Do we see multiple images in mirror?

[leonstavros]

When we look at a mirror we see our image but the light that reaches our eyes also reflects off our bodies back to the mirror again and back to our eyes. Do we see multiple images of our self but somehow ignore all but the primary reflection?

 

[256bits]

I would say theoretical yes, but NO for 2 reasons:
1 . the amount of absorption of light from clothing far out weighs any reflection - clothing is just not shiny.
2. If a ray of light from the mirror does hit your eye and is reflected, the angle of incidence = angle of reflection so only a very tiny amount of light would be at an angle sufficient so that the ray would be reflected back to the mirror and then back to your pupil - ie the reflected image would be something in between the size of your pupil and your iris, and again very faint, if not, completely unnoticable. due to absortion again.

 

[rcgldr]

Clothing reflects light (which is why you can see it in the first place, assuming the clothing isn't acting as a black body), but you don't see a reflective image because the clothing scatters the light it reflects, destroying any potential image with an extreme blurring effect. If the clothing approximated a mirror, then you could see an image reflected off that clothing, either directly or if viewing an image from a mirror.

 

[jtbell]

The extra light scattered off your clothing from your image in the mirror simply adds a tiny bit to the brightness of the image. It doesn't create a new image.

If you were to hang a mirror on your chest, and keep it parallel to the other mirror, then you would see an infinite regression of multiple images in the image of that mirror. I see something like this in a restaurant where I often eat, that has mirrors on walls facing each other, with my seat in between.

 

[leonstavros]

What if you're in a room with a mirror and the walls painted white and you're also clothed in white? would you still see a single image on the mirror? I would guess yes although there would be a whole bunch of reflections in the room.

 

[chrisbaird]

Yes, if all the other walls are white, they would reflect the light, but only diffusely, so that they would not create additional images. The image of the wall in the mirror just looks bright, but you don't see an additional image in the wall.

 

[leonstavros]

Yes, if all the other walls are white, they would reflect the light, but only diffusely, so that they would not create additional images. The image of the wall in the mirror just looks bright, but you don't see an additional image in the wall.

I know I'm beating this thing to death but I keep thinking of scenarios such as sitting in the middle of the white room with the mirror in front of me thinking about all those variable phase reflections of the original image creating variable interference patterns as they add or subtract on my retina. What do you think?

 

[chrisbaird]

No, you only get an interference pattern through a superposition of two or more coherent beams that are offset in some way. You don't get an interference pattern from a superposition of incoherent beams. All the random phase differences on average end up canceling out in non-coherent superpostions so that you get a sum of scalar sum of powers instead of a vector sum of fields. If you shine the beams from two different, isolated lasers at the same spot on the wall, you will not get an interference pattern. You have to lock the lasers together, or use one beam from one laser and split it into two beams that than interfere with each other (such as a Michelson interferometer).

The front of your eye is specularly reflective, so you may be able to see and image of yourself in your eyes in the mirror if you look really close, but it will just be a secondary image. There won't be interference fringes or anything.