# How can solve to calculate position of image for any random point and random paraxial rays?

I am a high school student and I have a confusion related to optics {image formation via spherical mirrors) in all my books they say to find location of image its convenient to use the following rays"

1. incident on pole

2. parallel to principal axis

3. passing through focus

4. passing through center of curvature

but when they are doing the derivation they already consider that for paraxial rays an image will form, i.e they already consider that paraxial rays from any point will almost converge or appears to converge at a single point .In other words can anyone derive the mirror formula by selecting any two random paraxial rays (other than mentioned above)?

because it will then convince me that we can treat any paraxial rays to be meeting at almost at single point and forming an image.

In short: please solve for location of image by considering any random point and random paraxial rays.

• In other words can anyone derive the mirror formula by selecting any two random paraxial rays (other than mentioned above)? Your question is unclear to me particular as an answer which has a ray diagram that you have accepted does not shown the incoming rays to be paraxial (parallel to the principle axis). . . . please solve for location of image by considering any random point and random paraxial rays. Any random point on what? Any point off the principle axis can only produce one paraxial ray as all other rays will be at an angle to the principle axis. Oct 17 '21 at 8:22

I try to "realize" your idea

• C center of spherical mirror Radius R
• F focus at $$~\frac R2$$

Image A'-B'

the ray B-1 is return to F, the intersection between the rays F-1 and C-B is the point B' so you obtain the Image A'-B'

Image A"-B"

you send a ray B-2 with angle $$~\alpha~$$ this ray is return back with the angle $$~\beta~$$ between B-2 and F-2 . where $$~\cos(\beta)=\cos(2\,\varphi+\alpha)~$$

you send another ray B_3 that return back with $$~\beta_2~$$ .

the intersection between those two rays is the point B", those you obtain the image A"-B".

is this what you wanted to do ??

Notice for $$~\alpha=0~$$ you obtain the image A'-B'

Edit

$$~\beta~$$ is the angle between the ray B-2 and F-2

with:

$$\vec B_2=\begin{bmatrix} \cos(\alpha) \\ -\sin(\alpha) \\ \end{bmatrix}\\ \vec F_2=\begin{bmatrix} \cos(2\varphi) \\ \sin(2\varphi) \\ \end{bmatrix}$$

and the cosine law

$$\vec B_2\,\cdot \vec F_2=|\vec B_2|\,|\vec F_2| \,\cos(\beta)\\ \Rightarrow\\\\ \cos(\beta)=\cos(2\varphi+\alpha)$$

Diagram for the mirror system.

$$x_1 = x_0 + t_0*\beta_0$$ with $$\theta´ = -\theta$$

$$\beta_0 - \beta_1 = \theta -\theta´= 2*\theta$$

$$\beta_1 = \phi - \theta$$

$$\phi = -\frac{x_1}{R_1}$$

combining these equations you obtain

$$\beta_1 = -\beta_0 - \frac{2*x_1}{R_1}\,$$ which is the reflection ray-tracing equation and propagation of the ray is

$$x_2 = x_1+ t_1*\beta_1 = x_0 + t_0*\beta_0 + t_1 * \beta_1$$

Reference:

FUNDAMENTALS OF GEOMETRICAL OPTICS, Virendra N. Mahajan, SPIE Press, 2014

Why do you want to make your calculations so tricky? Yes, it is possible to do it this way, but you would need to calculate the angles and make too many geometric constructions which would unnecessarily clutter your ray diagram. It is a lot easier to locate the points using the simple rules you employ for making the ray diagram, like the rays parallel to the axis passing through the focus for small aperture and so on. But if you are curious, you can use this app for simulations.

Edit: I have added a roughly drawn ray diagram on this website. The object is placed at the centre of curvature. The object and the image are of the same size and the image forms at the centre of curvature. The slight deviations can be attributed to minor construction errors and spherical aberration.

• I want to do this because that would convince me that for paraxial rays we can treat any paraxial ray from any point to be almost crossing a single point Oct 5 '21 at 8:39
• otherwise how would you explain that the image cannot be like this I have shown in the diagram? why it should be straight? Oct 5 '21 at 8:40
• Why don't you give it a try by constructing the ray diagram yourself first? Keep in mind that the aperture should be small compared to the radius of curvature; otherwise, you won't get the expected result due to spherical aberration. Oct 5 '21 at 9:05
• I tried but failed neither I found this anywhere that's why I am asking Oct 5 '21 at 13:39
• @ArunBhardwaj I have added a diagram. I hope it's helpful. Oct 5 '21 at 14:22