Astro ☆

Question 1

Convex/converging lens

Answer 1

Focuses incident light

Question 2

Concave/diverging lens

Answer 2

spreads out incident light

Question 3

Principal axis

Answer 3

the line passing through the centre of the lens at 90º to its surface

Question 4

Principal focus (F)

Answer 4

○ In a converging lens: the point where incident beams passing parallel to the
principal axis will converge.
○ In a diverging lens: the point from which the light rays appear to come from.
This is the same distance either side of the lens

Question 5

Focal length (f)

Answer 5

The distance between the centre of a lens and the principal focus
○ The shorter the focal length, the stronger the lens
M = fo/fe

Question 6

Real image

Answer 6

• formed when light rays cross after refraction
  ○ Real images can be formed on a screen

Question 7

Virtual image

Answer 7

formed on the same side of the lens. The light rays do not cross, so a
virtual image cannot be formed on a screen

Question 8

Lens formula

Answer 8

1/u + 1/v = 1/f
(Where u is the distance of the object from the centre of the lens, v is the distance of the image from the centre of the lens, and f is the focal length of the lens)

Question 9

Power of a lens

Answer 9

a measure of how closely a lens can focus a beam that is parallel to
the principal axis (in other words, how short the focal length is).
○ The shorter the focal length, the more powerful the lens.
○ In converging lenses this value is positive and in diverging lenses this value is negative.
○ Power is measured in Dioptres (D)
—> P = 1/u +1/v = 1/f

Question 10

Refracting telescopes are comprised of two converging lenses

Answer 10

➔ The objective lens - The role of this lens is to collect light and create a real image of a very distant object. This lens should have a long focal length and be large so as to
collect as much light as possible. The collecting power of a telescope is directly proportional to the square of the radius of the objective lens (more on this later).
➔ The eyepiece lens - This magnifies the image produced by the objective lens so that the observer can see it. This lens produces a virtual image at infinity since the light rays are parallel. This reduces eye strain for the observer as they do not have to refocus every time they look between the telescope image and the object in the sky

Question 11

Normal adjustment for a refracting telescope

Answer 11

when the distance between the objective lens and
the eyepiece lens is the sum of their focal lengths (fo + fe)
This means the principal focus (F) for these two lenses is in the same place

Question 12

Magnifying power

Answer 12

M =angle subtended by the image at the eye/angle subtended by the object at the unaided eye

Question 13

Reflecting telescope

Answer 13

Usually involves a concave primary mirror with a long focal length and a small convex secondary mirror in the centre
Mirrors are usually thin coatings of aluminium or silver atoms deposited on backing material
This allows mirrors to be as smooth as possible and minimise distortions

Question 14

Chromatic aberration

Answer 14

Focal length of red light is greater than blue light which means they are focused at different points
This causes a white object to produce an image with white fringing with the effect being most noticeable for light passing through the edges of the lens
Only occurs in eyepiece lens

Question 15

Spherical aberration

Answer 15

The curvature of a lens or mirror can cause rays of light at the edge to be focused in a different position to those near the centre, leading to image blurring and distortion
This effect is most pronounced in lenses with a large diameter, and can be avoided completely by using parabolic objective mirrors in reflecting telescopes

Question 16

Collecting power

Answer 16

A measure of the ability of a lens or mirror to collect incident EM radiation
Collecting power increases with size of objective lens/mirror
Directly proportional to area of objective lens
Greater the collecting power the brighter the images produced by the telescope

Question 17

Resolving power

Answer 17

The ability of a telescope to produce separate images of close-together objects

Question 18

Rayleigh criterion

Answer 18

Theta = lambda/D
“Two objects will not be resolved if any part of central maximum of either of the image falls within the first minimum diffraction ring of the other
This is the related to the fact that, as light enters the telescope, it is diffracted in a target like shaped called an airy disc
Central maximum is the bright white circle in the centre and each of the dark rings around it are the minimum rings

Question 19

Charged coupled devices

Answer 19

An array of light-sensitive pixels, which become charged when they are exposed to light by the photoelectric effect

Question 20

Luminosity

Answer 20

Rate of light energy release /power output of a star

Question 21

Intensity

Answer 21

This is the power received from a star (its luminosity) per unit area and has the unit,
W m-2

Question 22

Apparent magnitude (m)

Answer 22

How bright an object appears in the sky, this depends on star luminosity and distance from Earth

Question 23

Absolute magnitude (M)

Answer 23

The magnitude of an object if it were placed 10 parsecs away from Earth

Question 24

M and m

Answer 24

m-M = 5log(d/10)

Question 25

Parallax

Answer 25

Apparent change of position of nearer star in comparison to distant stars in the background, as a result of the orbit of the Earth around the sun
Greater the angle of parallax the closer the star is to Earth

Question 26

Astronomical unit (AU)

Answer 26

Avg distance between the centre of the Earth and and the centre of the sun
1AU = 1.50 x 10^11 m

Question 27

Parsec (pc)

Answer 27

The distance at which the angle of parallax is 1 arcsecond (1/3600th of a degree)
1pc = 2.06 x 10^5

Question 28

Lightyear (ly)

Answer 28

The distance that an EM wave travels in a year in a vacuum
1ly = 9.46 x 10^5

Question 29

A black body radiator

Answer 29

Perfect emitter and absorber of all possible wavelengths of
radiation.
➔ Stars can be approximated as black bodies

Question 30

Stefan’s law

Answer 30

The power output (luminosity/P) of a black body radiator is directly proportional to its surface area (A) and its (absolute temperature)4
P = deltaAT^4
This law can be used to compare the power output, temperature and size of stars

Question 31

Wien’s displacement law

Answer 31

The peak wavelength (λmax) of emitted radiation is inversely
proportional to the absolute temperature (T) of the object
Wein’s law shows that the peak wavelength of a black body decreases as it gets hotter meaning the frequency increases so the energy of the wave increases (as expected).
This law can be used to estimate the temperature of black-body sources

Question 32

Intensity of light emitted

Answer 32

I = P/4πd^2

Question 33

Hydrogen Balmer lines

Answer 33

Absorption lines that are found in the spectra of O, B and A type stars. They are caused by the excitation of hydrogen atoms from the n = 2 state to higher/lower energy levels
If the temperature of a star is too high, the majority of hydrogen atoms will become excited to higher levels than n = 2 or electrons might even become ionised, so hydrogen balmer lines will not be present
If the temperature of a star is too low, the hydrogen atoms are unlikely to become excited, or may not be present at all, so hydrogen balmer lines will not be present

Question 34

Star classes

Answer 34

O Blue 25 000 - 50 000 (He+, He, H Weak)
B Blue 11 000 - 25 000 (He, H Slightly stronger than O)
A Blue/White 7 500 - 11 000 (H, ionised metals Strongest)
F White 6 000 - 7 500 (Ionised metals Weak)
G Yellow/White 5 000 - 6 000 (Ionised and neutral metals None)
K Orange 3 500 - 5 000 (Neutral metals None)
M Red < 3 500 (Neutral atoms, Titanium Oxide None)

Question 35

1. Protostar

Answer 35

○ Clouds of gas and dust (nebulae) have fragments of varying masses that clump together under gravity.
○ The irregular clumps rotate and a gravity/conservation of angular momentum spins them inwards to form a denser centre – a protostar.
○ The protostar is surrounded by a disc of material (a circumstellar disc).
○ When the protostar gets hot enough, it begins to fuse elements, producing a strong stellar wind that blows away any surrounding material

Question 36

2. Main sequence

Answer 36

○ The inward force of gravity and the outward force due to fusion are in equilibrium – the star is stable.
○ Hydrogen nuclei are fused into helium.
○ The greater the mass of the star, the shorter its main sequence period because it
uses its fuel more quickly

Question 37

3. Red Giant (for a star < 3 solar
   masses)

Answer 37

○ Once the hydrogen runs out, the temperature of the core increases and begins fusing helium nuclei into heavier elements (E.g. Carbon, Oxygen and Beryllium).
○ The outer layers of the star expand and cool

Question 38

4. White Dwarf (for a star < 1.4 solar masses)

Answer 38

○ When a red giant has used up all its fuel, fusion stops and the core contracts as gravity is now greater than the outward force.
○ The outer layers are thrown off, forming a planetary nebula around the remaining core.
○ The core becomes very dense (around 108 - 109kg m-3).
○ A white dwarf will eventually cool to a black dwarf

Question 39

5. Red Supergiant (for a star > 3 solar masses)

Answer 39

○ When a high-mass star runs out of hydrogen nuclei, the same process for a red giant occurs, but on a larger scale.
○ The collapse of red supergiants in a supernova (see below) causes gamma ray bursts.
○ Red supergiants can fuse elements up to iron

Question 40

6. Supernova (for a star > 1.4 solar masses)

Answer 40

○ When all fuel runs out, fusion stops and the core collapses inwards very
suddenly and becomes rigid (as the matter can no longer be forced any closer together).
○ The outer layers of the star fall inwards and rebound off of the core, launching them out into space in a shockwave.
○ As the shockwave passes through surrounding material, elements heavier than iron are fused and flung out into space.
○ The remaining core depends on the mass of the star.
○ A defining characteristic of a supernova is its rapidly increasing absolute magnitude.
○ Supernovae may release around 1044 J of energy, which is the same amount of
energy as the sun outputs in its 10 billion year lifetime

Question 41

7. Neutron Star (for a star between 1.4 and 3 solar masses)

Answer 41

○ When the core of a large star collapses, gravity is so strong that it forces protons and electrons together to form neutrons.
○ A neutron star is incredibly dense – about 1017 kg m- 3 (the density of nuclear matter).
○ Pulsars are spinning neutron stars that emit beams of radiation from the magnetic poles as they spin (up to 600 times per second)

Question 42

8. Black Hole (for a star > 3 solar masses)

Answer 42

○ When the core of a giant star collapses, the neutrons are unable to withstand gravity forcing them together.
○ The gravitational pull of a black hole is so strong that not even light can escape.
○ The event horizon of a black hole is the point at which the escape velocity becomes greater than the speed of light .
○ The Schwarzchild radius is the radius of the event horizon, and can be calculated using the formula: Rs = 2GM/c^2

Question 43

Supermassive black holes can form from:

Answer 43

● The collapse of massive gas clouds while the galaxy was forming
● A normal black hole that accumulated huge amounts of matter over millions of years
● Several normal black holes merging together.

Question 44

Doppler effect

Answer 44

• The compression or spreading out of waves that are emitted or reflected
  by a moving source. As the source is moving, the wavelengths in front of it are compressed and the wavelengths behind are spread out
• The Doppler effect causes the line spectra of distant objects to be shifted either towards the blue
  end of the visible spectrum when they move towards the Earth (blue-shift) or towards the red
  end of the spectrum when they move away from the Earth (red-shift)

Question 45

Red shift

Answer 45

Red-shift is used as evidence for the expanding universe, as distant objects are red-shifted. The
more distant the object, the greater its red-shift
z = v/c = Δf/f
= − Δλ/λ

Question 46

Quasars

Answer 46

Objects which have very large red shifts, suggesting they are very far away,
however they are also extremely bright. Using the inverse square law, you can show that the
power output of a quasar must be around that of an entire galaxy

Question 47

Hubble’s law

Answer 47

States that a galaxy’s recessional velocity is directly proportional to its
distance from the Earth. It essentially states that the universe is expanding from a common starting point.
This can be summed up in the formula: v = Hd

Question 48

Quasars are characterised by the following features:

Answer 48

● Extremely large optical red-shifts
● Very powerful light output
● Their size being not much bigger than a star

Question 49

Exoplanets

Answer 49

Exoplanets are planets that are not within our solar system; they orbit other stars.
They can be difficult to detect directly as they tend to be obscured by the light of their host stars.

Question 50

Detection of exoplanets

Answer 50

1.➔ Radial velocity method -
This is very similar to the method of detecting spectroscopic binaries. The star and planet orbit a common centre of mass, which causes the star to ‘wobble’ slightly. This causes a Doppler shift in the light received from the star (this method is sometimes referred to as the ‘wobble effect’ for this reason). The effect is most noticeable with high-mass planetssince they have a greater gravitational pull on the star. The line spectrum of the star is blue-shifted when it moves towards the Earth, then red-shifted when it moves away. This shows that there is something else near the star that is exerting a gravitational force on it – the exoplanet. The time period (T) of the planet’s orbit is equal to the time period of the Doppler shift

2.➔ Transit method -
This involves observing the intensity of the light output of a star. If a planet crosses in
front of a star (‘transits’), the intensity dips slightly. If the intensity of a star dips regularly, it could be a sign that there is an exoplanet orbiting it. If there are variations in the regularity of the dips, there may be several planets orbiting the same star which have a gravitational effect on the transiting planet