Oxford Revise AQA A Level Physics | Chapter P5 answers
P5: Reflection, diffraction, and interference
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Question |
Answers |
Extra information |
Mark |
AO |
Spec reference |
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01.1 |
One with a constant/fixed phase relationship/difference |
1 |
1 |
3.3.2.1 |
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01.2 |
One with a single wavelength/frequency |
1 |
1 |
3.3.2.1 |
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01.3 |
Do not look directly at laser/do not point laser at anyone/do not look at reflection of laser light/wear safety goggles |
Allow any sensible suggestion |
1 |
1 |
3.3.2.1 |
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01.4 |
\( \begin{array}{l} w = \frac{8 \times 10^{-3}}{4}\ \rm{m}\ = 2\times10^{-3}\ \rm{m}\\ w = \frac{\lambda D}{s}\\ \lambda = \frac{ws}{D} = \frac{2 \times 10^{-3} \times 0.4 \times 10^{-3}}{1.5}\\ \lambda = 5.3 \times 10^7\ \rm{m} \end{array} \) |
1 |
2 |
3.3.2.1 |
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01.5 |
\( \begin{array}{l} \textrm{uncertainty in }D = \frac{0.001}{1.5} \times 100\% = 0.07\%\\ \textrm{uncertainty in }s = \frac{0.01}{0.40} \times 100\% = 2.5\%\\ \textrm{uncertainty in }w = \frac{0.1}{8.0} \times 100\% = 1.3\%\\ \textrm{uncertainty in }\lambda = 0.07 + 2.5 + 1.3 = 3.9\%\\ \end{array} \) |
1 |
2 |
3.3.2.1 |
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01.6 |
\(s\textrm{ and }D\textrm{ remain constant so } \lambda \propto w\)
Longer λ means the maxima would be further apart |
Can be expressed in words but must state s and D constant for this mark ignore ‘different colour’ |
1 |
3 |
3.3.2.1 |
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02.1 |
\( \begin{array}{l} n_1 \sin \ \theta_1 = n_2 \sin \ \theta_2\\ n_1 = 1\\ \sin \ \theta_2 = \frac{\sin \ \theta_1}{n_1} = \frac{\sin \ \ 60}{1.5}\\ \theta_2 = 35^{\circ} \end{array} \) |
1 |
2 |
3.3.2.3 |
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02.2 |
\( \begin{array}{l} \sin \ \theta_{\rm{C}}\ = \frac{n_1}{n_2} = \frac{1}{1.5}\\ \theta_{\rm{C}}\ = 42\ (41.8^{\circ}) \end{array} \) |
1 |
2 |
3.3.2.3 |
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02.3 |
Angle of incidence side KL = 55° Since this is > critical angle, ray is totally internally reflected |
Could be shown on sketch/on the diagram – possible e.c.f here from 02.1 |
1 |
3 |
3.3.2.3 |
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02.4 |
\( \begin{array}{l} \sin \ \theta_{\rm{C}} = \frac{1.4}{1.5}\\ \theta_{\rm{C}} = 59^{\circ} \end{array} \) |
1 |
2 |
3.3.2.3 |
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03.1 |
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Any mention of nodes/antinodes loses marks |
max 3 |
2 |
3.3.2.1 |
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03.2 |
f = 1500 Hz c = 340 m s−1 c = fλ \(\lambda = \frac{c}{f} = \frac{340}{1500} = 0.23\ \rm{m}\) |
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2 |
3.3.1.1 |
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03.3 |
\( \begin{array}{l} w = \frac{\lambda D}{s}\\ w = 0.23 \times \frac{20}{10} = 0.46\ \rm{m} \end{array} \) |
1 |
2 |
3.3.2.1 |
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03.4 |
\(f \propto \frac{1}{\lambda}\textrm{ so }\lambda\textrm{ halves}\)
s and D remain constant so \(\lambda \propto w\) Minima will be closer together |
ignore ‘higher pitched’ |
1 |
3 |
3.3.2.1 |
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04.1 |
Place the diffraction grating at a distance of 4 m (must be >1 m) from a screen Measure distance with a metre ruler or tape measure Shine laser directly onto grating (Identify the central maxima) and measure the distance of the first order maxima either side with a ruler Find the mean (or measure distance between 1st order and divide by 2) |
Correct names for measuring instruments should be given One point describing how to make results more accurate required for full marks |
max 4 |
2 |
3.3.2.2 |
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04.2 |
\(\frac{1 \times 10^{-3} \rm{m}}{330} = 3.0 \times 10^{-6}\ \rm{m}\) |
1 |
2 |
3.3.2.2 |
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04.3 |
nλ = d sin θ λ = 3.0 × 10−6 sin 12.5 = 6.5 × 10−7 m λ = 650 nm (649 nm) |
1 |
2 |
3.3.2.2 |
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04.4 |
Central white maxima Each of the orders is now a spectrum Violet closest to the centre/red furthest from centre \(\lambda\ \propto\ \theta \textrm{ so, as } \lambda \textrm{ increases, so does } \theta\) |
max 3 |
3 |
3.3.2.2 |
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05.1 |
\( \begin{array}{l} n_2 = \rm{air} = 1\\ \sin \ \theta_{\rm{c}} = \frac{n_2}{n_1} = \frac{1}{1.6}\\ \theta_{\rm{C}} = 39^{\circ} (38.7^{\circ}) \end{array} \) The angle of incidence = 50° (90 – 40) so angle of incidence < critical angle, so light will be totally internally reflected |
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2 |
3.3.2.3 |
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05.2 |
\( \begin{array}{l} n = \frac{c}{c_{\rm{S}}}\\ c_{\rm{s}} = \frac{3 \times 10^8}{1.5}\\ c_{\rm{s}} = 2 \times 10^8 \textrm{ m s}^{-1} \end{array} \) |
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2 |
3.3.2.3 |
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05.3 |
\( \begin{array}{l} \sin \ \ 40 = \frac{0.012}{x}\\ \textrm{Distance } = 2 x = \frac{2 \times 0.012}{\sin\ \ 40}\\ T = \frac{d}{c_s} = \frac{2 \times 0.012}{\sin\ 40 \times 2 \times 10^8} = 1.9\ (1.87) \times 10^{-10}\ \rm{s} \end{array} \) |
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2 |
3.3.2.3 |
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05.4 |
\( \begin{array}{l} \sin \ \theta_c = \frac{n_2}{n_1} = \frac{1.4}{1.6}\\ \theta_c = 61^{\circ} \end{array} \) |
1 |
2 |
3.3.2.3 |
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05.5 |
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max 2 |
1 |
3.3.2.3 |
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05.6 |
The reduced amplitude is due to absorption/scattering/attenuation of the signal in the fibre The broadening is due to modal dispersion/multipath dispersion/due to different distances travelled on different paths |
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3 |
3.3.2.3 |
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06.1 |
Intensity decreasing with distance from central max (at least 3) Have correct width (half width of central max)/are in correct positions |
Judge by eye – does not matter which side drawn on |
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2 |
3.3.2.2 |
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06.2 |
The central fringe would become narrower |
1 |
1 |
3.3.2.2 |
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06.3 |
The central fringe would become wider |
1 |
1 |
3.3.2.2 |
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06.4 |
Central fringe is white Other fringes show spectrum Red further from centre/ violet closer to centre |
1 |
2 |
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07.1 |
Single wavelength (or frequency) |
1 |
1 |
3.3.2.1 |
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07.2 |
\( \begin{array}{l} \sin \ \theta_c = \frac{1.30}{1.45}\\ \theta_c = 63.7^{\circ} (64^{\circ}) \end{array} \) |
1 |
2 |
3.3.2.3 |
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07.3 |
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max 3 |
3 |
3.3.2.1 |
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07.4 |
Different colours of white light have different wavelengths Constructive/destructive interference will happen for different thicknesses of oil Different wavelengths refract differently |
max 2 |
3 |
3.3.2.1 |
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08.1 |
Shows the wave-like nature of electrons |
1 |
1 |
3.2.2.4 |
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08.2 |
Diffraction patterns of electrons as they pass through lattice overlap Interference/superposition Constructive interference/reinforcement causes bright circles Path difference = nλ at maximum intensity |
max 3 |
2 |
3.2.2.4 |
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08.3 |
\( \begin{array}{l} V = \frac{W}{Q}\\ W = 1000 \textrm{ V} \times 1.6 \times 10^{-19}\ \rm{C} = 1.6\ \times 10^{-16}\ \rm{J} \end{array} \) |
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2 |
3.5.1.1 |
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08.4 |
\( \begin{array}{l} E_{\rm{k}} = \frac{1}{2}m{v^2}\\ v^2 = \frac{2 \times 1.6 \times 10^{-16}}{9.11 \times 10^{-31}}\\ v = 1.87 \times 10^7 \textrm{ m s}^{-1} \end{array} \) |
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2 |
3.4.1.8 |
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08.5 |
\( \begin{array}{l} \lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.11\ \times 10^{-31} \times 1.9 \times 10^7}\\ \lambda = 3.8 \times 10^{-11}\ \rm{m} \end{array} \) |
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2 |
3.2.2.4 |
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08.6 |
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max 4 |
2 |
3.2.2.4 |
Skills box answers
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Question |
Answer |
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1 |
\( \begin{array}{l} d =\frac{1}{500 \times 10^{-3}}\\ \textrm{tan }\theta = \frac{0.38}{1.25},\textrm{ so } \theta = 16.9^{\circ}\\ n = 1\\ \lambda = ? \end{array} \) \(\lambda = \frac{d\sin\ \theta}{n} = \frac{\left( \frac{10^{-5}}{5} \right) \times \sin \ 16.9}{1}\) λ = 5.8 × 10−7 m (580 nm) |
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2(a) |
\( \begin{array}{l} d \sin \ \theta = n\lambda\\ \textrm{so }\sin \ \theta = \frac{n \lambda}{d} = 1 \times 520 \times 10^{-9} \times 600 \times 10^{3}\\ \sin \ \theta = 0.312\\ \theta = 18.2^{\circ} \end{array} \) |
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2(b) |
\( \begin{array}{l} \sin \ \theta = \frac{n\lambda}{d} = 2 \times 520 \times 10^{-9} \times 600 \times 10^3\\ \sin \ \theta = 0.312\\ \theta = 38.6^{\circ} \end{array} \) |
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2(c) |
\( \begin{array}{l} \frac{d}{\lambda} = \frac{10^{-3}}{\left( \rm{600\ \times\ 520\ \times\ 10^{-9}} \right)} = 3.21\\ \textrm{rounds down to 3} \end{array} \) |
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3 |
\( \begin{array}{l} d \sin \ \theta = n\lambda\\ d = \frac{n\lambda}{\sin \ \theta} = \frac{1 \times 650 \times 10^{-9}}{\sin \ 40.5} = 1.00 \times 10^{-6}\textrm{ m}\\ \textrm{number per mm }N = \frac{10^{-3}}{d} = \frac{10^{-3}}{10^{-6}} = 1000\\ \end{array} \) |