Refracting/Reflecting Telescopes, Stefan's/Wien's Laws, Hertzsprung-Russell Diagram & Hubble's Law

1

Astronomical Refracting vs Cassegrain Reflecting Telescopes

1. Refracting telescope in normal adjustment (final virtual image at infinity, so focal points F_o and F_e coincide): consists of a long-focal-length, wide objective lens (f_o) and a short-focal-length eyepiece lens (f_e). Total tube length L = f_o + f_e, and Angular Magnification M = α / β = f_o / f_e. Suffers from chromatic aberration (glass has different refractive indices for different wavelengths, so blue focuses closer to the lens than red), heavy glass weight/sagging, and bubble absorption. 2. Cassegrain Reflecting Telescope: concave parabolic primary mirror (eliminates spherical aberration) with a central hole reflects parallel starlight onto a convex secondary mirror, which focuses it through the hole to the eyepiece/CCD. Free of chromatic aberration, lighter, and can be built with much larger diameters D using segmented mirrors.

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2

Collecting Power (∝ D²), Rayleigh Criterion (θ ≈ λ / D) & CCDs

1. Collecting Power is directly proportional to the cross-sectional area of the objective lens/mirror, so Collecting Power ∝ D² (doubling diameter D collects 4× more light energy per second, allowing fainter stars to be seen). 2. Minimum Angular Resolution (Rayleigh Criterion): two point sources are just resolved when the central maximum of the Airy diffraction disc of one image falls on the first minimum of the other: θ_min ≈ λ / D (in radians. Note: A smaller value of θ_min means better resolving power). 3. Charge-Coupled Devices (CCDs) vs Human Eye: CCDs have a quantum efficiency (percentage of incident photons that liberate a photoelectron) of ~70-80% compared to only ~1-4% for the human eye, can integrate light over long exposure times, detect wavelengths beyond visible (IR/UV), and have high spatial resolution.

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3

Black-Body Radiation: Wien's Law (λ_max T = 2.90 × 10⁻³) & Stefan's Law (P = σAT⁴)

Stars act as approximate black bodies (perfect absorbers and emitters of all EM wavelengths): 1. Wien's Displacement Law: the wavelength of peak intensity (λ_max) is inversely proportional to the star's absolute surface temperature T (in K): λ_max T = 2.90 × 10⁻³ m K (hotter stars have shorter λ_max, appearing blue-white; cooler stars appear red). 2. Stefan-Boltzmann Law: total power radiated (Luminosity L or P, in Watts) is given by P = σ A T⁴ = 4π r² σ T⁴, where r is the star's radius and σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

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4

Hipparcos Magnitude Scale (m, M), Parsecs & Standard Candles

1. Apparent Magnitude (m): brightness as seen from Earth; the scale is logarithmic such that a difference of 5 magnitudes corresponds to a brightness (intensity I) ratio of 100, so 1 magnitude step = 100^(1/5) = 2.512: I₁ / I₂ = 2.512^(m₂ - m₁) (smaller or more negative m means brighter). 2. Absolute Magnitude (M): the apparent magnitude a star would have if viewed from a standard distance of 10 parsecs (10 pc). 3. Distance Modulus Formula: m - M = 5 log₁₀(d / 10), where d must be in parsecs (1 parsec = distance at which 1 AU subtends a parallax angle of 1 arcsecond = 3.09 × 10¹⁶ m ≈ 3.26 light-years; d = 1 / p). Type Ia Supernovae (exploding white dwarfs at the 1.4 solar masses Chandrasekhar limit) have a known peak absolute magnitude (M ≈ -19.3) and act as 'standard candles' to measure vast extragalactic distances (revealing dark energy and accelerating expansion).

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5

Stellar Spectral Classes (O B A F G K M) & The Hertzsprung-Russell (H-R) Diagram

Spectral classes from hottest to coolest ('Oh Be A Fine Girl/Guy, Kiss Me'): O (blue, 25,000-50,000 K, He⁺/He lines), B (blue, 11,000-25,000 K, He/H), A (blue-white, 7,500-11,000 K, strongest Hydrogen Balmer absorption lines because n = 2 state is most populated), F (white, 6,000-7,500 K), G (yellow-white like the Sun, 5,000-6,000 K, ionised metals Ca²⁺), K (orange, 3,500-5,000 K), M (red, <3,500 K, TiO molecules). The H-R Diagram plots Absolute Magnitude M (+15 at bottom to -10 at top, or Luminosity L/L_sun on a log scale) against Temperature T (running backwards from 50,000 K on the left to 2,500 K on the right): Diagonal = Main Sequence (core H → He fusion); Top Right = Red Giants & Supergiants (cool T but huge radius r so high luminosity via P = 4πr²σT⁴); Bottom Left = White Dwarfs (very hot T but tiny Earth-sized radius r so very dim).

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6

Doppler Redshift, Hubble's Law (v = H₀d), Age of Universe & Big Bang Evidence

1. Redshift z: for v ≪ c, z = Δλ / λ = -Δf / f = v / c (where Δλ = λ_observed - λ_rest > 0 for recession). 2. Hubble's Law: the recession velocity v of a distant galaxy is directly proportional to its distance d from us: v = H₀ d (where H₀ ≈ 65-70 km s⁻¹ Mpc⁻¹; converting 1 Mpc = 3.09 × 10²² m and km to m gives H₀ in s⁻¹, and the approximate Age of the Universe = 1 / H₀ ≈ 14 billion years, assuming constant expansion rate). 3. Three Pillars of Evidence for the Big Bang: (a) Cosmological Redshift of distant galaxies (Hubble's Law showing space itself is expanding from a hot, dense initial point), (b) Cosmic Microwave Background (CMB) radiation (isotropic black-body microwave radiation at T = 2.7 K, which is the redshifted gamma/high-energy thermal radiation from recombination as the universe expanded and cooled), and (c) Relative Abundance of Hydrogen and Helium (3 : 1 by mass, or 12 : 1 by number of nuclei, formed during brief primordial nucleosynthesis in the early hot universe).

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1. Q1

   An astronomical refracting telescope in normal adjustment has an overall length of 1.26 m and an angular magnification of 20.0. Its objective lens has a diameter of 90.0 mm. Calculate: (a) the focal lengths of the objective lens and the eyepiece lens, (b) the minimum angular resolution of the telescope at a wavelength of 550 nm, and (c) the diameter of a radio telescope operating at a wavelength of 0.210 m that would have the exact same angular resolution as this optical telescope. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. (a) In normal adjustment, f_o + f_e = 1.26 m and f_o ÷ f_e = 20.0, so f_o = 20.0 f_e [1 mark]. 2. Substitute into the length equation: 20.0 f_e + f_e = 21.0 f_e = 1.26 m → f_e = 1.26 ÷ 21.0 = 0.0600 m (6.00 cm) and f_o = 20.0 × 0.0600 = 1.20 m [1 mark]. 3. (b) Use the Rayleigh Criterion θ ≈ λ ÷ D for the optical telescope: θ = (5.50 × 10⁻⁷ m) ÷ 0.0900 m = 6.11 × 10⁻⁶ rad [1 mark]. 4. (c) For the radio telescope to have the same angular resolution θ at λ_radio = 0.210 m, equate λ_radio / D_radio = λ_optical / D_optical [1 mark]. 5. Solve for D_radio: D_radio = 0.210 m ÷ (6.111 × 10⁻⁶ rad) = 3.44 × 10⁴ m = 34.4 km (explaining why single radio dishes have much poorer resolution than optical telescopes and must be linked in interferometry arrays) [1 mark].

2. Q2

   Betelgeuse is a red supergiant star with a surface temperature of 3500 K and a luminosity (power output) of 4.80 × 10³¹ W. The Sun has a surface temperature of 5800 K and a luminosity of 3.85 × 10²⁶ W (Stefan-Boltzmann constant σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴). Calculate the wavelength of peak emission (λ_max) for Betelgeuse, the radius of Betelgeuse in metres, and the ratio of the radius of Betelgeuse to the radius of the Sun. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Use Wien's Displacement Law (λ_max T = 2.90 × 10⁻³ m K) for Betelgeuse: λ_max = (2.90 × 10⁻³) ÷ 3500 K = 8.29 × 10⁻⁷ m (829 nm, in the near-infrared region) [1 mark]. 2. Use the Stefan-Boltzmann Law P = 4π r² σ T⁴ to find the surface area of Betelgeuse: A = P ÷ (σ T⁴) = (4.80 × 10³¹) ÷ (5.67 × 10⁻⁸ × 3500⁴) = (4.80 × 10³¹) ÷ 8.509 × 10⁶ = 5.641 × 10²⁴ m² [1 mark]. 3. Solve A = 4π r² for radius r of Betelgeuse: r = √(5.641 × 10²⁴ ÷ 4π) = √(4.489 × 10²³) = 6.70 × 10¹¹ m [1 mark]. 4. Calculate the radius of the Sun (or use the ratio P_B / P_Sun = (r_B / r_Sun)² (T_B / T_Sun)⁴): r_B ÷ r_Sun = √(P_B ÷ P_Sun) × (T_Sun ÷ T_B)² [1 mark]. 5. Substitute values: r_B ÷ r_Sun = √(4.80 × 10³¹ ÷ 3.85 × 10²⁶) × (5800 ÷ 3500)² = 353.1 × 2.746 = 970 (Betelgeuse's radius is ~970 times the radius of the Sun) [1 mark].

3. Q3

   The bright star Sirius A has an apparent magnitude of m = -1.46 and an absolute magnitude of M = +1.42. Its companion white dwarf star, Sirius B, is at the exact same distance from Earth and has an apparent magnitude of m = +8.44. Calculate the distance to the Sirius binary system in parsecs and in light-years (1 pc = 3.26 ly), and calculate the ratio of the brightness of Sirius A to Sirius B as seen from Earth. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Substitute m = -1.46 and M = +1.42 for Sirius A into the distance modulus formula m - M = 5 log₁₀(d / 10): -1.46 - 1.42 = -2.88 = 5 log₁₀(d / 10) [1 mark]. 2. Divide by 5 and take the inverse log (10^x): log₁₀(d / 10) = -2.88 ÷ 5 = -0.576 → d ÷ 10 = 10^(-0.576) = 0.2655 → d = 2.65 pc (2.66 pc) [1 mark]. 3. Convert parsecs to light-years: d = 2.655 pc × 3.26 ly pc⁻¹ = 8.65 light-years [1 mark]. 4. Difference in apparent magnitude between Sirius B and Sirius A: Δm = 8.44 - (-1.46) = 9.90 magnitudes; Brightness ratio I_A ÷ I_B = 2.512^(9.90) (or 100^(9.90 / 5)) = 9.12 × 10³ (Sirius A is ~9,120 times brighter than Sirius B!) [1 mark].

4. Q4

   In a laboratory on Earth, the Hydrogen-alpha Balmer spectral line has a wavelength of 656.3 nm. In the spectrum of a distant galaxy, this same absorption line is observed at a wavelength of 678.2 nm. Calculate the redshift z and the recession speed of the galaxy in km s⁻¹, calculate the distance to the galaxy in Mpc given that the Hubble constant is H₀ = 68.0 km s⁻¹ Mpc⁻¹, and estimate the age of the universe in years from H₀ (c = 3.00 × 10⁸ m s⁻¹, 1 Mpc = 3.09 × 10²² m). (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Calculate wavelength shift Δλ = 678.2 - 656.3 = +21.9 nm, and redshift z = Δλ ÷ λ = 21.9 ÷ 656.3 = 0.0334 [1 mark]. 2. Calculate recession velocity v = z × c = 0.03337 × (3.00 × 10⁸ m s⁻¹) = 1.00 × 10⁷ m s⁻¹ = 1.00 × 10⁴ km s⁻¹ (10,000 km s⁻¹) [1 mark]. 3. Use Hubble's Law (v = H₀ d) to find the distance d in Mpc: d = v ÷ H₀ = 10,010 km s⁻¹ ÷ (68.0 km s⁻¹ Mpc⁻¹) = 147 Mpc [1 mark]. 4. Convert H₀ from km s⁻¹ Mpc⁻¹ into s⁻¹: H₀ = (68.0 × 10³ m s⁻¹) ÷ (3.09 × 10²² m) = 2.201 × 10⁻¹⁸ s⁻¹ [1 mark]. 5. Estimate age of the universe T = 1 ÷ H₀ = 1 ÷ (2.201 × 10⁻¹⁸ s⁻¹) = 4.54 × 10¹⁷ s; divide by (365.25 × 86,400 = 3.156 × 10⁷ s yr⁻¹) to get T = 1.44 × 10¹⁰ years (14.4 billion years) [1 mark].

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