The Physics PHYA2 thread! 5th June 2013

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so how do you know how many sig figs answer you should give? cos it keeps changing in the markscheme

Rule of thumb is always 2 s.f. BUT, be very careful, if it asks to an appropriate number of s.f, you wanna look at the s.f of the numbers they've supplied in the question, and then adjust the s.f of your answer accordingly.

Reply 941

StalkeR47

Original post by masryboy94

so for the white light being diffracted, you don't get no dark fringes? just a continuous spectrum, right?

You do get dark fringes.

Reply 942

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

(edited 10 years ago)

ImageUploadedByStudent Room1370367512.162202.jpg141.6KB

Reply 943

masryboy94

Original post by SAS18

so how do you know how many sig figs answer you should give? cos it keeps changing in the markscheme

if they don't state to how many s.f. you should always have it to 2. if they say state to an appropriate s.f. then look at your question, if say they had blah blah blah 180N blah blah blah 46.5 degrees. you would know you need to have your answer to 3 s.f. (show full answer though in the step before)

Reply 944

StalkeR47

Original post by x-Sophie-x

For the optics topic, in the core of an optical fibre, if total internal reflection/refraction doesn't happen, then does the reflected ray always reflect at the same angle as the angle of incidence?

For example, question 4, June 2012.

Angle of incidence=Angle of refraction=85 degrees. Is this always the case?

Angle of incident is always the same as angle of reflection. Always. :smile:

Reply 945

masryboy94

Original post by StalkeR47

You do get dark fringes.

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. ?

Reply 946

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 947

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:

Reply 948

FLLF

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

Using w=fs, you've found out f, but need to multiply this by s (the distance), i.e. 1000m

Reply 949

x-Sophie-x

Original post by StalkeR47

Angle of incident is always the same as angle of reflection. Always. :smile:

Ah right, thanks :smile:

So how would you know when total internal reflection happens?
I know that TIR happens when the angle of incidence equals (?) the critical angle, but what if you're not given this?

Reply 950

StalkeR47

Original post by masryboy94

yeppp

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

Reply 951

Solivagant

Original post by x-Sophie-x

Ah right, thanks :smile:

So how would you know when total internal reflection happens?
I know that TIR happens when the angle of incidence equals (?) the critical angle, but what if you're not given this?

When I get questions on this, I never know why it TIR. I need some clarity as well.

Reply 952

StalkeR47

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 are given 1km distance which is a horizontal distance. so, Work done = force times distance in the direction of the force. By finding the horizontal force, you can use the equation. Is that ok? :smile:

Reply 953

x-Sophie-x

Original post by StalkeR47

You do get dark fringes.

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

Reply 954

kingmango

Anybody have a link to the Jan 2013 paper? Thanks

Reply 955

Jimmy20002012

Original post by StalkeR47

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

Posted from TSR Mobile

Reply 956

StalkeR47

Original post by x-Sophie-x

Ah right, thanks :smile:

So how would you know when total internal reflection happens?
I know that TIR happens when the angle of incidence equals (?) the critical angle, but what if you're not given this?

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:

Reply 957

masryboy94

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

P = Fvcos\theta

we already found out

Fcos\theta

in the previous question to be

(170\times10^3)(cos40) = 1.3\times10^5

so sub that value in with the velocity to give

P = (1.3\times10^5)(7) = 9.1\times10^5

Reply 958

SAS18

Original post by Raimonduo

Original post by masryboy94

Thanks so much guys!

Reply 959

Micheal123456

Pleaseeee can some one help me on the following question from the June 2010 paper , 3cii,5b,6ai,6aii, & 7d I really appreciate it thanks ( I'm going to fail this :nn: