The Ultimate AS Physics Thread

[Zenzenzen]

So as the CIE AS Physics theory papers are coming up in a few days, I thought it would be a good idea to make an organized thread which could be easy to navigate through for last minute revision.

Basically, if a normal person was just browsing this thread, he would have a lot to gain just by simply reading over the thread's questions and answers.

Rules:
To avoid making this a place where a bunch of people ask questions and get no answers: You must answer the question asked by the user above you, and THEN ask your question! This way you are helping everyone and also getting your question answered. Not to mention you are also getting a good review of your own concepts!


An Example Post:

Question of the previous user quoted

A: Your answer to the previous question (Try to be as detailed and easy to understand as possible!)


Q: Your own question


So to start things off:
Q: How do you find the phase difference between two waves?

 

[Zenzenzen]

Could someone atleast answer my question...?
How do you find the phase difference between two waves?

 

[Amna]

Could someone atleast answer my question...?
How do you find the phase difference between two waves?

i think you mean between two points in a wave...
if the points are both crests, or both troughs, i.e. the distance between them is equal to one wavelength, they are said to be "in phase". the phase difference is 2pi.
if the distance is half a wavelength, the phase difference is pi. i.e. the phase difference between a crest and trough of the same wave.
so in general, we get phase difference = 2(pi)(d)/(lambda) where d is the distance between the two points and lambda is the wavelength.
you can also remember it this way, i'm not sure if i can explain it well, though:
a wave represents the oscillation of a point, right? so if it oscillates and then comes back to its original position, it has traveled a whole wavelength. an since it's back where it was, the angle is 360 degrees. but we call the angle "phase difference" and give it in radians. does that make sense?

and my question is attached...

 

[Zenzenzen]

Well we know that the electric field goes from positive to negative.
Therefore, the force on P would be towards the right, and the force on N would be towards the left.

Now we must calculate torque. We know that torque = one force x perpendicular distance between forces

Force:
Charge is given to us (1.6 x 10^-19)
E = F/Q
F = EQ
F = (5.0 x 10^4)(1.6 x 10^-19)
F = 8 x 10^-15 N

Perpendicular distance:
using trigonometry:
D = (2.8 x 10^-10) x (sin 30)
D = 1.4 x 10^-10 m

Torque:
Torque = F x D
T = (8 x 10^-15) x (1.4 x 10^-10)
T = 1.12 x 10^-24 Nm^-1

Hope you understood that haha


My question: attached, it referes to a double-slit interference experiment

 

[Jaguar47]

Well, we know that x= λD/a, where x is the distance between the fringes (separation), D is the distance from the slits to the screen the fringes are projected on, a is the distance between the slits and λ is the wavelength.

1) λD/a = x so λ2D/a = 2x (double the separation)
The intensity is halved, brightness halved

2) again λD/a = x so 1.5λD/a = 1.5x (one and a half times the separation)
v=fλ so f= v/λ . λ is multiplied by 1.5, so f is decreased. Use Intensity= A^2 times f^2. Intensity is decrease, brightness is decreased

3) Nothing changes in the separation.
Intensity doubled, so brightness doubled

 

[Jaguar47]

Oh, I forgot to post my question

Question 4b in May/June 2010 Paper 21
I have no Idea how to attach the question, so here's the link to the paper: http://www.xtremepapers.me/CIE/Internat ... _qp_21.pdf

 

[AlyHamza]

@Zenzenzen could you just plz tell the year and paper of the question u have asked (the one involving separation and max. brightness)

 

[Hateexams93]

its may june 2003 ,the previous question about the brightness, lol now can some1 plz answer to my Q ?? http://www.xtremepapers.me/CIE/Internat ... 3_qp_2.pdf Question 3 (b) i

 

[Zenzenzen]

@Zenzenzen could you just plz tell the year and paper of the question u have asked (the one involving separation and max. brightness)

Its Question 4 in this http://www.xtremepapers.me/CIE/International A And AS Level/9702 - Physics/9702_s03_qp_2.pdf main problem was the second part, I didn't get how intensity increases the brightness.

Oh, I forgot to post my question

Question 4b in May/June 2010 Paper 21
I have no Idea how to attach the question, so here's the link to the paper: http://www.xtremepapers.me/CIE/Internat ... _qp_21.pdf

Well we know the formula for diffraction grating: dsinx=nλ
Basically you have to solve for d.

First you find the angle x. Consider the triangle made between X and P. The right side is (76 / 2) = 38cm.
The base is given (165), so:
tanx = 38/165
x = 12.96 = 13 degrees

Now put the value into the equation (wavelength is given as well)
d x sin(13) = (1)(632 x 10^-9)
d = 2.816 x 10^-6

Now this is the value of the diffraction grating. To find the number of lines per metre, you simply have to inverse d.
number of lines per metre = 1/d = 3.6 x 10^5


--

My question : #5 in http://www.xtremepapers.me/CIE/International A And AS Level/9702 - Physics/9702_w09_qp_22.pdf?

 

[Zenzenzen]

its may june 2003 ,the previous question about the brightness, lol now can some1 plz answer to my Q ?? http://www.xtremepapers.me/CIE/Internat ... 3_qp_2.pdf Question 3 (b) i

Well the diameter of the circle is given (3cm)
Circumference of circle = 2(pi)r = pi x diameter = 3pi

Since the pointer move 6.5 degrees:
distance = 3pi x (6.5/360)
distance = .17 cm

 

[hassam]

:Search: :Search: :Search:

 

[hassam]

SOMEBODY TELL ME PART 4 C

 

[AlyHamza]

@Jaguar47
Answer to Question 4(b) May/June 2010 Paper 21:

Considering the half triangle: Angle= tan^-1 (38/165)= 12.97 degrees

Now use "d x sin(angle)=n x wavelength" [d; slit separation , n: no. of order of spectrum (1st here)]
d * sin 12.97 = (1 x 632 x 10^-9)
d = 2.816 x 10^-6

N = (1/d) = 1/(2.816 x 10^-6) = 3.55 x 10^5 = 3.6 x 10^5 (2 sig. figures) [N: no of slits per unit length]



Hope u get this one. I'll be happy to explain if u have any queries about this.

 

[AlyHamza]

@Zenzenzen


Intensity is directly proportional to (amplitude)^2

so when intensity increases so does amplitude.
using principal of superposition : resultant amplitude has increased.
so obviously a light wave of more amplitude will be brighter

OR you can take it as: "Intensity is the amount of energy passing through per unit area per unit time". So increased intensity means more energy. The energy here is light energy. More light energy means more brightness.

 

[AlyHamza]

The (b) part please..............

 

[Hateexams93]

but why the brightness decreases when the wavelenght increases ?

 

[Jaguar47]

@Zenzenzen:

a) (i) f, the frequency is the same since its the same wave
(ii) A, the amplitude is the same since they're both the same distance away from an antinode in a stationary wave
b) I would have thought it would be 6/4 pi radians but the mark scheme says pi :/
c) (i) v = fλ where f = f and λ = L, so the answer is f times L.
(ii) I'm beat, my answer's got nothing to do with the mark scheme's

Same question again, 5 parts (b) and c(ii)
http://www.xtremepapers.me/CIE/Internat ... _qp_22.pdf

 

[Zenzenzen]

The (b) part please..............

@AlyHamza

I remember doing this problem before, I think it goes like this:
We know Voltage = Work/Charge (so voltage only depends on work and charge)

Then, we know that voltage is the same. So the work done by the electron is voltage x charge.
There is no mention of the uniformity of speed, so I guess the acceleration of the electron is independent of the uniformity of the field.

However, I think someone else should explain it as I don't really get it beyond this point. :/

I think the mark scheme answer was something along the lines of "only voltage affects a charged particles' speed"


@Jaguar47
This question was pretty weird in all honesty >_>

 

[yubakkk]

may 2003 q.4 i 1
and
5 b ii.
any one help???

 

[Zenzenzen]

may 2003 q.4 i 1
and
5 b ii.
any one help???

Question 4 part i 1: 1mm. And yeah, you have to memorize this

Question 5 b part ii:
At beginning of question it says the lamp operates normally at 6 volts. Looking at the graph you'll see the current for 6 volts is 40mA.

So basically, in the circuit with the resistor, you need to have a current of 40mA.

Resistance of the lamp:
V = IR
6 = (.040)R
R = 150

Resistance of resistor: 200

Total resistance: 150 + 200 = 350

So the total resistance of the circuit is 350, and the current has to be 40mA (.04 A)
E = V
V = IR
V = (350)(.04)
V = 14V

Hope you understood!