9. Astrophysics

Question 1

What is the axis of symmetry called?

Answer 1

The principal axis

Question 2

What is the principal focus?

Answer 2

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 3

Define focal length

Answer 3

The distance between the centre of the lens and the principle focus

Question 4

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

Answer 4

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

Question 5

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

Answer 5

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

Question 6

Draw the ray diagram for an object that is a long way from the lens (beyond 2F) and describe the image’s appearance

Answer 6

The image is real and inverted but smaller than the object - this is called diminished
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Question 7

Draw the ray diagram for an object that is very close to the lens and describe the image’s appearance

Answer 7

The image is virtual and upright but bigger than the object - this is called magnified
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Question 8

Draw the ray diagram for an object that is fairly close to the lens (Between f and 2f) to the lens and describe the image’s appearance

Answer 8

The image is real and inverted but bigger than the object - this is called magnified
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Question 9

Draw the ray diagram for an object that is at 2f and describe the image’s appearance

Answer 9

The image is real and inverted but the same size

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Question 10

What does normal adjustment in a telescope look like?

Answer 10

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Question 11

Give the formula for angular magnification in normal adjustment

Answer 11

M = angle subtended by image at eye ÷ angle subtended by object at unaided eye
Can also be written as M = 𝛼 / ß

Question 12

State the equation that relates M to the focal length for objective and eyepiece lenses

Answer 12

M = fo / fe

This can only be used if both angles from M = 𝛼 / 𝛽 are less than 10º

Question 13

How does an astronomical refracting telescope work?

Answer 13

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 14

What is chromatic aberration?

Answer 14

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 15

What is spherical aberration?

Answer 15

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 16

Describe a solution to chromatic and spherical aberration in lenses.

Answer 16

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 17

State 3 advantages of reflecting telescopes

Answer 17

• 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 18

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

Answer 18

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

Question 19

Define the Rayleigh Criterion

Answer 19

‘Two objects will be just resolved if the centre of the diffraction pattern of one image coincides with the first minimum of the other’.
θ ≈ λ / D

Question 20

What is apparent magnitude and absolute magnitude?

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 = σ A T^4, where A = surface area (m^2), 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.
λmax T = 2.898 x 10^-3 m k
(note: unit is metres Kelvin, not milliKelvin)

Question 26

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

Answer 26

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 27

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

Answer 27

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 (approximately -19.3) 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 28

Draw the light curve of a typical type Ia supernova

Answer 28

Light curves are graphs of absolute magnitude against time since peak magnitude. The peak magnitude of a type Ia supernova is -19.3.
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Question 29

What is the Doppler effect?

Answer 29

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.
∆λ / λ = - V / C = - ∆f / f

Question 30

What is red shift?

Answer 30

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.
The formula for red-shift can only be used when v is much smaller than c (the speed of light).

Question 31

State Hubble’s law

Answer 31

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

v = H。d

v = velocity of a retreating galaxy (km s-1)
d = Distance from Earth (Mpc)
H。= Hubble’s Constant = 65 km s-1 Mpc-1

Question 32

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

Answer 32

Time = distance / velocity = 1 / H。(since v = H。d)
The units of Hubble’s constant must be converted to SI units
65 km s-1 Mpc-1 x 10^3 gives H。 = 65,000 m s-1 Mpc-1
Divide by 1 Mpc (3.08 x 10^22 m) to get the units for H。as s-1
H。= 2.11 x 10^-18 s-1 so 1 / H。= 4.74 x 10^17 s
Age (convert to years) = 4.74 x 10^17 / 3600 / 24 / 365 = 1.5 x 10^10 years (or 15 billion years).

Question 33

Draw a simplified light curve for a binary star system.

Answer 33

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Question 34

What are exoplanets?

Answer 34

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 35

How can we detect exoplanets?

Answer 35

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 36

Draw the light curve for an exoplanet transiting in front of its star

Answer 36

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