grating of spacing 2.00×10−6 m2.00 \times 10^{-6}\,\text{m}. Find the angle of the first-order maximum.

    Hint

    Use dsin⁡θ=nλd\sin\theta = n\lambda with n=1n = 1. Change the wavelength to metres first.

    Next step

    Work out sin⁡θ=633×10−92.00×10−6\sin\theta = \dfrac{633 \times 10^{-9}}{2.00 \times 10^{-6}}, then take the inverse sine.

    Full working sin⁡θ=nλd=633×10−92.00×10−6=0.3165θ=18.5∘\begin{aligned} \sin\theta &= \frac{n\lambda}{d} = \frac{633 \times 10^{-9}}{2.00 \times 10^{-6}} = 0.3165 \\ \theta &= 18.5^\circ \end{aligned}

    The first-order maximum is at 18.5∘18.5^\circ from the straight-through direction.

    3. For the grating and light in question 2, find the highest order the grating can give.

    Hint

    The order is possible while sin⁡θ\sin\theta is no more than 1, which means nλ≤dn\lambda \le d.

    Next step

    Divide the slit spacing by the wavelength: dλ\dfrac{d}{\lambda}. The highest order is the largest whole number below that value.

    Full working dλ=2.00×10−6633×10−9=3.16\frac{d}{\lambda} = \frac{2.00 \times 10^{-6}}{633 \times 10^{-9}} = 3.16

    The order must be a whole number and nλn\lambda must be no more than dd. The largest such number is 3, so the highest order is 3. Order 4 would need sin⁡θ=1.27\sin\theta = 1.27, which is impossible.

    4. A grating of spacing 1.667×10−6 m1.667 \times 10^{-6}\,\text{m} gives its first-order maximum at 22.0∘22.0^\circ. Find the wavelength of the light.

    Hint

    Rearrange the grating equation to make the wavelength the subject.

    Next step

    With n=1n = 1, the wavelength is dsin⁡θd\sin\theta. Find sin⁡22.0∘\sin 22.0^\circ first.

    Full working dsin⁡θ=nλλ=dsin⁡θ=1.667×10−6×sin⁡22.0∘=1.667×10−6×0.3746=6.24×10−7 m\begin{aligned} d\sin\theta &= n\lambda \\ \lambda &= d\sin\theta = 1.667 \times 10^{-6} \times \sin 22.0^\circ \\ &= 1.667 \times 10^{-6} \times 0.3746 = 6.24 \times 10^{-7}\,\text{m} \end{aligned}

    The wavelength is 6.24×10−7 m6.24 \times 10^{-7}\,\text{m}, which is 624 nm.

    5. Plane-polarised light passes through an analyser whose axis is at 45∘45^\circ to the plane of polarisation. Find the fraction of the intensity that passes through.

    Hint

    The intensity follows the square of the cosine of the angle.

    Next step

    Work out cos⁡45∘\cos 45^\circ first, then square it.

    Full working II0=cos⁡245∘=(12)2=0.50\frac{I}{I_0} = \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = 0.50

    Half of the intensity passes through the analyser.

    6. Explain why light can be polarised but sound cannot. Include the words transverse and longitudinal in your answer.

    Hint

    Polarisation is about the direction in which the vibration happens, compared with the direction of travel.

    Next step

    Say what the particles do in each type of wave, and which direction a filter can pick out.

    Full working

    Light is a transverse wave, so its electric field vibrates at right angles to the direction of travel. That vibration has a direction across the beam, and a polarising filter can select the component along one line. Sound is a longitudinal wave. Its particles vibrate along the direction of travel, so there is no crossways direction for a filter to select. Polarisation is therefore possible only for transverse waves.

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