Astrophysics (A-level only)

Explanation

• An astronomical telescope in normal adjustment uses a long-focal-length converging objective and a short-focal-length converging eyepiece. Parallel rays from a distant object are focused by the objective to form a real, inverted intermediate image.
• This image lies in the focal plane shared by the two lenses, so the eyepiece sends parallel rays to the relaxed eye and the final image is virtual at infinity.
• The lens separation is therefore $f_o+f_e$.
• Angular magnification compares the angle subtended at the eye through the telescope with that at the unaided eye; its magnitude is $M=f_o/f_e$.
• A complete ray diagram must show the intermediate image, both focal points and parallel emergent rays.

Explanation

• A Cassegrain reflecting telescope uses a large parabolic concave primary mirror and a small convex secondary mirror.
• Parallel incident rays reflect from the primary and begin to converge; before reaching the primary focus, they strike the secondary and are reflected back through a central hole in the primary towards the eyepiece.
• A parabolic primary brings on-axis parallel rays to one focus, avoiding the spherical aberration produced by a spherical mirror.
• Reflectors also avoid chromatic aberration because reflection does not depend on refractive index.
• Compared with a large refracting objective, a mirror can be supported across its back and made with a larger practical aperture, although the secondary obstructs part of the incoming beam.

Explanation

• A single-dish radio telescope resembles an optical reflector: a concave dish collects waves and directs them to a receiver near the focus, and its collecting power is proportional to $D^2$.
• Radio wavelengths are much longer than visible wavelengths, so $\theta\approx\lambda/D$ means that a much larger dish is needed for comparable angular resolution.
• Radio telescopes can operate through cloud and in daylight, but require sites with little human-made radio interference.
• Infrared observations favour high, dry sites and cooled detectors because atmospheric water vapour and thermal radiation interfere.
• Most ultraviolet and X-ray astronomy is performed above the atmosphere, usually on satellites, because those wavelengths are strongly absorbed before reaching the ground.

Explanation

• A larger telescope diameter improves both resolution and sensitivity. The Rayleigh criterion gives the minimum angular separation as $\theta\approx\lambda/D$, where $\theta$ is in radians, so increasing $D$ allows two closer sources to be resolved at the same wavelength.
• Collecting power is proportional to the aperture area and therefore to $D^2$, so a larger telescope detects fainter sources in a given exposure. A CCD has greater quantum efficiency than the eye: it detects a greater proportion of the incident photons.
• It can also accumulate charge during a long exposure and gives a permanent digital record that is convenient to process.
• Detector resolution depends on pixel size, while the telescope aperture still sets the diffraction limit.
• No knowledge of CCD internal structure is required.

Explanation

• Apparent magnitude $m$ describes how bright a star appears from Earth on the Hipparcos scale. The order is reversed: a smaller or more negative magnitude means a brighter object, while the dimmest stars visible to the unaided eye have $m\approx+6$.
• The scale is logarithmic and subjective.
• A difference of one magnitude corresponds to an intensity ratio of $2.51$, so a magnitude difference $\Delta m$ corresponds to an intensity ratio $2.51^{\lvert\Delta m\rvert}$, with the lower-magnitude object being more intense.
• Apparent magnitude alone does not give luminosity because distance changes the detected intensity.
• Examiners expect the direction of the comparison as well as the calculated ratio.

Explanation

• Absolute magnitude $M$ is the apparent magnitude a star would have if placed $10\,\text{pc}$ from Earth, so it permits intrinsic luminosities to be compared without the effect of differing distances. The distance modulus is $m-M=5\log_{10}(d/10)$, where $d$ must be in parsecs.
• A more negative value of $M$ represents a more luminous star.
• One parsec is approximately $3.26$ light years; both are units of distance, although a light year is defined using the distance light travels in one year.
• In calculations, the logarithm is base ten and the distance must not be entered in metres.
• A valid conclusion distinguishes apparent brightness $m$ from intrinsic brightness $M$.

Explanation

• A star may be modelled as a black body. Its continuous spectrum has a single peak; increasing temperature moves the peak to shorter wavelength and increases the area under the curve.
• Wien's displacement law, $\lambda_{\max}T=2.9\times10^{-3}\,\text{m K}$, estimates surface temperature from the peak wavelength. Stefan's law is $P=\sigma AT^4$ and, for a spherical star, $A=4\pi R^2$, so $P\propto R^2T^4$.
• Received intensity follows $I=P/(4\pi d^2)$ when emission is isotropic and absorption is negligible.
• Calculations require wavelength in metres and temperature in kelvin.
• Comparisons should state which quantities, such as radius or distance, are held constant.

Explanation

• The spectral sequence OBAFGKM runs from hottest to coolest.
• O and B stars are blue, A stars blue-white, F white, G yellow-white, K orange and M red.
• Prominent lines also change: O spectra show ionised helium, helium and hydrogen; B show helium and hydrogen; A have the strongest hydrogen Balmer lines and ionised metals; F show ionised metals; G show ionised and neutral metals; K show neutral metals; M show neutral atoms and TiO.
• Balmer absorption requires hydrogen atoms already in the $n=2$ state.
• Cooler stars have too few atoms excited to $n=2$, while hotter stars have much hydrogen ionised, so Balmer lines are strongest near class A rather than in the hottest stars.

Explanation

• A Hertzsprung-Russell diagram plots absolute magnitude vertically from about $-10$ at the top to $+15$ at the bottom, against surface temperature decreasing from about $50000\,\text{K}$ on the left to $2500\,\text{K}$ on the right; spectral classes OBAFGKM may replace temperature.
• The main sequence runs diagonally from hot, luminous stars at upper left to cool, faint stars at lower right.
• Giants occupy the upper-right region, while white dwarfs are hot but faint and lie at lower left.
• The Sun is a class G main-sequence star at about $5800\,\text{K}$ and $M\approx+5$.
• A Sun-like star progresses from formation to main sequence, red giant and finally white dwarf.

Explanation

• A supernova shows a rapid increase in luminosity and a slower decline. Type Ia supernovae have standardisable peak absolute magnitudes, so comparing their known luminosity with observed brightness gives distance; their light curves helped reveal an accelerating Universe, interpreted using dark energy.
• A neutron star is an extremely dense compact remnant composed mainly of neutrons.
• Collapse of a supergiant to a neutron star or black hole can produce a gamma-ray burst whose brief energy output is comparable with the Sun's total output over a very long time.
• For a black hole, escape velocity exceeds $c$ within the event horizon and $R_s\approx2GM/c^2$.
• Supermassive black holes occur at galactic centres.

Explanation

• Relative motion along the line of sight changes observed wavelength and frequency. For speeds much smaller than $c$, $\Delta f/f=-v/c$ and the redshift $z=\Delta\lambda/\lambda=v/c$, with positive $v$ for recession.
• A receding source has a lower observed frequency and longer wavelength; an approaching source has a higher frequency and shorter wavelength.
• The denominator is the unshifted laboratory frequency or wavelength.
• In a binary system viewed in the plane of its orbit, each star alternately approaches and recedes, so its spectral lines shift periodically between blue and red.
• Calculations must confirm that $v\ll c$ before relying on the non-relativistic approximation.

Explanation

• Hubble's law, $v=Hd$, states that a distant galaxy's recession speed is proportional to its distance.
• A straight-line graph of $v$ against $d$ has gradient $H$, supporting the interpretation that space is expanding rather than that galaxies left one central point through fixed space.
• If $H$ is assumed constant, extrapolation gives an age estimate $t\approx1/H$; $H$ must first be converted from $\text{km s}^{-1}\text{Mpc}^{-1}$ to $\text{s}^{-1}$.
• Big Bang evidence also includes cosmological microwave background radiation, interpreted as cooled relic radiation, and the relative abundance of hydrogen and helium produced in the early Universe.
• The constant-$H$ age is an estimate, not an exact expansion history.

Explanation

• Quasars were first identified as unusually bright radio sources. Their spectra show large optical redshifts, placing them among the most distant measurable objects, yet their received flux can still be substantial.
• This combination implies an enormous power output. A quasar is an active galactic nucleus powered by matter accreting onto a supermassive black hole, where gravitational energy is converted into radiation.
• For a sufficiently small redshift, $v\approx zc$ and Hubble's law gives $d=v/H$.
• If emission is assumed isotropic, the luminosity follows $P=4\pi d^2F$.
• These estimates require explicit low-redshift, isotropic-emission and inverse-square assumptions; observed flux is not the same quantity as emitted power.

Explanation

• Direct exoplanet detection is difficult because a planet is much fainter than its host star and the angular separation is very small. The radial-velocity method detects periodic red and blue Doppler shifts in the star's absorption lines as the star orbits the system's centre of mass; it measures only line-of-sight motion.
• The transit method detects a repeated fall in brightness when an aligned planet crosses the stellar disc.
• A typical light curve has a steady baseline, ingress, a low region and egress, repeated once per orbital period.
• Approximately, transit depth satisfies $\Delta I/I=(R_p/R_*)^2$.
• Matching periods from both methods provide stronger evidence, while a missing transit does not rule out a planet because alignment may be unsuitable.