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So the question goes as follows "A reflecting surface is represented by the equation x² + y² =a². A ray travelling in negative x-direction is directed towards positive y-direction after reflection from the surface at point P. Then co-ordinates of point P are" enter image description here

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These options were apparently ambiguous because a can be positive or negative, and depending upon that, the answer could be option B or D.

My doubt is a bit towards mathematics, but i am asking it here, because i don't know if there could be a difference in the interpretation of this question in physics. Do we not take a as positive only because it represent the radius of the circle here?

Ham Lemon
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2 Answers2

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No, it is ambiguous because a was merely described as being part of an equation.

Had the question said "radius a", that would be what constrains a to what "radius" means -- a radius is always positive.

Alex K
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  • So in x² + y² = r², r does not mean the radius of the circle? – Ham Lemon Jan 06 '23 at 14:00
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    If I was given just that equation, and someone asked me for the radius, I would say |r| not r. – Alex K Jan 06 '23 at 14:04
  • No actually, if you were given that "it is a curve given by the equation x²+y²=r²" then would you take r as the radius or |r|? – Ham Lemon Jan 06 '23 at 14:54
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    |r|. Consider this example. Let's say you know that in some rotating system, the distance from the center is inversely proportional to the angular velocity. Then you might write x^2 + y^2 = 1/w^2. If someone asks what the radius of the circular path taken by something with a given angular velocity is, you'd need to say 1/|w| because it might be negative – Alex K Jan 06 '23 at 23:20
  • Ok, i understood what you are saying, thanks. – Ham Lemon Jan 07 '23 at 03:59
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Do we not take a as positive only because it represent the radius of the circle here?

$a$ IS positive only, cause it's a radius. You simply mis-understood something.

These options were apparently ambiguous

No, they are not. Reflection point $P$ on the surface is described by a pair of coordinates $x,y$ such that $x \land y \in [-a, +a]$. I.e. $x,y$ are projections of circle radius from a point $P$ into $x$ or $y$ axis. Hence it's perfectly valid to be them positive or negative. And it's not ambiguous, because $P_1=\left(+\frac {a}{\sqrt 2},+\frac {a}{\sqrt 2}\right)$ and $P_2=\left(-\frac {a}{\sqrt 2},-\frac {a}{\sqrt 2}\right)$ are totally different points on the surface of sphere (here one of points is a mirror reflection of the other point). All answers are valid.

Agnius Vasiliauskas
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  • I am sorry if I misunderstood you, but are you saying that the answer to this question is B and D? But you said a is positive? If a is positive and the ray is travelling in negative x direction, then if it intersects at (-a/√2, -a/√2), then how would it be reflected back in the positive y direction? – Ham Lemon Jan 06 '23 at 14:52
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    You misunderstood the question, i.e. you need to find just ONE point from the given list which covers problem conditions. I will not say to you which point is an answer, you'll find it yourself. What I wanted to say that you are messing up, radius $a$ (which is always positive) with a projection of radius $a$ onto $x$ or $y$ axis, which can be $\pm a$ (min,max). So to speak, point on a surface is a vector and vector projection on $x$ or $y$ can be positive or negative. Hence these two points $P_1, P_2$ are on a different parts of sphere. Hope that helps. – Agnius Vasiliauskas Jan 06 '23 at 15:08
  • Ok i think i got your point, just to confirm 2 things, 1. is the answer B? 2. If this was a multiple correct question (the question language allows more than 1 option to be correct), then too the the answer would remain only B, right? – Ham Lemon Jan 06 '23 at 15:11
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    Can't you see that if B would be the case, then ray should pierce through the sphere ? Thus B is invalid. Yes, a pair of answers here are right. I have drawn the answer rays in red, map them to coordinates given in list. – Agnius Vasiliauskas Jan 06 '23 at 15:31
  • Oh yeah sorry I mistyped my answer I meant D only, and yeah that was my concern regarding option B, thank you so much for clearing my doubt! – Ham Lemon Jan 06 '23 at 15:38
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    Yes, D is an answer and A is partly valid too,- depends on exact interpretation of "ray travelling in negative x-direction", does it mean that ray is strictly co-linear to $x$ or can be slightly out of phase. – Agnius Vasiliauskas Jan 06 '23 at 15:51
  • For the question, it means strictly collinear to x, very exact, even along positive y means strictly along positive y, no approximations (as far as this question goes). – Ham Lemon Jan 06 '23 at 15:53
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    Then yes, it's only D. – Agnius Vasiliauskas Jan 06 '23 at 15:55
  • Thank you so much, i have been debating about this question with my friends, and they just don't seem to understand. – Ham Lemon Jan 06 '23 at 15:57
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    You are welcome :-) – Agnius Vasiliauskas Jan 06 '23 at 15:59