Proof of image formation property of spherical mirrors

In summary, the author is seeking a proof that a concave spherical mirror will form images of objects in front of it. He has already done some work on this and is now looking for help with simplification. He has tested the rays he has with math and they converge to a point, but he is missing something. He needs help simplifying the equation.
  • #1
Gabriel Maia
72
1
Hi. I'm trying to proof the image formation property of a concave spherical mirror. I know you can do this easily with a particular choice of rays (namely one that hits the vertex and one that passes through the center of the sphere) but I would like to show that a generic ray yields the same result because

1) Any two non-parallel rays will cross paths sonner or later so to use only two rays gives you a result but not a satisfactory proof.

2) The rays chosen for this easy proof pass through special points of the problem's geometry and could be a particular result.

I know there is a purely geometrical approach to this general proof I'm seeking but since I want to keep consistency with the work I've already done, I'm doing it with Analytical Geometry.

The Mirror is given by the equation

[itex]y=R-R\sqrt{1-\left(\frac{\displaystyle x}{\displaystyle R}\right)^2}[/itex],

where [itex]R[/itex] is the radius of the mirror and its axis is parallel to the y-axis. The object in front of the mirror has coordinates [itex](x_{1},y_{1})[/itex].

The ray reflected at the mirror's vertex has a line equation given by

[itex]y=-\frac{\displaystyle y_{1}}{\displaystyle x_{1}}\,x[/itex]

and the one passing through the mirror's center has a line equation

[itex]y=R+\frac{\displaystyle y_{1}-R}{\displaystyle x_{1}}\,x{\displaystyle }[/itex].

These rays cross paths at [itex]x_{2}=\frac{\displaystyle R}{\displaystyle R-2y_{1}}\,x_{1}[/itex] and showing that

[itex]y_{2}=-\frac{\displaystyle R}{\displaystyle R-2y_{1}}\,y_{1}[/itex],

we obtain the known amplification factor [itex]x_{2}=-\frac{\displaystyle y_{2}}{\displaystyle y_{1}}\,x_{1}[/itex].

Now... let's consider a ray that hits the mirror at a point [itex](x_{0},y_{0})[/itex]. When this ray is reflected it will have a line equation

[itex]y=y_{0}+\tan\left(2\arctan\left(-\frac{\displaystyle R}{\displaystyle x_{0}}\sqrt{1-\left(\frac{\displaystyle x_{0}}{\displaystyle R}\right)^2}\right)-\arctan\left(\frac{\displaystyle y_{1}-y_{0}}{\displaystyle x_{1}-x_{0}}\right)\right)(x-x_{0})[/itex].

Under the paraxial approximation (which allows us to consider [itex]x_{0}\ll R[/itex]) and using the property of arctan addition we can rewrite it as

[itex]y=y_{0}+\tan\left(\arctan\left(\frac{\displaystyle 2x_{0}R}{\displaystyle R^2-x^2_{0}}\right)-\arctan\left(\frac{\displaystyle y_{1}-y_{0}}{\displaystyle x_{1}-x_{0}}\right)\right)(x-x_{0})[/itex].

I know we can simplify it a bit more by using the [itex]\tan(a+b)[/itex] formula but the way I kept it above is easier to contemplate. I have tested all these rays with mathematica and they converge to the same point so I know everything up to here is ok. My problem is:

If I equal this last ray to one of the others (the one that reflects at the mirror's vertex, for instance) I'll obtain the x at which they cross paths as a function of [itex]x_{0}[/itex] and [itex]y_{0}[/itex]. Since all the rays converge to the same point (given the paraxial approximation) I would not expect them to depend upon the point at which they touch the mirror.

What am I missing here? There is some simplification I did not do? Could you lend me a hand?

Thank you very much.
 
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  • #2
I haven't attempted to follow all of your equations, but since no one has responded yet I'll ask one question - you derived a general equation and then applied the paraxial approximation. So why do you still have trig functions in that equation? For instance, if x0 ≪ R then R2 - x02 ≈ R2 so the argument of the first arctan can be written 2x0/R which << 1 in the paraxial approximation. Etc.

What I'm guessing is that if you rewrite the equation fully making use of the paraxial approximation then either x0 and y0 won't appear, or if they do they would be with other expressions that they are much less than.
 
Last edited:

1. What is the proof of image formation property of spherical mirrors?

The proof of image formation property of spherical mirrors is based on the laws of reflection. According to the laws of reflection, the incident ray, reflected ray, and the normal to the surface at the point of incidence, all lie in the same plane. This property can be observed by conducting experiments with spherical mirrors.

2. How can we experimentally prove the image formation property of spherical mirrors?

To prove the image formation property of spherical mirrors, we can conduct experiments such as placing an object in front of a spherical mirror and observing the position and characteristics of the reflected image. We can also use geometric constructions and measurements to verify the accuracy of the image formation property.

3. What is the difference between concave and convex spherical mirrors in terms of image formation property?

The image formation property of concave and convex spherical mirrors is based on the same principle of reflection. However, the position and characteristics of the reflected image differ between the two types of mirrors. In a concave mirror, the image formed can be real or virtual, depending on the position of the object. In a convex mirror, the image formed is always virtual and smaller than the object.

4. Can the image formation property of spherical mirrors be applied in real-life situations?

Yes, the image formation property of spherical mirrors is applied in various real-life situations such as in telescopes, car side mirrors, and cosmetic mirrors. The property allows us to see objects and their reflections with clarity and accuracy.

5. Are there any limitations to the image formation property of spherical mirrors?

The image formation property of spherical mirrors is limited by the size and curvature of the mirror. If the mirror is too small or too flat, the reflected image may appear distorted. Additionally, the image formed may also be affected by the quality of the mirror's surface and the angle of incidence of the light rays.

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