The Physics PHYA2 thread! 5th June 2013

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Does anyone have a few steps that can help me when working a question says to show the path of the ray until it returns to air, the seem to always get me! Thanks!!

think of the light ray as a car, and when it refracts going into a denser medium, the side of the car that hits the boundary first slows down first, allowing the other side to catch up, so the light bends towards the normal and vice versa

Reply 841

frankieeeeeee

question 4 b i) on the june 2012 paper? thanks so much!!

Reply 842

StalkeR47

Original post by frankieeeeeee

question 4 b i) on the june 2012 paper? thanks so much!!

Q4ai right? Since the angle is 85 degrees to the left, it is also 85 degrees inside the ray that forms the triangle. (Note: The angle of reflection is always the same as angle of incident). So, to find i, you add 85 + 30 which gives 115 degrees. Since triangle is 180, 180 - 115 gives 65 degrees. Any more help? Ask me

Reply 843

Raimonduo

Original post by StalkeR47

Alright!

How confident are you feeling for the physics exam?

Mmmm, not too sure. I know the content pretty well but I always have the feeling they're gonna ask something which I'll struggle to do. Otherwise, I'm pretty confident I guess.

Reply 844

dark3city

when do you use f cos theta or f sin theta its confusing me some say horizontal or vertical others say line of action

Reply 845

frankieeeeeee

Original post by StalkeR47

thanks so much!
i wasn't sure you could use 85 as the angle inside the triangle!
good luck for tomorrow :smile:

Reply 846

StalkeR47

Original post by dark3city

when do you use f cos theta or f sin theta its confusing me some say horizontal or vertical others say line of action

On the inclined, well.... Someone have already drawn a nice diagram.

Reply 847

RU486

Is there an easy way to calculate work done on a Load/extension graph? There's a question for this on June 10 5b which I can't do.
Thanks!

Reply 848

StalkeR47

Original post by dark3city

when do you use f cos theta or f sin theta its confusing me some say horizontal or vertical others say line of action

Here!

Reply 849

Substitution

the ms says 1.4 my calc gives 0.6 -.-

Reply 850

SortYourLife

Original post by StalkeR47

Jan 13 last q? Common! It was the easiest question. But you are doing really well! Keep it up! :smile:

I know aha! Just didn't like it :')

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Reply 851

StalkeR47

Original post by frankieeeeeee

thanks so much!
i wasn't sure you could use 85 as the angle inside the triangle!
good luck for tomorrow :smile:

no probs! Good luck to you too for tomorrow! You will be fine! :smile:

Reply 852

BigBadJFly

Original post by RU486

Is there an easy way to calculate work done on a Load/extension graph? There's a question for this on June 10 5b which I can't do.
Thanks!

Area under the graph

Reply 853

StalkeR47

Original post by RU486

Is there an easy way to calculate work done on a Load/extension graph? There's a question for this on June 10 5b which I can't do.
Thanks!

The area is wd. Count the number of blocks and times the number of blocks by the area of each block. :smile:

Reply 854

Sukura

Can you please help me on a question?

Question 6di from June 2010!

I'd truly appreciate it

Reply 855

henpen

Original post by frankieeeeeee

Can anyone explain Q. 4. b) (i) from the June 2012 paper?
It's on fibre optics and I really don't understand, the fibre is bent at 30 degrees and it asks you to calculate the angle of incidence.

What?

Thank you.

The laser beam at first undergoes total internal reflection, so this angle

is also 85 degrees, thus i=180-85-30=65 degrees.

Let

x

be the cladding and

y

be the core, and applying snell's law:

\frac{n_x}{n_y}= \frac{ \sin( \theta_y)}{ \sin ( \theta _x)}

(edited 10 years ago)

Reply 856

iamsocoollol

hi guys can some one please upload the MS AND QP jan 2013 unit 2 pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

my stupid teacher forgot to email :frown:

Reply 857

Raimonduo

Original post by Sukura

Can you please help me on a question?

Question 6di from June 2010!

I'd truly appreciate it

6(d)(i) is simply 1/d, where d is the slits per mm. Although, be VERY careful with your s.f and standard form here. Mark scheme suggests that an incorrect s.f is no mark.