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PHYA5 ~ 20th June 2013 ~ A2 Physics

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Reply 420
I mean you can kind of swing the telescope lens around to look at other objects without the eye having to refocus, because ANY object you choose to look at through the telescope is at infinity. This is a situation where an astronomer is using the telescope to examine many objects, over a long period of time - not just one.

I dug up this:
For comfortable viewing, the final image formed by the eyepiece (ocular) should be at infinity. Although
the human eye can focus on objects which are 25 cm away to infinity, the eye muscles are most relaxed
when viewing objects which are very far away.

from here: http://n.ethz.ch/~geric/download/4.%20Semester/Physik%20II/FS12/Blatt11_2012_Lsg.pdf

Which is, weirdly. a paper used by the Swiss Federal Institute of Technology in Zurich.

I haven't been able to find any other reasons that the image is formed at infinity, so I'm just going with that.
Reply 421
Original post by bugsuper
I mean you can kind of swing the telescope lens around to look at other objects without the eye having to refocus, because ANY object you choose to look at through the telescope is at infinity. This is a situation where an astronomer is using the telescope to examine many objects, over a long period of time - not just one.

I dug up this:
For comfortable viewing, the final image formed by the eyepiece (ocular) should be at infinity. Although
the human eye can focus on objects which are 25 cm away to infinity, the eye muscles are most relaxed
when viewing objects which are very far away.

from here: http://n.ethz.ch/~geric/download/4.%20Semester/Physik%20II/FS12/Blatt11_2012_Lsg.pdf

Which is, weirdly. a paper used by the Swiss Federal Institute of Technology in Zurich.

I haven't been able to find any other reasons that the image is formed at infinity, so I'm just going with that.


I agree with your explanation. Because the image rays are parallel to the construction line which goes through the centre of the eye piece lens, the image will appear at infinity.
Reply 422
Original post by bugsuper
I mean you can kind of swing the telescope lens around to look at other objects without the eye having to refocus, because ANY object you choose to look at through the telescope is at infinity. This is a situation where an astronomer is using the telescope to examine many objects, over a long period of time - not just one.

I dug up this:
For comfortable viewing, the final image formed by the eyepiece (ocular) should be at infinity. Although
the human eye can focus on objects which are 25 cm away to infinity, the eye muscles are most relaxed
when viewing objects which are very far away.

from here: http://n.ethz.ch/~geric/download/4.%20Semester/Physik%20II/FS12/Blatt11_2012_Lsg.pdf

Which is, weirdly. a paper used by the Swiss Federal Institute of Technology in Zurich.

I haven't been able to find any other reasons that the image is formed at infinity, so I'm just going with that.


for this sort of telescope (two converging lens), do you think we would need to know any other scenarios other than normal adjustment where the two focal points at the same place?
Reply 423
I certainly don't think so. The textbook doesn't mention any other kind of arrangement - plus, if it's not in normal arrangement, I think some of the equations (like magnification = fo/fe) don't apply. I doubt it.
Reply 424
For a cassegrain telescope, how is the focal length of the objective (concave) mirror increased by using a convex mirror as the secondary mirror?

surely the focal length of a mirror is intrinsic to it? why does another mirror have an effect on this?

EDIT, I get it now :biggrin:
(edited 10 years ago)
Reply 425
by the way, what are you guys using the revise astrophysics?

is it just me, or the the online nelson thrones stuff seems to go into a lot of detail? it seems to be compared to the the collins student purple unit 5 book
Reply 426
I've been using my own notes which are mainly from the collins book. I didn't even know there was Nelson Thornes online stuff... :/
What is this online stuff?
Reply 428
Could someone please explain standard candles to me? I get that they are like a set 'measurement' from the peak of a type 1a supernovae but how do we actually determine distances from them? I know that we know he absolute magnitude but how do we work out the absolute magnitude to be able to determine the distance? And also, for how long does a type 1a supernova stay at an absolute magnitude of -19.2? Is it for substantial period of time for it to be observed?

Finally, I have query to the big bang theory. Is the increased rate of expansion evidence for the big bang theory? It just noted down as that in my book, but I don't actually understand why :/
Reply 429
The absolute magnitude of these stars - that is, how bright they REALLY are at a distance of 10Pc - is known. We can look at them through telescopes, realise they're standard candles, and determine the apparent magnitude by observing them from Earth.

If you compare apparent and absolute magnitude, when they're both known, you can use the formula m - M = 5log(d/10) - for d in parsecs - to determine the distance. I think that the type-1 supernovas peak magnitude remains there for a day or so, but it really doesn't matter - in reality, they'd observe and record THE ENTIRE LIGHT CURVE throughout the supernova event (which can last for weeks) and compare that to the known light curve. So they use lots of values to determine the distance.

The increased rate of expansion is not evidence for the Big Bang Theory; it's kind of a seperate issue to do with the possible existence of dark energy. Evidence for the big bang theory that we need to know includes:

- The cosmic microwave background, which corresponds to the radiation that was formed in the universe shortly after the Big Bang which has been redshifted due to the Universe's expansion
- The relative abundance of elements in the Universe (hydrogen / helium ratios) which fits with the Big Bang model
- The redshifts in distant galaxies which seems to show that they're all moving away from us, implying expansion

Additionally, when we observe faroff stars and galaxies - so far away that the light from them arrived when the Universe was young - we find that conditions are different to closer parts of the Universe. This isn't evidence for the Big Bang theory per se, but if the Universe was always the same size etc, "steady state theory", then we'd expect to see things looking very much the same billions of years ago compared to know

Hope it helps
Reply 430
This is going to be the easiest 100 UMS, without a doubt.

The easiest option by far is applied physics.
After all my maths exams, ill be glad to have a nice easy one to end it.
Reply 431
Original post by JoshL123
Could someone please explain standard candles to me? I get that they are like a set 'measurement' from the peak of a type 1a supernovae but how do we actually determine distances from them? I know that we know he absolute magnitude but how do we work out the absolute magnitude to be able to determine the distance? And also, for how long does a type 1a supernova stay at an absolute magnitude of -19.2? Is it for substantial period of time for it to be observed?

Finally, I have query to the big bang theory. Is the increased rate of expansion evidence for the big bang theory? It just noted down as that in my book, but I don't actually understand why :/


Type 1a supernovae can be used as standard candles as they reach the same peak value of M (absolute magnitude) of around -19.3. The -19.3 is the value for M taken after around 20 days after the start of increase in brightness. From these measurements and using m-M=5log(d/10), we can measure the distance to these galaxies. However, by using red shift measurements and Hubble's Law (which assumes a constant rate of expansion), these supernovae are dimmer than predicted. This must mean that the expansion of the Universe is accelerating and some form of 'dark energy' is driving this expansion.
(edited 10 years ago)
Original post by OmegaKaos
This is going to be the easiest 100 UMS, without a doubt.

The easiest option by far is applied physics.
After all my maths exams, ill be glad to have a nice easy one to end it.


Do you mean 120ums?

Posted from TSR Mobile
Reply 433
Original post by amish123
Type 1a supernovae can be used as standard candles as they reach the same peak value of M (absolute magnitude) of around -19.3. The -19.3 is the value for M taken after around 20 days after the start of increase in brightness. From these measurements and using m-M=5log(d/10), we can measure the distance to these galaxies. However, by using red shift measurements and Hubble's Law (which assumes a constant rate of expansion), these supernovae are dimmer than predicted. This must mean that the expansion of the Universe is accelerating and some form of 'dark energy' is driving this expansion.


But how do we determine the apparent magnitude? As in, how do we measure it?
Reply 434
I think they measure the brightness of the star - i.e. the watts per square m of light energy from the star that reaches the Earth
and then they convert it to the magnitude scale using m = -2.5logb where b is the brightness in watts per m^2 and m is the magnitude
Reply 435
Original post by bugsuper
I think they measure the brightness of the star - i.e. the watts per square m of light energy from the star that reaches the Earth
and then they convert it to the magnitude scale using m = -2.5logb where b is the brightness in watts per m^2 and m is the magnitude


Thank you :smile:
Reply 436
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Just got this question from a friend of mine and couldn't get the right answer. I keep getting 10^-12 which is definitely wrong. It should be 10^12. I used the idea of Power(S)*d(s)^2 = Power (Q)* d(Q)^2 and then rearranged. By the way S represents for the sun and Q for the quasar. Thank you
A little help here guys please :smile:

thermal help.png

The answer to part b) ii 808K .... well anyway for part iii ... they add 35 to it... it just seems a bit too simple. Could someone please explain why this is :smile: I mean the water was also 15 degrees celcius to start with, so I hope some you understand where my confusion lies. But in general that part of the question on a whole makes no sense to me :frown:

Thanks in avance
That's because you've worked out the change which is 808, so the copper is going to decrease from whatever temperature it was in the Bunsen burner to 35. So if you add the temperature difference you should get the temp in the Bunsen burner. Hope it helps not that good at explain things over the internet.
Reply 439
Original post by JoshL123
Untitled.png

Just got this question from a friend of mine and couldn't get the right answer. I keep getting 10^-12 which is definitely wrong. It should be 10^12. I used the idea of Power(S)*d(s)^2 = Power (Q)* d(Q)^2 and then rearranged. By the way S represents for the sun and Q for the quasar. Thank you


Inverse sqaure law ----> P = k/(x2) where P is the power output and x is the distance.

Let Ps be the power of the Sun and x1 be the distance to the Sun. Pq is the power of the quasar and x2 the distance to the quasar.

Then, Ps/Pq = x22/x12.

Plug in the values and you'll find your power ratio to be 1012 :smile:.
(edited 10 years ago)

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