[PhoenixTear]

(Original post by Sagacious)
What's everyone's favourite part of this unit? Waves? Electricity? Light?

Light, I think the photoelectric stuff is all really interesting

[Sagacious]

(Original post by youlostme)
looks like i wont get any sleep tonight, im trying to finish waves(the whole topic) today

why do you guys solve pastpapers now? leave it for the day before the exam, by then you would be completely done with studying and can test yourself...no?

I have written up all my notes; and I've completed all the past-papers. I just keep going over the notes.

[KamilS]

Does anyone know the difference between the equations hf=f and hf=1/2mv^2 + f where f is the work function?

[avipsita]

(Original post by Sagacious)
What's everyone's favourite part of this unit? Waves? Electricity? Light?

i love electricy... thats my favourite

[Sagacious]

(Original post by KamilS)
Does anyone know the difference between the equations hf=f and hf=1/2mv^2 + f where f is the work function?

hf=workfunction is just simply saying that there will be enough energy to extract the electron from the metal, but it wont move after due to Ek = 0.

Whereas hf=1/2 mv^2 + work function is the photoelectric equation. It allows you to determine the velocity of the photoelectron etc.

(Original post by avipsita)
i love electricy... thats my favourite

Oh my, someone who shares my passion for electricity! *high five*. I was thinking everybody hated it for a moment..

[youlostme]

(Original post by KamilS)
Does anyone know the difference between the equations hf=f and hf=1/2mv^2 + f where f is the work function?

in the first equation, the frequency is equal to the threeshold frequency therefore the energy given to the electrons is equal to the workfunction(so in this case, photoelectrons released wont have any kinetic energy after being emitted, and kinetic energy is 0)
the second equation denotes the frequency being larger than the threeshold frequency, therefore larger amounts of energy are given to the electrons so when photoelectrons are emitted they still have enough energy to keep moving(kinetic energy > 0)

[youlostme]

(Original post by Sagacious)
Oh my, someone who shares my passion for electricity! *high five*. I was thinking everybody hated it for a moment..

on the contrary! all my classmates love electricity, im the only one who doesnt haha

[avipsita]

(Original post by Sagacious)
hf=workfunction is just simply saying that there will be enough energy to extract the electron from the metal, but it wont move after due to Ek = 0.

Whereas hf=1/2 mv^2 + work function is the photoelectric equation. It allows you to determine the velocity of the photoelectron etc.



Oh my, someone who shares my passion for electricity! *high five*. I was thinking everybody hated it for a moment..

lol .. nope i wanna be an electrical engineer .. i obvious have to love electricity

[KamilS]

(Original post by youlostme)
in the first equation, the frequency is equal to the threeshold frequency therefore the energy given to the electrons is equal to the workfunction(so in this case, photoelectrons released wont have any kinetic energy after being emitted, and kinetic energy is 0)
the second equation denotes the frequency being larger than the threeshold frequency, therefore larger amounts of energy are given to the electrons so when photoelectrons are emitted they still have enough energy to keep moving(kinetic energy > 0)

Alright, I get it now. So when photoelectrons are emitted at threshold frequency, what happens to them since they don't have the energy to move any further?

[youlostme]

they dont move at all, they remain sudpended just above the metal sheet

[KamilS]

(Original post by youlostme)
they dont move at all, they remain sudpended just above the metal sheet

Ok thank you! That helped a lot

[Sagacious]

(Original post by youlostme)
on the contrary! all my classmates love electricity, im the only one who doesnt haha

all my class are light-lovers. Oh, and then there's one wave-fanatic.

(Original post by avipsita)
lol .. nope i wanna be an electrical engineer .. i obvious have to love electricity

Haha yeah, it helps :-)

[Liver2029]

Are polarizing filters always set to absorb light polarized on the horizontal plane?

[Princess Aysha]

(Original post by youlostme)
looks like i wont get any sleep tonight, im trying to finish waves(the whole topic) today

why do you guys solve pastpapers now? leave it for the day before the exam, by then you would be completely done with studying and can test yourself...no?

Hi! I attend Tauheedul Islam Girls High school. I will sit the GCSE Physics exam this Thursday. For revision the school has provided us with revision cards, mind maps, extended questions and answers and obviously past papers and mark schemes. As well as 1.5 hour revision sessions 2 days before the exam which will be very useful and organised, colourful notes.

I've revised key topics and will also leave past papers for last 2 days.

[Princess Aysha]

(Original post by Liver2029)
Are polarizing filters always set to absorb light polarized on the horizontal plane?

I'm freaking out! We haven't learnt any of this stuff!

[youlostme]

(Original post by Liver2029)
Are polarizing filters always set to absorb light polarized on the horizontal plane?

errm...polarizing filters dont work that way! a polarizer would eliminate ALL waves that dont oscillate in the same plane as the plane of polarization(i.e. if a polarizer is set vertically then ONLY light waves that propagate along the vertical plane would pass, all other waves are blocked)

[youlostme]

(Original post by Princess Aysha)
Hi! I attend Tauheedul Islam Girls High school. I will sit the GCSE Physics exam this Thursday. For revision the school has provided us with revision cards, mind maps, extended questions and answers and obviously past papers and mark schemes. As well as 1.5 hour revision sessions 2 days before the exam which will be very useful and organised, colourful notes.

I've revised key topics and will also leave past papers for last 2 days.

lucky you! my sorry excuse of a school shoved pastpapers and a few worksheets in our faces and thats it...its only because i absolutely love physics and dont mind spending hours doing my own research that ive been doing well until now

[Liver2029]

(Original post by youlostme)
errm...polarizing filters dont work that way! a polarizer would eliminate ALL waves that dont oscillate in the same plane as the plane of polarization(i.e. if a polarizer is set vertically then ONLY light waves that propagate along the vertical plane would pass, all other waves are blocked)

I thought if the polarizing filter such as sunglasses were set to a vertical plane, they would absorb polarizing light oscillating on the vertical plane, which would reduce the light intensity?

[youlostme]

(Original post by Liver2029)
I thought if the polarizing filter such as sunglasses were set to a vertical plane, they would absorb polarizing light oscillating on the vertical plane, which would reduce the light intensity?

nope its the other way round, if sunglasses were set to a vertical plane then only light oscillating along the vertical plane would pass...which is why the intensity decreases.
think of it logically: light; being a transverse wave, propagates in ALL directions, if only ONE out of many planes is eliminated it wouldnt have a very significant effect on the intensity right?

[Liver2029]

(Original post by youlostme)
nope its the other way round, if sunglasses were set to a vertical plane then only light oscillating along the vertical plane would pass...which is why the intensity decreases.
think of it logically: light; being a transverse wave, propagates in ALL directions, if only ONE out of many planes is eliminated it wouldnt have a very significant effect on the intensity right?

Yes, but we're talking about polarized light. When light is reflected off a surface it is typically plane polarized to a horizontal plane. The polarizing filter works by absorbing polarized light on the horizontal plane and allowing the passage of polarized light on the vertical plane. Which reduces the light intensity.