# Paths of red and blue light at boundary - Question from AQA Alevel past paper

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#1
Hi,

I am struggling to get my head around the answer to a question on a past paper I have just completed. I am sure the explanation is simple and that I am being stupid, but I cant seem to understand what's going on.

The question is 03.3 on this paper https://filestore.aqa.org.uk/sample-...1-QP-JUN17.PDF

"The source of green light is changed for one that contains only red and blue light.
For any material red light has a lower refractive index than green light, and blue light has a higher refractive index than green light. The angle of incidence at the glass–liquid interface remains at 58°. Describe and explain the paths followed by the red and blue rays immediately after the light is incident on the glass–liquid interface. "

The mark scheme says that the blue light undergoes TIR as it has a lower critical angle, and that red light is refracted. How do you work out that the critical angle is more or less in this case though?

The mark scheme also says that you can't predict the paths as you don't know if the refractive indices scale by the same factor and so can't know the effect on critical angles. How can both answers be true?

Can someone explain this to me? I've got myself in a fiddle here and I'm sure it's incredibly obvious.

Any help is appreciated, thanks!
0
1 year ago
#2
This is not the answer you're looking for, just to note something.

I remember answering this a while ago in class, and the answer that was apparently right according to my teacher (but evidently incorrect according to the mark scheme given, so, apparently not right) was that it was a safe assumption that the refractive indices scaled roughly in accordance with wavelength, i.e. for this instance:

Red light has a lower index. For the optically rarer medium, the index will decrease less than that of the denser medium, and thus the critical angle will increase. The blue light has a larger index, and thus the critical angle will (by similar argument) decrease. That gives the answer.

That's apparently not true, though, according to the markscheme. That makes sense (the markscheme that is) now since the dispersion relations aren't going to be that simple and by that argument, if you were to decrease wavelength by some fact of a thousand, you wouldn't get refraction that curved into some infinite mobius strip or anything... so there's probably some deeper meaning in the syllabus for this kind of thing.

There's probably some random equation out there mocking us right now.

Anyway, I've looked through (literally) five textbooks: two from the 1980's, two of the course books I used last year, and a book that combines electrodynamics and wave optics, and there isn't any question like it. There is however a section that does say that there is nonlinear scaling of the critical angle based on media, which is a plus. But that just makes it random as to the scaling and throws the equation out the window when it comes to guessing.
1
1 year ago
#3
The mark scheme gives three separate answers that you may give, all of which apparently get 3 marks. You are correct to be befuddled.

Last edited by Sinnoh; 1 year ago
1
1 year ago
#4
Blue light has a higher refractive index, meaning you can use SinC = 1/n where C is the critical angle and n is the refractive index to calculate what exactly will happen to the critical angle. If you raise the refractive index, the critical angle decreases, and both conditions for TIR are reached. The blue light can TIR because the light is travelling from a higher refractive index material to a lower refractive index material (glass to water) and since the critical angle has been decreased, it is much easier for it to TIR.

This can be explained for the red light too (if refractive index decreases, critical angle increases, thus it's harder to meet the conditions for TIR), but I think in your answer you were supposed to make a lot of assumptions.

I know the markscheme has many different answers, but all seem correct. It's a bit unfair because it looks like a suggest question to me.
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