Astro

Question 1

What is the axis of symmetry called?

Answer 1

The principal axis

Question 1

What is the principal focus?

Answer 1

A point on the axis which is the same distance from the optical centre as the focal length. This is where light rays travelling parallel to the principal axis prior to refraction converge.

Question 2

Define focal length.

Answer 2

The distance between the centre of the lens and the focal point

Question 3

What does ‘u’ represent in lens diagrams and equations?

Answer 3

The distance between the object and the centre of the lens, u is always positive

Question 4

What does ‘v’ represent in lens diagrams and equations?

Answer 4

The distance between the image and the centre of the lens, v is positive for real images and negative for virtual images

Question 5

How does an astronomical refracting telescope work?

Answer 5

There are two converging lenses, the objective lens and the eyepiece lens. The role of the objective lens is to collect light and create a real image of a distant object. This image is then magnified by the eyepiece lens, which produces a virtual image (formed at infinity so as to reduce eye strain when looking between the object and the telescope image).

Question 6

How does a cassegrain telescope work?

Answer 6

There is a concave primary mirror with a long focal length and a small convex secondary mirror in the centre. The light is collected by the primary mirror and focused onto the secondary mirror, which then reflects it onto an eyepiece lens.

Question 7

What is chromatic aberration?

Answer 7

When a lens refracts different colours of light by different amounts as they have different wavelengths. This causes the image for each colour to form in a slightly different position, causing coloured fringes around the image

Question 8

What is spherical aberration?

Answer 8

When light is focused in different places due to the curvature of a lens or mirror, causing image blurring. This can be resolved in reflecting telescopes by using a parabolic mirror.

Question 9

Describe a solution to chromatic and spherical aberration in lenses.

Answer 9

Using an achromatic doublet brings all rays of light into focus in the same position by using a convex lens and a concave lens of different types of glass cemented together.

Question 10

State 3 advantages of reflecting telescopes

Answer 10

• There is very little chromatic aberration (only in the
  eyepiece lens, but this can be resolved by using an achromatic doublet)
• Simpler to increase the size of the objective since mirrors can be supported from behind and are lighter than lenses
• Using parabolic mirrors stops spherical aberration

Question 11

What happens when you increase the size of the objective lens/mirror?

Answer 11

Increasing the diameter of the objective means you can observe fainter objects. This is because collecting power is proportional to (objective diameter)2

Question 12

Define the Rayleigh Criterion

Answer 12

Two objects will be just resolved if the centre of the diffraction pattern of one image coincides with the first minimum of the other’.

Question 13

Explain the structure, positioning and uses of a single dish radio telescope.

Answer 13

Structure: Large parabolic dish that focuses radiation onto a receiver
Positioning: can be ground-based but must be in isolated locations
Uses: observing things such as galaxies, stars and black holes

Question 14

Why do radio telescopes need to be larger than optical telescopes?

Answer 14

Since radio waves have a much larger wavelength than visible light, in order to achieve the same resolving power as an optical telescope, the objective diameter must be much larger in accordance with θ ≈ λ/D .

Question 15

Explain the structure, positioning and uses of an infrared telescope

Answer 15

Structure: Large concave mirror focusing light onto a detector. Must be cooled with cryogenic fluids to avoid interference.
Positioning: Must be in space as infrared light is blocked by the atmosphere
Uses: observing cooler regions in space (from a few tens to 100K)

Question 16

Explain the structure, positioning and uses of an ultraviolet telescope

Answer 16

Structure: Cassegrain configuration that focuses radiation onto solid state devices Positioning: Must be in space as ultraviolet light is blocked by the ozone layer
Uses: observing the interstellar medium and star formation regions

Question 17

Explain the structure, positioning and uses of an x-ray telescope

Answer 17

Structure: combination of hyperbolic and parabolic mirrors to focus radiation onto a CCD
Positioning: Must be in space as x-rays are blocked by the atmosphere
Uses: observing high-energy events and areas such as active galaxies, black holes and neutron stars

Question 18

Compare the quantum efficiency of a CCD to the eye

Answer 18

Quantum efficiency: the percentage of incident photons that liberate an electron in the photoelectric effect. This can be upwards of 80% for a CCD, compared to 4-5% for the human eye.

Question 19

Compare the convenience of a CCD to the eye

Answer 19

The CCD is more convenient for accessing data remotely (like retrieving data from space telescopes such as Hubble). It is easier to analyse CCD data on computers, and CCDs have a wider spectral range, allowing them to perceive wavelengths that cannot be detected by the human eye. That being said, looking down a telescope is not an inconvenient task.

Question 20

What is apparent magnitude and absolute magnitude? What equation links them?

Answer 20

Apparent magnitude (m): how bright the star appears from Earth
Absolute magnitude (M): how bright the star would appear if it were placed 10 parsecs from Earth.
m-M=5log(d/10) (where d = distance from Earth)

Question 21

What is the Hipparcos scale?

Answer 21

The Greek astronomer Hipparchus catalogued stars, defining their brightness in terms of apparent magnitudes (m), with brightest stars a magnitude of 1 and the faintest a magnitude of 6.
The scale has since been extended to include brighter objects (like the Sun, with an m of -26.47) and fainter objects that were discovered with the invention of the telescope.

Question 22

Define parsec

Answer 22

The distance to an object that subtends an angle of one arcsecond (1/3600th of a degree) to the line that runs from the centre of the Earth to the centre of the Sun

Question 23

Define light year

Answer 23

A light year is the distance travelled by light in a vacuum in one year. In metres this is 9.46 x 1015 m (speed of light multiplied by the number of seconds in a year).

Question 24

State Stefan’s law

Answer 24

The power output of a star is directly proportional to its surface area and it’s (absolute temperature)4.

P = σAT4, where A = surface area (m2), T = temperature (K) and σ = the Stefan constant = 5.67 x 10-8 W m-2 K-4

Question 25

State Wien’s displacement law

Answer 25

The wavelength of a star’s emission at peak intensity is inversely proportional to its absolute temperature.

λmaxT = 2.898 x 10-3 m k
(note: unit is metres Kelvin, not milliKelvin)

Question 26

What is a black body?

Answer 26

A black body absorbs electromagnetic radiation of all wavelengths and can emit electromagnetic radiation of all wavelengths. A black body does not reflect any radiation – it absorbs all radiation incident on it.

Question 27

What are Hydrogen Balmer Lines

Answer 27

Hydrogen Balmer lines are 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.

Question 28

O Spectral Class?

Answer 28

Blue

25 000k - 50 000k

He+, He, H (absorption lines)

Weak (prominence o f hydrogen Balmer lines)

Question 29

B Spectral Class?

Answer 29

Blue

11 000k - 25 000k

He, H (absorption lines)

Slightly stronger than O (prominence o f hydrogen Balmer lines)

Question 30

A Spectral Class?

Answer 30

Blue/White

7 500 - 11 000k

H, ionised metals (absorption lines)

Strongest (prominence o f hydrogen Balmer lines)

Question 31

F Spectral Class?

Answer 31

White

6 000 - 7 500k

Ionised metals (absorption lines)

Weak (prominence o f hydrogen Balmer lines)

Question 32

G Spectral Class?

Answer 32

Yellow/White

5 000 - 6 000k

Ionised and neutral metals (absorption lines)

None (prominence of hydrogen Balmer lines)

Question 33

K Spectral Class?

Answer 33

Orange

3 500 - 5 000k

Neutral metals (absorption lines)

None (prominence of hydrogen Balmer lines)

Question 34

M Spectral Class?

Answer 34

Red

< 3 500k

Neutral atoms, Titanium Oxide (absorption lines)

None (prominence of hydrogen Balmer lines)

Question 35

What are supernovae? Describe how type Ia and type II form

Answer 35

A supernova is the explosion of a star, which causes it to very suddenly and rapidly increase in absolute magnitude. Type Ia Supernova: The result of a white dwarf core accumulating too much matter from its binary partner and exploding above a critical mass Type II Supernova: A single star (for example a red giant) that collapses rapidly under its own gravity, causing its outer layers to be ejected.

Question 36

Explain why Type II supernovae cannot be used as standard candles whereas Type Ia supernovae can.

Answer 36

A standard candle is an astronomical object that has a known absolute magnitude so astronomers can calculate the distance using m - M = 5log(d/10). All Type Ia supernovae explosions have the same peak absolute magnitude as they all have the same critical mass (thus have consistent light curves) so they can be used as standard candles. Type II supernovae are not as predictable, so they cannot be used as standard candles.

Question 37

What is a black hole?

Answer 37

When the core of a star larger than 3 solar masses collapses, it forms a black hole. The escape velocity of a black hole is greater than the speed of light – light cannot escape it, which is where black holes earned their name.

Question 38

What is the event horizon?

Answer 38

The boundary at which the escape velocity equals the speed of light

Question 39

What is the Schwarzschild Radius?

Answer 39

the distance from the centre of the black hole to the event horizon

Question 40

What is dark energy?

Answer 40

• When astronomers calculated the distance to some Type Ia supernovae, they discovered them to be dimmer than expected. This suggested the expansion of the universe is accelerating, which has been attributed to dark energy.
• Dark energy is thought to be energy that has an overall repulsive effect throughout the universe

Question 41

What is the Doppler effect?

Answer 41

The change in wavelength and frequency of a wave as the source moves away or towards the observer.

As the source moves towards the observer, the waves are compressed and wavelength decreases. As the source moves away from the observer, the waves spread out and the wavelength increases.

Question 42

What is red-shift?

Answer 42

Red shift (z) is the shift in wavelength and frequency of waves from a retreating source towards/beyond the red end of the electromagnetic spectrum. Cosmological red shift is evidence for the Big Bang.

Question 43

State Hubble’s law

Answer 43

The velocity of receding galaxies is proportional to their distance from Earth.

v = Hd

v = velocity of a retreating galaxy (km s-1)
d = Distance from Earth (Mpc)
H = Hubble’s Constant

Question 44

Use Hubble’s law to estimate the age of the universe

Answer 44

Time = distance / velocity = 1 / H0 (since v = H0d)

Question 45

What are quasars?

Answer 45

A quasar is a nucleus of an active galaxy; a supermassive black hole surrounded by a disc of matter. As matter falls into the black hole, jets of radiation are emitted from the poles of the quasar.

Question 46

What suggests that they are extremely distant objects?

Answer 46

Large optical red shift shows quasars are the most distant observable objects. From the inverse square law for intensity we know they are extremely powerful, with the same energy output as several galaxies. They were initially found to be powerful radio sources but with further telescope developments we now know they emit all wavelengths of EM radiation.

Question 47

What is the Big Bang theory? State evidence that lead us to believe this is true.

Answer 47

Scientists believe that, 13.8 billion years ago, the universe exploded from an extremely hot and dense point and is still expanding now.

CMBR (Cosmological Microwave Background Radiation) is the heat signature left behind from the big bang. The EM radiation released in the explosion shifted from extremely high energy waves into the microwave region as the universe expanded, stretching out the waves. CMBR has a black body distribution with a peak that corresponds to a temperature of 2.7K. There was nuclear fusion of hydrogen into helium which explains the large abundance of helium in today’s universe.

Question 48

How is the relative abundance of Hydrogen and Helium used as evidence for the Big Bang?

Answer 48

During the Big Bang, Hydrogen was converted into Helium via nuclear fusion. The expansion of the universe caused it to cool, meaning it was no longer hot enough to fuse elements. About ¼ of all Hydrogen was converted into Helium, resulting in a ratio of H:He of 3:1. The observed distribution of matter now is 73% hydrogen, 25% helium and 2% everything else, which follows the predicted ratio.

Question 49

What are exoplanets?

Answer 49

Exoplanets are planets that are not in our solar system. Direct observations of exoplanets are difficult as their light is often obscured by the stars they orbit. Also they tend to be too close together for the telescope to resolve them.

Question 50

How can we detect Exoplanets?

Answer 50

1. Radial Velocity Method: A star and a planet will orbit their common centre of mass, this means the star will have tiny variations in its distance from Earth, shown by tiny red and blue shifts in its spectrum.
2. Transit method: As a planet moves between the star it orbits and the Earth, the star’s brightness appears to decrease slightly. We can detect this and use it to calculate the diameter of the planet. Unfortunately there is a low chance of this orbit being in the right place for us to measure this, so it is mostly only useful for detecting planets with small orbits (they are more likely to cross the star’s disc)

Question 51

Equation for intensity?

Answer 51

I2/I1 = 2.51^m1-m2