Competition - post a photo you've taken of nature >>

PHYSICS AS - Help!!

https://5505b400f275fd9f29b9b4679ca512236789cf2e.googledrive.com/host/0B1ZiqBksUHNYZEhDNnU3TUxGaG8/June%202010%20QP%20-%20Unit%202%20AQA%20Physics.pdf

Please help me with Question 7 part C... I cannot understand how to calculate that without knowing the refractive index of the liquid. :/ Is there something I'm missing?

Reply 1

Spelly456

Original post by Someboady

Read the last line of the question :smile:

Reply 2

Someboady

At this point the angle of refraction is 27º and the ray is totallyinternally reflected at P for the first time.

"for the first time".... does that mean that the critical occurs when the angle of refraction is 27 degrees :/ ?

Original post by Spelly456

Read the last line of the question :smile:

Reply 3

samb1234

Original post by Someboady

No, you need to use a triangle to find the critical angle

Reply 4

Spelly456

Original post by Someboady

Think about which angle you will be measuring as the critical angle when you consider the water-glass boundary. You're almost right!

Reply 5

hellomynameisr

So what is the calculation required for the answer? its only a 1 marker? lol

Reply 6

Spelly456

Original post by hellomynameisr

So what is the calculation required for the answer? its only a 1 marker? lol

90-27 = 63 degrees
That is the answer
Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle

Reply 7

hellomynameisr

Original post by Spelly456

90-27 = 63 degrees
That is the answer
Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle

Oh right, I see. Thanks

Reply 8

Someboady

Original post by Spelly456

90-27 = 63 degrees
That is the answer
Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle

still slightly confuzzled sorry.... So I've drawn a new diagram... part a says to continue the ray that's been totally internally reflected..So I drew the ray reflected off the bounday and coming out of the glass. I didn't do any calculations.

Am I correct in saying the ray of light, when striking the boundary between liquid and glass, strikes at the angle of 63 degrees... this can be figured out using 90-27 = 63.... But I don't understand how that is the critical angle. My understanding of the critical angle is that the angle of refraction must be 90 degrees.. Does this mean my diagram in part A is wrong and that the ray should continue at 90 degrees from the normal and continue travelling along the boundary :/

(edited 8 years ago)

Reply 9

Someboady

Original post by Spelly456

90-27 = 63 degrees
That is the answer
Remember the critical angle is the angle between the boundary and the normal. You can draw a right-angled triangle to show the required angle

I just read the examiners report :
For part (c), candidates needed to realise the incident angle had just passed the critical angleand therefore the critical angle would be 63° to two significant figures.

Why has it "just" passed the critical angle... why can't it have passed the critical angle by a large amount? what tells us that its only passed the critical angle "slightly"?

Reply 10

Spelly456

Original post by Someboady

In the question it says the ray is totally internally reflected for the first time so 63 degrees is just after the critical angle. You are right that at the critical angle the angle of refraction is 90 degrees so the critical angle would be slightly below 63 degrees.
Hope this makes sense.

Reply 11

Spelly456

Original post by Someboady

The reason you quote the answer as 63 degrees is due to rounding of 62.99999

Reply 12

Someboady

Original post by Spelly456

Yep it makes sense thanks! However when it comes to generalising... If a similar question was to come up and it stated that it was "totally internally reflected for the first time"... would this again mean that the critical angle is slightly below the angle of incidence (in ANY situation)?