Notes · PhysicsUK · A-Levels

Astrophysics (optional)

This is one of the five optional A-Level modules (AQA Paper 3, Section B); a student studies just one option. Astrophysics applies core physics to the cosmos: the optics of telescopes and their resolving power, the classification of stars by temperature and luminosity through the Hertzsprung-Russell diagram, the black-body laws that reveal a star's temperature and size, and the Doppler and Hubble evidence for an expanding universe.

5 sections~14 min reading time3 competenciesLevel Standard 2 · Advanced 3

T·0999 / 13

Exam profile

AO1 · Describe refracting and reflecting telescopes, stellar classification, the black-body laws and Big Bang cosmology (optional module)AO2 · Calculate angular magnification, resolving power, magnitudes, stellar temperature and radius, Doppler shift and Hubble distancesAO3 · Interpret HR diagrams, black-body spectra and redshift data as evidence for stellar evolution and an expanding universe

Operators:describecalculatedetermineexplainestimatecompare

basic level

This is an optional A2 module examined in Paper 3 Section B; it is studied only if chosen as the school's option. It requires telescopes, stellar classification and the basics of cosmology.

higher level

The full option demands quantitative treatment of resolving power, magnitudes, the black-body laws applied to stars, and the Doppler and Hubble analysis of the expanding universe.

Depth

Text

Contents · 5 sections

Astrophysics (optional)

Lenses and optical telescopes#

StandardAQA 7408 3.9.1.1DfE GCE Physics - astrophysics option

Refracting telescope in normal adjustment

Fig. 1Parallel light from a distant star is focused by the objective in its focal plane, then viewed by the eyepiece; the two lenses share a focal point and are separated by

f_0 + f_e

A converging lens brings parallel light to a focus at its principal focus, a distance

f

(the focal length) from the lens; its power is

P = 1/f

in dioptres. For an object at distance

u

forming an image at distance

v

, the thin-lens equation is

\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}

(in the real-is-positive convention). Astronomical objects are effectively at infinity, so their light arrives as parallel rays and is focused in the focal plane.

An astronomical refracting telescope uses two converging lenses: a long-focal-length objective forms a real image of the distant object in its focal plane, and a short-focal-length eyepiece, acting as a magnifying glass, views that image. In normal adjustment the telescope is set so that the final image is at infinity (relaxed eye), which requires the two lenses to be separated by

f_0 + f_e

, the sum of their focal lengths, with their focal points coinciding.

The angular magnification is the ratio of the angle subtended by the image seen through the telescope to the angle subtended by the object with the naked eye. In normal adjustment this equals the ratio of the focal lengths,

M = \dfrac{f_0}{f_e}

. A large objective focal length and a small eyepiece focal length therefore give high magnification - which is why refracting telescopes are long.

Reflecting telescopes replace the objective lens with a concave mirror, which has decisive advantages for large instruments: a mirror can be made very large and supported from behind (a big lens sags under its own weight and can only be held at its rim), it suffers no chromatic aberration (a lens brings different colours to slightly different foci), and it is cheaper to make. The Cassegrain arrangement folds the light path with a small secondary mirror, so all large modern telescopes are reflectors.

\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}

Thin-lens equation

Relates focal length, object and image distances.

M = \dfrac{f_0}{f_e}

Angular magnification

Objective focal length divided by eyepiece focal length, in normal adjustment.

Worked example

Magnification and length of a telescope

A refracting telescope has an objective of focal length $900\,\text{mm}$ and an eyepiece of focal length $30\,\text{mm}$ . Find the angular magnification and the telescope's length in normal adjustment.

Angular magnification
$M = f_0/f_e = 900/30 = 30$ .
Length in normal adjustment
The lenses are separated by the sum of the focal lengths.
$L = f_0 + f_e = 900 + 30 = 930\,\text{mm}$

Result: The magnification is $\times 30$ and the telescope is $930\,\text{mm}$ long.

Typical mistakes

Confusing the objective (long focal length) with the eyepiece (short focal length) in the magnification.
Forgetting that in normal adjustment the lenses are separated by the sum of their focal lengths.

Active revision

A refracting telescope has an objective of focal length $1.20\,\text{m}$ and an eyepiece of focal length $25\,\text{mm}$ . Find its angular magnification and the length of the telescope in normal adjustment.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Resolving power, CCDs and non-optical telescopes#

StandardAQA 7408 3.9.1.2AQA 7408 3.9.1.3DfE GCE Physics - resolving power

The resolving power of a telescope is its ability to distinguish two close objects as separate. Because light diffracts as it passes through the circular aperture, each point source produces a small diffraction disc; two sources can just be resolved when the centre of one disc falls on the first dark ring of the other. This is the Rayleigh criterion,

\theta \approx \dfrac{\lambda}{D}

, where

\theta

is the minimum resolvable angle (in radians),

\lambda

the wavelength and

D

the aperture diameter. A larger aperture resolves finer detail.

A larger aperture also collects more light: the light-gathering power is proportional to the area, and so to

D^2

. Doubling the diameter quadruples the light collected, revealing fainter objects. This is a second reason - alongside resolution - why astronomers build ever-larger telescopes.

Modern telescopes record images with a charge-coupled device (CCD) rather than the eye or film. A CCD is a silicon chip of millions of pixels; incoming photons free electrons by the photoelectric effect, and the accumulated charge in each pixel is read out to form a digital image. Its quantum efficiency (the fraction of incident photons detected) is far higher than the eye's or film's - typically over

70\%

- so it captures faint objects in a fraction of the time.

Not all astronomy is optical. Objects emit across the electromagnetic spectrum, so radio, infrared, ultraviolet, X-ray and gamma-ray telescopes each reveal different physics - cool gas in radio and infrared, hot accreting matter in X-rays. Some wavelengths are absorbed by the atmosphere and must be observed from space. Radio dishes can be very large (giving good collecting power) but their long wavelength limits resolution, which is improved by combining many dishes as an interferometer.

\theta \approx \dfrac{\lambda}{D}

Rayleigh criterion

Minimum resolvable angle in radians; smaller is better.

\text{collecting power} \propto D^2

Light-gathering power

Proportional to the area of the aperture.

Worked example

Resolving power of a telescope

A telescope has an aperture of $0.15\,\text{m}$ and observes light of wavelength $500\,\text{nm}$ . Estimate the smallest angle it can resolve.

Apply the Rayleigh criterion
$\theta \approx \lambda/D$ , with both in metres.
Substitute
$\theta = \dfrac{500\times10^{-9}}{0.15}$ .
$\theta = 3.3\times10^{-6}\,\text{rad}$

Result: It can just resolve objects about $3.3\times10^{-6}\,\text{rad}$ apart (roughly $0.7$ arcseconds).

Typical mistakes

Using the wavelength or aperture in inconsistent units - both must be in metres to give $\theta$ in radians.
Confusing resolving power (smaller angle is better) with magnification.

Active revision

A telescope of aperture $0.20\,\text{m}$ observes at a wavelength of $550\,\text{nm}$ . Estimate the smallest angle it can resolve, and state how the collecting power changes if the aperture is doubled.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Classification of stars and the HR diagram#

AdvancedAQA 7408 3.9.2.1AQA 7408 3.9.2.4DfE GCE Physics - stellar classification

The Hertzsprung-Russell diagram

Fig. 2Luminosity (here as log of luminosity relative to the Sun) against surface temperature. Most stars lie on the main sequence; red giants sit upper right and white dwarfs lower left. (Temperature is conventionally drawn decreasing to the right.)

Stars are classified by their spectra into the sequence O, B, A, F, G, K, M, from hottest (O, blue, over

25\,000\,\text{K}

) to coolest (M, red, about

3\,000\,\text{K}

). The spectral class is read from the absorption lines in the star's spectrum, which depend on temperature: hydrogen Balmer lines are strongest in A-type stars, where the temperature is just right to populate the relevant electron level. The Sun is a G-type star.

Brightness is described by magnitude. Apparent magnitude

m

is how bright a star looks from Earth; absolute magnitude

M

is how bright it would look at a standard distance of

10\,\text{parsecs}

, and so measures true luminosity. The two are linked by the distance

d

(in parsecs):

m - M = 5\log\left(\dfrac{d}{10}\right)

. The scale is inverted and logarithmic - a smaller (or more negative) magnitude is brighter, and five magnitudes correspond to a factor of

100

in brightness.

The Hertzsprung-Russell (HR) diagram plots luminosity (or absolute magnitude) against temperature (or spectral class), conventionally with temperature decreasing to the right. Most stars, including the Sun, lie on a diagonal band called the main sequence, where they spend the stable hydrogen-burning phase of their lives. Off the main sequence lie the cool but very luminous red giants and supergiants (upper right) and the hot but faint white dwarfs (lower left).

The HR diagram is a map of stellar evolution. A main-sequence star exhausts its core hydrogen, swells into a red giant, and then - for a Sun-like star - sheds its outer layers and leaves a white dwarf. A much more massive star ends in a supernova, leaving a neutron star or, if massive enough, a black hole. A star's position and path on the diagram are set principally by its mass.

m - M = 5\log\!\left(\dfrac{d}{10}\right)

Distance modulus

Links apparent and absolute magnitude to distance in parsecs.

Worked example

Absolute magnitude from distance

A star has apparent magnitude $+9.0$ at a distance of $500\,\text{parsecs}$ . Find its absolute magnitude.

Distance modulus
$m - M = 5\log(d/10) = 5\log(500/10) = 5\log(50)$ .
Evaluate the log
$\log 50 = 1.699$ , so $m - M = 5 \times 1.699 = 8.50$ .
Solve for M
$M = m - 8.50 = 9.0 - 8.50$ .
$M = +0.50$

Result: The absolute magnitude is $+0.5$ - the star is intrinsically far brighter than it appears.

Typical mistakes

Forgetting that the magnitude scale is inverted - a smaller magnitude means a brighter star.
Reading the HR diagram's temperature axis the wrong way - temperature conventionally decreases to the right.

Active revision

A star has apparent magnitude $+7.0$ and lies at a distance of $200\,\text{parsecs}$ . Calculate its absolute magnitude.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Black-body radiation and stellar properties#

AdvancedAQA 7408 3.9.2.2AQA 7408 3.9.2.3DfE GCE Physics - black-body radiation

Black-body curves for two temperatures

Fig. 3As the temperature rises, the peak of the black-body curve shifts to shorter wavelengths (Wien's law) and the total power radiated increases steeply (Stefan's law).

A star radiates approximately as a black body - an ideal emitter whose spectrum depends only on its temperature. As the temperature rises, the whole spectrum brightens and its peak shifts to shorter wavelengths, which is why hot stars look blue-white and cool stars red. This continuous curve, overlaid with absorption lines, carries a star's temperature and size.

Wien's displacement law gives the peak wavelength:

\lambda_{\max} T = 2.90\times10^{-3}\,\text{m K}

, a constant. Measuring the wavelength at which a star's spectrum peaks therefore gives its surface temperature directly - a hotter star has a shorter peak wavelength. This is the standard route to a stellar temperature from its spectrum.

Stefan's law gives the total power radiated:

P = \sigma A T^4

, where

\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}

is the Stefan constant and

A = 4\pi r^2

the star's surface area. The fourth-power dependence is dramatic - doubling the temperature increases the power sixteenfold. Combining Wien's and Stefan's laws lets astronomers deduce a star's radius from its temperature and luminosity, even though the star is a mere point of light.

Together these black-body laws turn a spectrum into physics: the peak wavelength gives

T

(Wien), and

T

with the measured luminosity gives the radius

r

(Stefan). This is how we know that Betelgeuse is a cool but vast supergiant and that a white dwarf is hot but tiny - conclusions drawn entirely from the light they emit.

\lambda_{\max}\,T = 2.90\times10^{-3}\,\text{m K}

Wien's displacement law

The peak wavelength is inversely proportional to temperature.

P = \sigma A T^4 = \sigma (4\pi r^2) T^4

Stefan's law

Total power radiated; the fourth power of temperature is decisive.

Worked example

Temperature and radius of a star

A star's black-body spectrum peaks at $580\,\text{nm}$ and its luminosity is $3.8\times10^{26}\,\text{W}$ . Find its surface temperature and radius. Take $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ .

Temperature from Wien's law
$T = \dfrac{2.90\times10^{-3}}{580\times10^{-9}} = 5.0\times10^{3}\,\text{K}$ .
Rearrange Stefan's law for radius
$P = \sigma 4\pi r^2 T^4$ , so $r = \sqrt{\dfrac{P}{4\pi\sigma T^4}}$ .
Substitute
$r = \sqrt{\dfrac{3.8\times10^{26}}{4\pi(5.67\times10^{-8})(5.0\times10^{3})^4}}$ .
$r = \sqrt{7.66\times10^{17}} = 8.8\times10^{8}\,\text{m}$

Result: The star has a surface temperature of $5000\,\text{K}$ and a radius of about $8.8\times10^{8}\,\text{m}$ (a little larger than the Sun).

Typical mistakes

Forgetting to convert the peak wavelength to metres before using Wien's law.
Using the radius instead of the surface area $4\pi r^2$ in Stefan's law, or forgetting the fourth power of temperature.

Active revision

A star's spectrum peaks at $480\,\text{nm}$ . Find its surface temperature, and hence its radius if its luminosity is $4.0\times10^{27}\,\text{W}$ . Take $\sigma = 5.67\times10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ .

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Doppler shift, Hubble's law and cosmology#

AdvancedAQA 7408 3.9.3DfE GCE Physics - cosmology

Hubble's law

Fig. 4The recession speed of galaxies is proportional to their distance; the gradient of the line is the Hubble constant

H_0

When a source of waves moves relative to an observer, the observed wavelength changes - the Doppler effect. For a star or galaxy moving at speed

v

much less than the speed of light, the fractional change in wavelength equals the fractional speed:

z = \dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}

, where

z

is the redshift. A galaxy moving away shows its spectral lines shifted to longer (redder) wavelengths; one approaching shows a blueshift.

In the 1920s Hubble found that almost all galaxies are redshifted, and that the recession speed is proportional to distance:

v = H_0 d

, where

H_0

is the Hubble constant (about

70\,\text{km s}^{-1}\,\text{Mpc}^{-1}

). This is the observational cornerstone of modern cosmology: the universe is expanding, with every galaxy receding from every other, faster the further apart they are.

Running the expansion backwards implies that everything was once together in an extremely hot, dense state - the Big Bang. A rough estimate of the age of the universe follows from Hubble's law: if galaxies have receded at constant speed since the beginning, the time elapsed is

t \approx \dfrac{1}{H_0}

, giving about

14\,\text{billion years}

. The decisive supporting evidence is the cosmic microwave background radiation, the cooled afterglow of the hot early universe, detected as a near-perfect black-body spectrum at about

2.7\,\text{K}

The same Doppler technique reveals objects and worlds beyond our galaxy. Quasars are extremely distant, extremely luminous active galactic nuclei, identified by their enormous redshifts. Exoplanets are detected by the tiny periodic Doppler wobble a planet induces in its parent star's spectrum (the radial-velocity method), or by the small dip in brightness when a planet transits across its star - techniques that have now found thousands of planets.

z = \dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}

Doppler redshift

For speeds much less than $c$ , redshift equals the fractional speed.

v = H_0 d

Hubble's law

Recession speed is proportional to distance.

Worked example

Redshift, recession speed and distance

A galaxy's calcium line at $393\,\text{nm}$ is observed at $402\,\text{nm}$ . Find the recession speed and the distance. Take $c = 3.0\times10^{5}\,\text{km s}^{-1}$ and $H_0 = 70\,\text{km s}^{-1}\,\text{Mpc}^{-1}$ .

Redshift
$z = \Delta\lambda/\lambda = (402 - 393)/393 = 9/393 = 0.0229$ .
Recession speed
$v = zc = 0.0229 \times 3.0\times10^{5} = 6.87\times10^{3}\,\text{km s}^{-1}$ .
Distance from Hubble's law
$d = v/H_0 = 6870/70$ .
$d = 98\,\text{Mpc}$

Result: The galaxy recedes at about $6900\,\text{km s}^{-1}$ and lies roughly $98\,\text{Mpc}$ away.

Typical mistakes

Mixing units in Hubble's law - keep $v$ in $\text{km s}^{-1}$ and $d$ in $\text{Mpc}$ , or convert consistently.
Confusing redshift $z$ (a ratio) with the wavelength change $\Delta\lambda$ itself.

Active revision

A galaxy's hydrogen line, normally at $656\,\text{nm}$ , is observed at $665\,\text{nm}$ . Find the recession speed and, using $H_0 = 70\,\text{km s}^{-1}\,\text{Mpc}^{-1}$ , its distance.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Astrophysics (optional)

5 sections~14 min reading time3 competenciesLevel Standard 2 · Advanced 3

Astrophysics (optional)

Key points

Magnification and length of a telescope

Key points

Resolving power of a telescope

Key points

Absolute magnitude from distance

Key points

Temperature and radius of a star

Key points

Redshift, recession speed and distance

Sources

Astrophysics (optional)

Key points

Magnification and length of a telescope

Key points

Resolving power of a telescope

Key points

Absolute magnitude from distance

Key points

Temperature and radius of a star

Key points

Redshift, recession speed and distance

Sources