[CD315]

(Original post by North Irelandman)
... Really want to see this question you're all talking about.

A wave was radiated towards 2 layers of ground (or rock or whatever), first layer A and 2nd layer B. It goes through layer A. We are told that it refracts, but it just reflects when the angle of incidence from A into B is 35 degrees, so therefore the critical angle is 35 degrees. Now, it gives you the equation that the speed of sound in A divided by the speed of sound in B or whatever it was, is equal to the refractive index from A to B. (aNb), and asks you to find speed in B. Usually you would use 1/sinC=n, however thats only if you're given the refractive index from air into the material. But this time, it told you that the refractive index was from A to B. Hence why I think the refractice index was just sin35 and not 1/sin35.

[Fladam]

Ok so speed in A over speed in B is the refractive index from A to B... Speed in A was 5000 metres each second


Let's say for the purpose of making this more common to us that A is a glass block and B is the air...
So when 35 degrees TIR occurs but let's say that it refracts at 90 degrees along the medium of A to B for the sake of letting 35 be the critical angle... So aNb= sin 35/ sin 90 so aNb = sin 35

Then sin 35 will be equal to 5000/x

5000/sin 35 = x

X = 8717 ms^-1


You're right CD if that was how the question was worded If it was a sound wave however I'd be questioning if CCEA were right on this one... But if it was just a wave then you're certainly right by taking sin35


This was posted from The Student Room's iPhone/iPad App

[BLUERUN]

It's one good question , my physics teacher said that in the department they where all debating what the answer was : D

[CD315]

(Original post by Fladam)
Ok so speed in A over speed in B is the refractive index from A to B... Speed in A was 5000 metres each second


Let's say for the purpose of making this more common to us that A is a glass block and B is the air...
So when 35 degrees TIR occurs but let's say that it refracts at 90 degrees along the medium of A to B for the sake of letting 35 be the critical angle... So aNb= sin 35/ sin 90 so aNb = sin 35

Then sin 35 will be equal to 5000/x

5000/sin 35 = x

X = 8717 ms^-1


You're right CD if that was how the question was worded If it was a sound wave however I'd be questioning if CCEA were right on this one... But if it was just a wave then you're certainly right by taking sin35


This was posted from The Student Room's iPhone/iPad App

Hate physics haha.

[North Irelandman]

Cheers, looks like CCEA had fun designing that question! Snell's Law was not taught well in my class last year so that would have been marks lost for a number of us. But yes, it would not have been 1/sinC, I would need to have a proper think but Fladam's solution makes sense to me. Definitely a good question to catch people out though, wouldn't worry too much about dropping a mark or two.

[Fladam]

Was a very nice question for thinking through for sure! Caught me out in the exam, tried to rush it :P Generally snell's law isn't too bad so I wasn't expecting any tricks, well spotted CD! Should be fine, I would say Maybe a little bit of scaling on that paper too?


This was posted from The Student Room's iPhone/iPad App

[CD315]

(Original post by Fladam)
Was a very nice question for thinking through for sure! Caught me out in the exam, tried to rush it :P Generally snell's law isn't too bad so I wasn't expecting any tricks, well spotted CD! Should be fine, I would say Maybe a little bit of scaling on that paper too?


This was posted from The Student Room's iPhone/iPad App

North Irelandman knows how much I love my scaling physics generally gets scaled up. I was hoping for at least 100/111 UMS there (90%) but I'm doubting it.

[calender1994]

does anione know the grade boundaries for this?

[CD315]

(Original post by calender1994)
does anione know the grade boundaries for this?

No idea, but I got 66/75 which got scaled up to full UMS..

[calender1994]

i got 106 so like 62 or there abouts then? and thanks for the reply