Test your understanding of the electromagnetic spectrum, wave properties, intensity, interference, diffraction and polarisation. Covers Lessons 1–5.
1. Which region of the electromagnetic spectrum has the highest photon energy?
2. A radio station broadcasts at 100 MHz. What is the wavelength?
3. The intensity of radiation from a point source at distance $r$ is $I$. At distance $3r$, the intensity is:
4. A 200 W light bulb emits uniformly in all directions. What is the intensity at 5.0 m?
5. Electromagnetic waves are produced by:
6. In a double slit experiment, the fringe spacing is 2.0 mm using light of wavelength 500 nm. If the wavelength is changed to 600 nm without changing the apparatus, the new fringe spacing is:
7. Two coherent sources produce a bright fringe when the path difference is:
8. A diffraction grating has 500 lines per mm. The grating element $d$ is:
9. A single slit of width 0.20 mm produces a diffraction pattern. The first minimum occurs at angle $\\theta$. If the slit width is halved, the new angle of the first minimum:
10. Which statement about diffraction gratings is correct?
11. Unpolarised light of intensity $I_0$ passes through a single polarising filter. The transmitted intensity is:
12. Plane-polarised light of intensity 36 W/m² passes through an analyser at 60° to the polarisation direction. The transmitted intensity is:
13. Which observation provides the most direct evidence that light is a transverse wave?
14. A person measures an intensity of 12 W/m² at 2.0 m from a point source. What is the intensity at 6.0 m?
15. In a double slit experiment using white light, the central fringe is white because:
SA1. (a) Explain why electromagnetic waves are classified as transverse waves. (b) A radio transmitter emits 25 kW uniformly in all directions. Calculate the intensity at 5.0 km. (c) Calculate the maximum distance at which a receiver requiring $1.0\\times10^{-6}$ W/m² can operate. (4 marks)
SA2. In a Young's double slit experiment, light of wavelength 550 nm passes through slits separated by 0.25 mm. A screen is placed 1.8 m from the slits. (a) Calculate the fringe spacing. (b) Calculate the distance from the central maximum to the 3rd bright fringe. (c) Explain what would happen to the pattern if the slit separation were halved. (4 marks)
SA3. A diffraction grating has 600 lines per mm. (a) Calculate the grating element. (b) Monochromatic light produces a first-order maximum at 22.0°. Calculate the wavelength. (c) Determine the highest order maximum observable for this wavelength. (4 marks)
SA4. (a) State Malus's Law and explain what each symbol represents. (b) Unpolarised light of intensity 64 W/m² passes through two polarisers whose transmission axes are at 30° to each other. Calculate the final transmitted intensity. (c) Explain why inserting a third polariser at 45° between two crossed polarisers allows some light to pass through. (4 marks)
SA5. Discuss how observations of interference, diffraction and polarisation collectively provide evidence for the wave model of light. In your answer, explain why each phenomenon specifically supports the wave model and cannot be explained by a simple particle model. (6 marks)
(a) In an electromagnetic wave, the electric and magnetic fields oscillate perpendicular to each other and both are perpendicular to the direction of propagation. This makes EM waves transverse. (1 mark)
(b) $I = P/(4\\pi r^2) = 25000/(4\\pi \\times (5000)^2) = 25000/(3.14\\times10^8) = 7.96\\times10^{-5}$ W/m² $\\approx$ $8.0\\times10^{-5}$ W/m². (1 mark)
(c) $r = \\sqrt{P/(4\\pi I_{min})} = \\sqrt{25000/(4\\pi \\times 1.0\\times10^{-6})} = \\sqrt{1.99\\times10^9} = 4.46\\times10^4$ m $\\approx$ 45 km. (2 marks — correct formula 1, correct answer 1)
(a) $\\Delta x = \\lambda L/d = (550\\times10^{-9})(1.8)/(0.25\\times10^{-3}) = 3.96\\times10^{-3}$ m $\\approx$ 4.0 mm. (1 mark)
(b) $x_3 = 3\\Delta x = 3 \\times 3.96$ mm = 11.9 mm $\\approx$ 12 mm. (1 mark)
(c) Halving $d$ doubles $\\Delta x$ because $\\Delta x \\propto 1/d$. The fringes become wider and more widely spaced. The pattern becomes easier to measure but dimmer because less light passes through each slit. (2 marks — relationship stated 1, effect on pattern explained 1)
(a) $d = 1/N = 1/(600\\times10^3) = 1.67\\times10^{-6}$ m $\\approx$ $1.67 \\mu$m. (1 mark)
(b) $\\lambda = d\\sin\\theta/n = (1.67\\times10^{-6})(\\sin 22.0°)/1 = 6.25\\times10^{-7}$ m = 625 nm. (1 mark)
(c) $n_{max} \\leq d/\\lambda = (1.67\\times10^{-6})/(6.25\\times10^{-7}) = 2.67$. Highest observable order: n = 2. (2 marks — correct method 1, correct answer 1)
(a) Malus's Law: $I = I_0 \\cos^2\\theta$, where $I_0$ is the incident polarised intensity, $\\theta$ is the angle between the transmission axes of polariser and analyser, and $I$ is the transmitted intensity. (1 mark)
(b) After first polariser: $I_1 = I_0/2 = 32$ W/m². After second: $I_2 = 32 \\cos^2 30° = 32 \\times 0.75 = $ 24 W/m². (1 mark)
(c) Crossed polarisers ($90°$) block all light because $\\cos^2 90° = 0$. A third polariser at $45°$ re-polarises the light at an intermediate angle. Light passes through the middle polariser at $I_0/2 \\times \\cos^2 45° = I_0/4$, then through the final polariser at $I_0/4 \\times \\cos^2 45° = I_0/8$. The intermediate polariser provides a "stepping stone" that prevents complete cancellation. (2 marks — explanation of re-polarisation 1, mathematical reasoning 1)
Interference (2 marks): The alternating bright and dark fringes in Young's double slit experiment result from the superposition of waves from two coherent sources. Constructive interference (bright fringes) occurs when path difference = $n\\lambda$; destructive interference (dark fringes) when path difference = $(n+1/2)\\lambda$. A particle model predicts only two bright patches (one per slit), so interference is strong evidence for waves.
Diffraction (2 marks): When light passes through a narrow slit, it spreads into a pattern of maxima and minima. This bending is characteristic of wave behaviour — particles would travel in straight lines and cast a sharp shadow. The single slit pattern with its wide central maximum cannot be explained by particles.
Polarisation (2 marks): Polarisation restricts oscillations to a single plane perpendicular to propagation. This is only possible for transverse waves. Longitudinal waves cannot be polarised because their oscillations are parallel to the direction of travel. Since light can be polarised, it must be a transverse wave.
Collectively, these three phenomena form an overwhelming body of evidence that light exhibits wave properties — specifically, transverse wave properties that a particle model cannot account for.