Question

In concave mirrors how to show that the reflected rays are parallel when the object is placed at focus ?
Please see part b of the attached pic

Answer 1

I thought focus was only defined for parabolae...

Answer 2

Or is it "I thought foci were only defined for parabolae"? "I thought focus was only defined for parabola"?

Answer 3

here focus is very much different...
it has nothing to do with the parabola or other conics

Answer 4

I believe the only concave surface that reflects rays in a parallel fashion is parabola. Other surfaces might reflect it really close to parallel though.

Answer 5

Probably some sort of small angle approximation, presumably by Taylor expanding some trigonometric function to first-order?

Answer 6

`We can assume the rays coming from a source large distance are parallel. When these parallel rays reach a concave mirror, those near the central axis are reflected through a common point F; two of these reflected rays are shown in the figure. If we placed a (small) card at F, a point image of the infinitely distant object O would appear on the card. (This would occur for any infinitely distant object.) Point F is called the focal point (or focus) of the mirror, and its distance from the center of the mirror c is the focal length f of the mirror.`

Answer 7

yeah, focus is not really a point. it is a small region

Answer 8

Probably something to do with small angle approximation and the Lensmaker's equation.

Answer 9

No it is the thin lens formula not the Lensmaker's equation.

Answer 10

What equations do you have?

Answer 11

You could use the Gaussian mirror equation which I copied straight from Wikipedia.
\[
\frac{1}{p}+\frac{1}{i}=\frac{1}{f}
\]
If \(p=f\) then \(\dfrac{1}{i}=0\) so \(i=\infty\) sloppily.

Answer 12

So far, the textbook has covered one law and one result :

1) angle of incidence = angle of reflection
2) f = r/2

Answer 13

Of course the equation is equipped with a sign convention that nobody remembers and it is totally horrible.

Answer 14

Oh that equation is next. For now, I am trying to get convinced using simple geometry, if psble.... .

Answer 15

I wish I still had my physics book, I never really did learn optics. My guess nowadays would be to parametrize the reflecting surface, use calculus to find the normal vector, and from that look at angle of incidence = angle of reflection to try to sort out where the point was that reflected a parallel set of rays... But easier said than done.

Answer 16

The proof in essence boils down to deriving the thin lens equation, which is highly non-trivial to do it rigorously.

https://physics.stackexchange.com/questions/83751/thin-lens-formula

Answer 17

I'm pretty sure there is some neat way like that... For some reason, it reminds me of deriving kepler laws using the differential eqn :
\[mr'' = \dfrac{GmM}{|r|^3}r\]

Answer 18

Yes there is the non-rigorous similar triangle method that looks like this.
http://hirophysics.com/Anime/thinlenseq.html

Answer 19

I shall side track this question. There is a saying that "You miss 100% of the shots you don't take.", purported said by Wayne Gretzky. Is it mathematically true that you also hit 100% of the shots you don't take because the statement is vacuous?

Answer 20

@thomas5267 I guess it's kind of like how \(f(x)=0\) is both an even and an odd function. In a sense, Wayne Gretzky's statement is like the "zero vector" of some vector space of shots you do take. :O

Answer 21

It is in some sense even worse than that because shots that you don't take is an empty set of shots? So you could say all shots you don't take are broccolis!

Answer 22

I think that's better though, I like broccoli... lol

Answer 23

Me too. Broccoli do not deserve the hate it has now. Celery however taste horrible and somehow it is more highly rated than broccoli...