Astrophysics Revision

Question 1

Converging Lenses?

Answer 1

Lenses that are thicker in the middle than the edges and cause rays of light to bend towards each other. They can produce virtual and real images

Question 2

Axial and Non-Axial Rays into converging lenses

Answer 2

Axial rays are parallel to each other and perpendicular to the lens and will have the principle focus on the principle axis. Whereas non axial rays are still parallel to each other but not perpendicular to the lens and will have their principal focus below the principle axis

Question 3

Object Ray Diagrams?

Answer 3

Unless told otherwise. Draw a line from the top of the object that passes through the centre of the lens. Draw another line that goes from the top of the object to the lens in a straight line and then bends after the lens axis, where it cuts the principal axis is the focal length. The intersection point of these two lines if the top of the image.

Question 4

Virtual Images on ray diagrams?

Answer 4

This is identified by the two rays drawn never intersecting. This means the image is formed behind the lens meaning it’s consequently a virtual image. If the distance from the object to the lens is smaller than the distance to the image it is virtual, if the distance from the object to the lens is larger than the focal length it is a real image

Question 5

Real images compared to virtual images?

Answer 5

Real images are where light rays from an object pass through another point in space. Virtual images are when light rays from a point on an object appear to have come from somewhere else meaning the image can’t be captured on a screen

Question 6

Lens Equation?

Answer 6

1/f = 1/u + 1/v. Where “u” is the distance between the object and the lens axis, “v” is the distance between the image and the lens and is negative if virtual and positive if real, “f” is the focal length

Question 7

Refracting Telescopes?

Answer 7

Two converging lenses with an objective lens which is the narrow lens and converges the rays to form a real image inside the telescope. The eyepiece lens is the wider lens and creates a magnified virtual image the observer can view

Question 8

Normal Adjustment?

Answer 8

The adjustment of a refracting telescope when the principle focus of the eyepiece lens is in the same position of the focal point of the objective lens. It has non-axial rays with the lowest one having a straight line to the eyepiece lens and the other one parallel to it before the lens cuts through the middle of the lens. The focal length of the objective lens is always greater than the focal length of the eyepiece lens

Question 9

Angular Magnification?

Answer 9

The angle subtended by the image at the eye divided by the angle subtended by the object at the unaided eye. It can also be the ratio of focal lengths of the objective lens divided by the focal length of the eyepiece lens

Question 10

Cassegrain Reflecting Telescopes?

Answer 10

Telescopes that use mirrors to reflect and focus light from a parabolic concave mirror to a principle focus forming a real image. This arrangement has a secondary convex mirror to reflect the light from this focal point to outside the telescope when an eye lens magnifies the image.

Question 11

CCD?

Answer 11

Stands for charge coupled devices and are light sensitive detectors that are used to capture images digitally

Question 12

CCD function?

Answer 12

A chip with a grid of millions of pixels that when photons hit a pixel it creates a free electron which builds charge. This is measured and creates a digital signal. This describes where the light hits and the intensity. The charge on each pixel varies depending on the number of photons which hit it and this allows a digital image of an object to be created

Question 13

CCD’s Compared to the human eye?

Answer 13

The quantum efficiency of a CCD is about 80% whereas the eye’s is about 1% meaning they detect more light. CCD’s can detect a wider spectrum of light as they can detect not just visible light and instead wavelengths like UV and infrared. CCD’s have 500 megapixels to avoid pixelation whereas the eye only needs 50. The spatial resolution of a CCD is around 10µm whereas for the eye the minimum resolvable distance is 100µm, these two mean the ccd is better for fine detail but the eye is better for tracking small movements. CCD’s are less convenient as they require additional equipment but they do have the benefit of storing images digitally

Question 14

Quantum Efficiency?

Answer 14

The proportion of incident photons detected by a light detector

Question 15

Spatial Resolution?

Answer 15

How far different parts of the object need to be viewed in order to be distinguishable

Question 16

Megapixels?

Answer 16

The more megapixels the more detail typically but the spatial resolution needs to be considered as the lower this is the more detailed the image is even if the number of megapixels is lower

Question 17

Resolving Power?

Answer 17

A measure of how much detail a telescope can see and the higher this is the less blurry an image will be. It is dependent on the minimum angular resolution

Question 18

Minimum Angular Resolution?

Answer 18

The smallest angular separation an instrument can have to be able to distinguish two points. The smaller this angle the better the resolving power

Question 19

Distinguishing Objects?

Answer 19

Resolution is limited by diffraction. Light diffracts through a circular aperture has a diffraction pattern with maxima and minima rings and a central maximum central circle called an Airy disc. Two light sources are distinguished if the airy disc’s centre of one sources is at least the first minimum away from the other sources airy disk

Question 20

Rayleigh Criterion?

Answer 20

θ = λ/D where “θ” is the minimum angular resolution and “D” is the diameter of the aperture in meters otherwise the diameter of the objective lens or mirror

Question 21

Chromatic Aberration?

Answer 21

Where glass refracts different colours of light by different amounts so each colour is in a slightly different position and this blurrs the image

Question 22

Disadvantages of Refracting Telescopes?

Answer 22

They suffer from chromatic aberration. Any impurities in the glass cause light to scatter meaning faint objects aren’t seen. Have to be supported from the edges so their shape doesn’t distort. For large magnifications the telescopes are very long due to a long focal length being needed. They are expensive due to large lenses being needed and building needed to store them

Question 23

Benefits of reflecting telescopes compared to refracting telescopes?

Answer 23

Large mirrors of good quality are cheaper to build than large lenses and can be supported from underneath and they don’t distort as much as lenses

Question 24

Spherical Aberration?

Answer 24

If a mirror isn’t quite parabolic parallel rays reflecting off different parts of the mirror don’t have the same focal point

Question 25

Reflecting Telescope disadvantages?

Answer 25

Suffer from spherical aberration. Some incoming light will be blocked by the secondary mirror and mirror supports. Some light reflected by the primary mirror diffracts around the secondary mirror which like the blocking of light decreases the detail in the image

Question 26

Radio Telescopes function?

Answer 26

Have a parabolic wire mesh, due to long wavelengths of radio waves not noticing gaps, that reflect the waves into an antenna used as a detector

Question 27

Radio Telescope Advantages and Disadvantages?

Answer 27

Advantages- Manoeuvrable meaning the source waves can be tracked. Cheaper manufacturing costs as its dish is made of wire mesh and impurities have less of an effect due to radio waves having a long wavelength Disadvantages- Waves need to be amplified due to signals being weak which adds noise and errors to faint signals. Resolving power is low so signals can be combined from different telescopes but since more are needed this brings additional cost

Question 28

IR Telescopes?

Answer 28

Has a parabolic mirror that focuses IR radiation onto a CCD. The longer wavelength means it's not affected as much by impurities in the parabolic mirror. However the telescopes heat up and produce its own IR radiation which means they need to be cooled to have more accurate readings

Question 29

U-V Telescopes?

Answer 29

They have a parabolic mirror that focuses UV radiation onto a CCD. UV radiation cannot pass through ozone so these telescopes are used in space but stellar bodies like white dwarfs emit a lot of UV which means they are useful at getting information from astronomical objects. A disadvantage is they have a short wavelength which means they need to be precisely made

Question 30

X-Ray Telescopes?

Answer 30

Have grazing mirrors that angle x-rays so they focus on a detector. This has to be done as usually x-rays are absorbed or pass through a material. A disadvantage is these mirrors are expensive due to their material and they are difficult to repair as the telescope has to be in space. An advantage is very hot astronomical bodies emit x-rays allowing information to be collected from stellar objects like neutron stars

Question 31

Positioning Telescopes?

Answer 31

Visible and radio waves can pass through to the surface allowing optical and radio telescopes to be set up on the surface. Some IR is transparent to Earths atmosphere but IR is absorbed by moisture so high up dry places like mountains are suitable locations for these telescopes. UV, X-Ray and other infrared are absorbed higher up so using weather balloons, aeroplanes or having it orbit Earth or set up elsewhere is space can allow these wavelengths to be detected

Question 32

Resolving Power Factor?

Answer 32

The Rayleigh Criterion affects the resolving power with it being dependent on the wavelength of radiation and diameter of the objective mirror or dish. The resolving power of the telescope can be limited by the quality of the detector

Question 33

Collecting Power?

Answer 33

Otherwise energy collected per second, it is proportional to the collecting area otherwise the dish diameter squared. For most telescopes it is the area of the objective mirror or dish but for x-ray telescopes its the opening x-rays enter the telescope. X-ray telescopes have a much lower collecting power than other telescopes

Question 34

Parallax?

Answer 34

A measure of how much a nearby object appears to move in relation to a distant background due to the observers motion

Question 35

Distance to nearby stars?

Answer 35

The distance to nearby stars can be calculated by observing how far they move relative to distant stars that don't appear to move. This is done by comparing the position of nearby stars in relation to background stars at different parts of Earth's orbit. This has the formula d = r/θ where "r" is the radius of Earth's orbit, "θ" is the angle of parallax in radians and "d" is the distance to the star

Question 36

Parsec?

Answer 36

Otherwise "pc" and a star is one parsec away if the angle of parallax is one arc second otherwise 1/3600 degrees. It has a value of 1 parsec = 3.08x10 16 m

Question 37

Light Year?

Answer 37

Otherwise "ly", it is the speed at which electromagnetic waves travel through a vacuum in one year. 1 light year is 9.46x10 15m

Question 38

Measuring Distances?

Answer 38

2θ = r /d -> Where "θ" is half the angle subtended by the object at the eye, "r" is the radius of the object and "d" is the distance to the object. Measuring distances can be done with this method of the parallax to find the distance but the size of the object needs to be known to use it, other methods to measure distance are standard candles, red shift and quasars

Question 39

Intensity?

Answer 39

The intensity of an object is the power received per unit area at Earth and is the effective brightness of an object

Question 40

Luminosity?

Answer 40

The total amount of energy emitted in the form of EM radiation per second otherwise the rate of energy transfer. Luminosity is the power output of the star

Question 41

Luminosity compared to brightness?

Answer 41

Luminosity is the power emitted by a star but brightness is the power received at Earth per unit area

Question 42

Brightness?

Answer 42

A subjective scale based on the power output of the star and its distance away from Earth. The brightest stars will have a high luminosity and be close to Earth

Question 43

Apparent Magnitude?

Answer 43

Has the symbol "m" and is the measure of brightness or intensity of an object

Question 44

Apparent Magnitude Scale?

Answer 44

Originally invented by Hipperarchus where dim stars had an apparent magnitude of 6 and the brightest stars had an apparent magnitude of 1. On the modern scale which is continuous the lower otherwise the more negative the apparent magnitude the brighter the star. This formula is use: I2 / I1 = 2.51 ^ (m1-m2)

Question 45

Absolute Magnitude?

Answer 45

An objects apparent magnitude 10 parsecs from Earth. The absolute magnitude is based on the power output of the object not its distance from Earth. It has the the symbol "M" in the equation: m-M = 5log(d/10)

Question 46

Standard Candle?

Answer 46

An object where the absolute magnitude is known otherwise can be calculated directly and is a fixed value like the peak absolute magnitude of a type 1A supernovae

Question 47

Temperature and Radiation?

Answer 47

All objects hotter than absolute zero emit electromagnetic radiation as a result of their temperature. The wavelengths of radiation emitted depend on the temperature of the object

Question 48

Black Body?

Answer 48

A blackbody is a body of radiation that absorbs all electromagnetic radiation of all wavelengths and can emit all wavelengths of electromagnetic radiation. It is an object with a pure black surface and emits radiation strongly in a continous spectrum in a well defined way and absorbs all light incident on it

Question 49

Black Body Curves?

Answer 49

A graph of radiation power output on the y-axis and wavelength on the x-axis. The higher the peak power the steeper the gradient of the curve to 0m wavelength and the closer the wavelength of the peak power is to 0m wavelength. Stars are a good approximations tp behave as black bodies which means these curve can allow estimations to be made on their temperature and other properties

Question 50

Properties of stars from black body curves?

Answer 50

The higher the surface temperature of the star the higher its power output and the shorter the peak wavelength of intensity of raditation. This is identified through Wein's displacement Law: (λmax)(T) =2.9 x 10-3 where "T" is the temperature in kelvin. A hotter star may not appear as bright as the radiation it emits may not mostly come from the visible part of the spectrum meaning a cooler star could be brighter as it could emit more radiation from the visible part of the spectrum

Question 51

Stefan's Law?

Answer 51

Where the power output of a star is proportional to the 4th power of the stars surface temperature and is directly proportional to the surface area. This produces the equation P = σAT^4 where "σ" is Stefan's Constant of 5.67x10-8

Question 52

Intensity Inverse Square Law?

Answer 52

Intensity is the power of radiation per square metre. As radiation spreads out the intensity decreases and this produces the equation: I = P/4πd^2 where "d" is the distance from the star. The assumptions of this inverse square law are the star is spherical and it gives out an even amount of power in every direction

Question 53

Absorption Lines in space?

Answer 53

Absorption lines are produced when radiation from the star passes through a cooler gas in the stars atmosphere. The low temperatures of space mean most of the electrons in the cool gas are at ground state. When the radiation passes through the photons which excite the atom are absorbed. This reduces the intensity of radiation at particular wavelengths and is identified by the darker the absorption line the more of that wavelength that has been absorbed

Question 54

Balmer Series?

Answer 54

The absorption line in an absorption line spectrum caused by electrons moving between the first excitation level and higher energy levels in atomic hydrogen

Question 55

Hydrogen absoprtion line in visible spectrum?

Answer 55

For this to occur the electrons in the hydrogen atom need to be in the first excitation level of n=2 due to this being the level where photons emitted are the wavelengths in the visible spectrum. This happens at high temperatures where collisions between atoms give electrons the addtional energy. If the temperature is too high most will be in the n=3 level. This means there will be less Balmer transitions so this shows temperature impacts the intensity of Balmer lines

Question 56

Spectral Classes?

Answer 56

The groups at which stars are classified into due to their temperature and relative strength otherwise intensity of certain absorption lines. The order in decreasing temperature is OBAKGKM

Question 57

Spectral Class O?

Answer 57

Blue in colour and have a temperature range of 25,000 to 50,000 kelvin. Has absoprtions of helium plus ion, helium and weak hydrogen

Question 58

Spectral Class B?

Answer 58

Blue in colour and have a temperature range of 11,000 to 25,000 kelvin. Has Helium and Hydrogen absorptions

Question 59

Spectral Class A?

Answer 59

Blue-white in colour and have a temperature range of 7,500-11,000 kelvin and there are some metal iron absorptions for this spectral class