# What would I see if I locate myself in one of the focus of an ellipsoid whose surface is mirrored? [closed]

Imaging a 3D ellipsoid whose surface is completely mirrored. According to the optical laws of the ellipsoid, all the rays that start in one of the focus, end up forming an image on the other. I was wondering:

1. What would I see if I put something, e.g. an apple in one of the focus. Would I see another apple reflected on the other focus? Would it be upside down?
2. What would I see if I locate myself in one of the focuses? Would I see the image of myself formed on the other focus? Would I be upside down?
• What are your thoughts about these questions? What is your difficulty? Commented May 6, 2018 at 15:01
• wether it is right what I'm saying or not. It seems specially weird to me point 2 and I'm not sure my reasoning is right Commented May 6, 2018 at 15:21
• You aren't saying anything. You are just asking questions, and you have not provided any reasoning. Commented May 6, 2018 at 15:37
• sorry if you didn't get it ;-) Commented May 6, 2018 at 17:08

The formation of image will be pretty much the same as image due to a convex or concave mirror which is a paraboloid. Lets consider the image formation in 2D, then:

But for objects like an apple or a $human$, something weird will happen. It is well explained in this example from this source

This kind of image formation will take place for 3D objects in all directions (consider an apple as a set of 2D objects of different heights and located at different points) And yes, the image will be injerted .

• Your ray diagram only shows 2 rays, emanating from the top of the object, striking a parabolic (not ellipsoidal) mirror. What about rays in other directions, striking other parts of the mirror? What about rays from other points of the object? Commented May 6, 2018 at 17:29
• I already explained about it. Kindly read the full answer. Its not possible for me to show a 3D version here. Commented May 6, 2018 at 17:36
• I'm not asking for a 3D version. You can consider other rays in the 2D version. If other rays from the top of the object converge at a different point, you will have a smeared-out image of this point. Other points on the object will also be smeared out, with no in-focus image anywhere. eg The base of the object at $F_1$ will be focussed at $F_2$, not where you have placed it. You are only using a small portion of the mirror, not the whole mirror shown in your sketch. Commented May 6, 2018 at 17:55
• We only need 2 rays of light for finding the image of an object. Can you correct my answer with another figure ? Commented May 6, 2018 at 17:57
• You only need 2 rays if all rays from a point converge to the same point. This happens approximately for paraxial rays (the paraxial approximation). If another pair of rays converge at a different point, the image of the point object is not in focus. Commented May 6, 2018 at 18:01

1 If the apple is small (compared to the size of the ellipsoid), it can be considered to be a point object. The rays coming from the apple will be concurrent at this focus and the reflected rays will passes through the second focus. It is a real image and lies on the same side as the object and is therefore, inverted. The reflected rays when pass through your cornea you will be able to see a inverted apple at the other focus.

2 Just like the apple your image is formed at the other focus - real and inverted. When the rays from this strike your cornea, you will see yourself inverted facing towards yourself.

In both the cases the are multiple reflections taking place and in the second case you might see a unclear upright image of yourself - the one that is formed at your position. But only your back will be visible in this unclear image.

• Not clear. You need to consider more than one point on the object to say whether the image is inverted : point objects don't point in any direction. You also need to consider where non-paraxial rays will be focussed, if you are using a complete 3D ellipsoid as a mirror. Commented May 6, 2018 at 18:09
• A point right above one focus has its image right below the other focus. So, the image is inverted. I started with the assumption that the object is 'small (compared to the size of the ellipsoid)', so the image will be inverted and focused at the other focus. Commented May 7, 2018 at 10:08
• By not clear, I meant the rays possibly won't get focused at the retina. This image is formed right at the point you are standing. So, you either won't see the image at all and even if you do, it will be an unclear picture of your back. Commented May 7, 2018 at 10:12
• I agree I impetuously assumed that the distance between the foci is greater than the least distance of distinct vision of the observer. Commented May 7, 2018 at 10:17