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The Physics PHYA2 thread! 5th June 2013

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You don't get dark fringes with a continuous spectrum?!?!

Interference pattern produces with a white light has min max. But, each fringe is a continuous spectrum. You do not get dark with in the fringes do you? :smile:

If I am correct. HAHA

Reply 961

Linked

In an experiment, a narrow beam of white light from a filament lamp is directed at normal incidence at a diffraction grating. Complete the diagram in Figure 5 to show the light beams transmitted by the grating, showing the zero-order beam and the first-order beams.

can anyone help me here please!

Reply 962

Raimonduo

Original post by Jimmy20002012

Cannot believe I am stuck on the first question :frown:

. I don't get it, got the answer but got 10 to the power 6 instead of 10 to the power 5??

Posted from TSR Mobile

You got 1 mark, for working out the horizontal component of the tension. However, the question asks for the work done, so then, you need to use

w = f \times s

. This should give you the correct s.f as the distance is

1.0\times10^{3}m

Reply 963

masryboy94

Original post by StalkeR47

Oh I know what you mean. You do get max and min for the interference pattern. However, each fringe is a continuous spectrum. :smile:

so do you call it bright and dark fringes or a continuous spectrum?

Reply 964

mazzah

what do you think the 6 mark question will be on?

and what topics are likely to come up?

Reply 965

username984585

Original post by pushkin_

Does anybody have notes going through everything we need to know?
Yup, I'm in panic, haven't done any solid revision for physics, just past papers and I would be VERYYYYYYYYYYYYY thankful If somebody could share with their revision notes.

p.s. have anybody posted all unit 2 experiments in this thread ? pleaaase , somebody help, I feel so miserable in first place because of todays chemistry.

:banghead:

I feel pretty screwed for physics as well, had bio on Monday and Chem today. And I'm seeing that you had the pleasure of sitting the chem paper today as well?

I'm looking for a list with all the experiments as well, I think the ones we need to know are just: Hookes law, Youngs Modulus, Diffraction (single and double slit)

:smile:

Reply 966

fuzzybear

seems like common sense, but why would air resistance have a bigger effect on a lighter object?

I thought the acceleration 'g' was constant for everything

this is for question 6b on page 131, it basically asks why the measured time for lead to fall would be less than a tennis ball

Reply 967

Solivagant

Original post by StalkeR47

No probs! So, TIR happens if the angle of incident is bigger than the critical angle. Or, when the incident substance has a larger refractive index than the emergent substance. If you are not given the either, you will have to use one of the equations. sini(critical)=n2/n1 :smile:

Which one is the angle of incidence?

Reply 968

looloo4

Hi, would anybody be able to list the experiments we have to know, I'm having trouble knowing what to revise in terms of practicals.
Thanks!

Reply 969

RU486

Original post by posthumus

Time period is time taken for 1 oscillation... you want it for quarter of an oscillation.

T= 1/frequency

then divide T by 4

Why do we need a quarter of an oscillation? If it moves from maximum displacement back up to zero displacement isn't it half a cycle?

Reply 970

StalkeR47

Original post by Jimmy20002012

Ohhhh? Thanks, what would I do without you :smile:

Posted from TSR Mobile

Thanks for the compliment m8. :smile:

But, do not say that. You are doing really well! :smile:

I just knew it cause I did the question so I remembered really well. :smile:

Reply 971

FLLF

Should we write out the 4sf answer then give the answer to 2sf? Because this allows us to use the 4sf figure in later parts of the question.

Original post by masryboy94

so do you call it bright and dark fringes or a continuous spectrum?

Look at the question 3e from jan 2011. The mark scheme says each maxima (fringe)indicates different colours. So, each fringe (maxima) itself is a spectrum.

Reply 974

posthumus

Original post by masryboy94

but in that picture you posted, there was no dark fringe, it was the rainbow continuously you went from violet to red and straight back to violet and so on. ?

Your correct, you don't get any dark fringes.

Usually only with coherent light sources (monochromatic).

Original post by RU486

Why do we need a quarter of an oscillation? If it moves from maximum displacement back up to zero displacement isn't it half a cycle?

The point of zero displacement is at the equilibrium line... remember displacement is a vector (the reference is the equilibrium line - hence the name).

Reply 975

StalkeR47

Original post by CJG21

Which one is the angle of incidence?

Angle of incident is where the ray is incident into the more dense material and the angle to the normal is the angle of incident. Is that what you were asking? :smile:

I am a bit confused sorry. :smile:

Reply 976

FLLF

Original post by fuzzybear

I suppose that because it is more dense, the weight of the lead per unit surface area is greater than that of the tennis ball. Acceleration due to gravity is the same for every object (pretty much), but different objects fall at different rates depending on the resistive forces too.

Reply 977

RU486

Original post by posthumus

The point of zero displacement is at the equilibrium line... remember displacement is a vector (the reference is the equilibrium line - hence the name).

Ohh that makes much more sense, thanks a lot :smile:

Reply 978

StalkeR47

Original post by masryboy94

so do you call it bright and dark fringes or a continuous spectrum?

Like this m8.

Reply 979

Solivagant

Original post by StalkeR47

Angle of incident is where the ray is incident into the more dense material and the angle to the normal is the angle of incident. Is that what you were asking? :smile:

I am a bit confused sorry. :smile:

Erm... too many words I don't get there.

I still don't understand TIR when it's not in an optical fibre.

I'm going to fail this. I can't do anything.