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Rays and Geometry

https://isaacphysics.org/questions/maths_ch7_8_q1?stage=a_level#accchi_m_of_n\

I thought that part a of this question should be relatively straightforward but for some reason I've been having trouble comprehending how you get from C to the final angle.

Any help greatly appreciated.
Reply 1
Working (and calling alpha and beta as a and B for short):
From A to C, with clockwise as negative and anticlockwise as positive, I got -(a-B)+2B+pi = 3B-a+pi, and then from here im just stuck. I've tried working out the angle using co-interior angles etc. and trying some way but it just doesnt seem to work.
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(edited 2 months ago)
Reply 2
Original post by mosaurlodon
Working (and calling alpha and beta as a and B for short):
From A to C, with clockwise as negative and anticlockwise as positive, I got -(a-B)+2B+pi = 3B-a+pi, and then from here im just stuck. I've tried working out the angle using co-interior angles etc. and trying some way but it just doesnt seem to work.
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Try thinking "outside the circle" so extend the lines on the right and chase the angle where they intersect. In hindsight, its fairly clear that (two identical angles) gives what youre after.
(edited 2 months ago)
Reply 3
I cant lie im still a little stuck.
I changed the focus to finding 'y' since if I can find that I can find chi.
I tried chasing the angle by using alternate angles (and other methods which didnt work) and am trying to figure out the key.
I know vaguely that the key is to do with the identical angles of beta - (atleast I think thats what you meant I could be totally wrong) but im probably overcomplicating the solution.
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Reply 4
Original post by mosaurlodon
I cant lie im still a little stuck.
I changed the focus to finding 'y' since if I can find that I can find chi.
I tried chasing the angle by using alternate angles (and other methods which didnt work) and am trying to figure out the key.
I know vaguely that the key is to do with the identical angles of beta - (atleast I think thats what you meant I could be totally wrong) but im probably overcomplicating the solution.
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I meant extending (right direction) the lines passing through A (horizontal), B and C until they meet. You should have a couple of congruent triangles and it should be easy to get the angle(s) and see how it relates to chi.

Alternatively extend BO past the centre (slightly) and chase the angle it makes with the horizontal, then due to symmetry chi is ... Or extending your original diagram slightly, the missing angle at C is alpha-beta (symmetry) which gives ...

The hint also suggests considering the rotation at each vertex which is another way to do it.
(edited 2 months ago)
Reply 5
Oh I get what you mean! so y is just a-B
so 3B-a+pi-(a+B)=pi+chi
so 4B-2a = chi - thank you!

On a more sour note though, im now having trouble on part B.
I set up 1) sin(B)=3/4 sin(a)
and 2) 4B-2a = chi
but tbh I dont really know what it means by set up dB/da "at the maximal χ"
does this mean that I have to work out dχ/da 4arcsin(3/4 sin(a)) -2a =0
and then sub this value of alpha into 1)?
I tried simply working out the derivative of 1) but that didnt seem to work either
(edited 2 months ago)
Reply 6
Original post by mosaurlodon
Oh I get what you mean! so y is just a-B
so 3B-a+pi-(a+B)=pi+chi
so 4B-2a = chi - thank you!
On a more sour note though, im now having trouble on part B.
I set up 1) sin(B)=3/4 sin(a)
and 2) 4B-2a = chi
but tbh I dont really know what it means by set up dB/da "at the maximal χ"
does this mean that I have to work out dχ/da 4arcsin(3/4 sin(a)) -2a =0
and then sub this value of alpha into 1)?
I tried simply working out the derivative of 1) but that didnt seem to work either

The question part is a bit obfuscated, but you just need the result from part a) and differentiate wrt alpha. Part c) brings in sin/snells.

Note if you have the time, it could be a good exercise to do part a) in the different ways. The recommended summing local deflections to get the global isnt really done explicitly, but its a standard thing in several domains. Similarly, the other ones have some interesting bits.
(edited 2 months ago)
Reply 7
So X = 4B-2a
dX/da = 4*dB/da -2?
dX/da = 0 if X is maximum so dB/da = 1/2? Thank you again.

Sure! I'd love to try the other methods.
So I already (kind of?) did the rotation method, so I'll try the "extend BO" method.
I got that angle as 2B-a, and since X is also measured from the horizontal, and that the light is "halfway" through its route so still has another half of the amount of deflections left to go due to the symmetry, the total deflection must be 2(2B-a) - my justification could be wrong for this one but I believe this is the argument.
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For the alpha-beta symmetry method, since alpha is the angle between the light and the normal from air and beta is the angle while entering the droplet, and if beta is the angle while exiting the droplet, that must mean that alpha is the angle between the light and the normal in the air again, thus y = a-B, I definitely worded that a bit weirdly but I hope the basic essence of the argument comes across.

I used the congruent triangles method to get y, and then used rotation/deflection angles for each vertex to get chi, but if the congruent triangles method gets chi by itself, then I'll quickly try that as well.
(edited 2 months ago)
Reply 8
I managed to do the rest of the questions, they werent too bad actually, thanks for all your help :smile:

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