Checkpoint 2: Wave Nature of Light

(a) $f_1 = v/(2L) = 520/(2 \\times 0.65) = 400$ Hz. (1 mark)

(b) $f_4 = 4f_1 = 4 \\times 400 = $ 1600 Hz. (1 mark)

(c) Touching the midpoint forces a node at that point. This eliminates all harmonics that do not have a natural node at the midpoint — i.e., all odd harmonics. Only even harmonics ($n = 2, 4, 6, ...$) survive because they already have a node at the midpoint. (2 marks)

(a) A $\\pi$ phase change occurs when light reflects from a boundary where $n$ increases. For a film in air, the top reflection (air→film) has a $\\pi$ change; the bottom (film→air) does not. This creates a net $\\pi$ phase difference, swapping the conditions for constructive and destructive interference. (1 mark)

(b) With one $\\pi$ change, constructive: $2nt = (m+1/2)\\lambda$. $2 \\times 1.33 \\times 150 = 399 = (m+1/2)\\lambda$. For $m=0$: $\\lambda = 798$ nm (near IR, barely visible as deep red). For $m=1$: $\\lambda = 266$ nm (UV). Closest visible: ~798 nm (deep red). (2 marks)

(c) As the film drains, its thickness varies across the surface. Different locations have different optimal reflection wavelengths due to the $2nt$ condition, producing swirling colours. Where $t \\to 0$, destructive interference occurs for all wavelengths and the film appears black. (1 mark)

(a) Two point sources are just resolvable when the central maximum of one diffraction pattern falls on the first minimum of the other. The minimum resolvable angle is $\\theta_{min} = 1.22\\lambda/D$. (1 mark)

(b) $\\theta_{min} = 1.22 \\times 550\\times10^{-9}/(5.0\\times10^{-3}) = 1.34\\times10^{-4}$ rad $\\approx$ $1.3\\times10^{-4}$ rad (about 28 arcseconds). (1 mark)

(c) Since $\\theta_{min} \\propto \\lambda/D$, radio telescopes need much larger $D$ because radio wavelengths ($\\sim$0.1–1 m) are millions of times longer than optical wavelengths ($\\sim$500 nm). To achieve the same $\\theta_{min}$, a radio telescope needs an aperture millions of times larger than an optical telescope. This is why radio astronomers use interferometry. (2 marks)

(a) $z = \\Delta\\lambda/\\lambda_0 = (670.0 - 656.3)/656.3 = 13.7/656.3 = $ 0.0209. (1 mark)

(b) $v = zc = 0.0209 \\times 3.00\\times10^8 = 6.27\\times10^6$ m/s $\\approx$ 6270 km/s. (1 mark)

(c) Edwin Hubble observed that distant galaxies show redshifted spectral lines, with more distant galaxies having larger redshifts. This implies galaxies are receding, and more distant galaxies recede faster ($v \\propto d$). This is the observational basis for the expanding universe. (2 marks)

(a) Key wave evidence: (2 marks)

• Interference (Young's double slit): alternating bright/dark fringes from superposition — impossible for particles.
• Diffraction (single slit, gratings): light bends and spreads — particles would cast sharp shadows.
• Polarisation: restriction to one plane of oscillation — only transverse waves can be polarised.

(b) The wave model predicts that any frequency should eventually eject electrons given sufficient intensity, and that $K_{max}$ should increase with intensity. Observations show a threshold frequency and that $K_{max}$ depends on frequency, not intensity. Einstein's photon model: light consists of quanta of energy $E = hf$. One photon interacts with one electron. Ejection only occurs if $hf > \\phi$, and $K_{max} = hf - \\phi$. (2 marks)

(c) $\\phi = hc/\\lambda_0 = (4.14\\times10^{-15})(3.00\\times10^8)/(450\\times10^{-9}) = 2.76$ eV. (1 mark)

$E_{photon} = hc/\\lambda = (4.14\\times10^{-15})(3.00\\times10^8)/(300\\times10^{-9}) = 4.14$ eV.

$K_{max} = 4.14 - 2.76 = $ 1.38 eV. (1 mark)