Reflectivity, Light and The Mirror

[LJP07]

There are 3 mirrors:

Mirror 1, Mirror 2 and Mirror 3 . Or M1, M2, M3.

 

If I reflect light onto M1 and it reflects onto M2, but the light intensity decreases as it passes by each mirror, therefore when the lower light intensity reaches M2 theres not enough light to make the route to M3. What happens that light as it passes M2 but can't reach M3, do the particles fall, go straight, or go up, this is assuming the light is shone at level.

 

It's probably an elementary problem but I can't seem to understand it.

 

Another problem I want to understand it staring at a light bulb or sun. I stress that nobody tries to look directly at the sun or light bulb.

However, when we do look at a reasonable light intensity like the sun or light bulb even for an instant and when we don't mean to, we always get these shapes with colours on the outside and colours on the inside of the shape, usually circular, also these shapes continue to appear even after closing the eye for several seconds or blinking several times, I was wondering what is physically happening with the light and although biological, what's physiologically happening in these instances.

[eric l]

Another problem I want to understand it staring at a light bulb or sun. I stress that nobody tries to look directly at the sun or light bulb.

However, when we do look at a reasonable light intensity like the sun or light bulb even for an instant and when we don't mean to, we always get these shapes with colours on the outside and colours on the inside of the shape, usually circular, also these shapes continue to appear even after closing the eye for several seconds or blinking several times, I was wondering what is physically happening with the light and although biological, what's physiologically happening in these instances.

I'm not sure I have the right answer here, but I think I can help you.

 

For observing light, we have "rods" and "cones" in the retina of our eye. The rods respond to the total intensity of the light, the cones come in three types with different sensitivities as function of the wavelenght. The explanation in Wikipedia is good enough, I do not feel the need to expand on it.

 

http://en.wikipedia.org/wiki/Colour_perception

 

Whenever you look into a light source - wether intentionally or by accident - rods and cones go in overload, and will continue to give a signal to the brain even after the light no longer reaches them. Because the cones have different sensitivities, the different signals will not stop at exactly the same nanosecond. Which colour will remain seen the longest will depend on the actual spectrum of the light source.

 

You see something very similar if you point a webcam in the direction of a candle burning in the dark. Point the webcam in an other direction, and you'll see the same kind of retention on your screen.

 

If you want to get rid of the retention, point your eyes to some surface of the colour you keep seeing. Don't ask me why, but this works.

[InfiniteNow]

I had a similar question once... check out:

 

Hypography Thread: A Reflection on a reflection of a reflection

 

 

Eric's point above is more or less accurate, and the neuroreceptors just cannot keep firing... go into a point of exhaustion. But, the quick answer is adaptation. Kind of like when someone comes near you wearing too much cologne. At first, it's nearly unbearable, but after a few moments it's barely noticable. Same with light. The body adjusts, adapts, or "acclimates" to the stimuli.

[cwes99_03]

As to the first question, which is a simple optics question, you only lose a percentage due to the amount of light absorbed, or not reflected, by the material in the mirror and scattered by the molecules in the air.

 

Most glass panes reflect 10% of any light that hits them. This means that if that light then hits another piece of glass only 10% of the original light is reaching it and 90% of that light (9% of the original light) passes through and 10% (1% original) is again reflected and this continues on and on and on. Of course molecules in the air also scatter light which means that some of it eventually will just be dispersed into the air or absorbed by other molecules as energy.

Light doesn't just mysteriously decrease in intensity. It goes somewhere (second law of thermodynamics?).

[LJP07]

I haven't done Thermodynamic Laws or Physics yet, in a year I will.

[cwes99_03]

Sorry I meant that it had nothing to do with biology, thus was more simple. ;) I didn't mean that it was simple.

[cwes99_03]

Oh, I should also point out that light is not made of massed particles. They don't fall, they pass through the air largely unaffected by earth's simple gravity. The only way to decrease light along a certain path is through absorption and scattering.

[Turtle]

There are 3 mirrors:

Mirror 1, Mirror 2 and Mirror 3 . Or M1, M2, M3.

 

Without specifying 'front surface-silvered' or 'back surface-silvered' mirrors, the rest that follows is posed out of context. ;) :) :coin:

[Turtle]

There is also the matter of the reflectivity index of the material used to 'silver' a mirror, whether front-surface or back-surface. Here's a link on high reflectivity designs:

http://www.kruschwitz.com

http://www.kruschwitz.com/HR's.htm

 

:hyper:

[Jay-qu]

Apart from the fact that there is not 100% reflection, there is also the fact that it scatters, it doesnt concentrate into a beam from the mirror, so the particles will all end up going in lots of different directions

[Turtle]

Apart from the fact that there is not 100% reflection, there is also the fact that it scatters, it doesnt concentrate into a beam from the mirror, so the particles will all end up going in lots of different directions

 

Unless the light is LASER light. :D Nonetheless you are correct to introduce the variability of the light. The color for instance is a factor in light's reflection from a given mirror.

Another thought springs to mind, which is the flatness of the mirror, which directly affects the amount of dispersion of the reflected rays. :hyper: ;)

[Jay-qu]

LASER light can still be scattered by uneven surfaces of the silver - ie it wont be completely flat, so the angle of refraction of that photon wont be equal to the angle of that other photon.

[Turtle]

LASER light can still be scattered by uneven surfaces of the silver - ie it wont be completely flat, so the angle of refraction of that photon wont be equal to the angle of that other photon.

 

Ah contraire. Completely "flat" relative to what? Why the frequency of the light of course. If the irregularities in the mirror are smaller than the wavelength of the light, then the mirror won't scatter the light. The wavelengths of light vary roughly from 700 nanometers for red to 400 nanometers for violet.

 

Optically flat mirrors to 1/20 wave (1/20 wave = 632.8nm) are just a phone call away:

http://www.edmundoptics.com/onlinecatalog/DisplayProduct.cfm?productid=1905

[Jay-qu]

toche, but they are not your average mirror! tell me what kind of applications require such precise optics?

[Turtle]

toche, but they are not your average mirror! tell me what kind of applications require such precise optics?

 

LASER applications abound for these mirrors. Not my average mirror? They are everyones' average mirror & they just don't know it. Every checkout scanner reading bar codes has these mirrors in them & without such precision mirrors the devices simply wouldn't work or work poorly because of the scattering you mention.

We do have two sides to the coin however because there is an "ideal" symbolic representation of reflection & then there is the real world; where the silver meets the substrate so to speak. :hihi:

[Jay-qu]

you have an answer for everything dont you Turtle! by average I meant my bathroom mirror..

[LJP07]

Without specifying 'front surface-silvered' or 'back surface-silvered' mirrors, the rest that follows is posed out of context. :hihi: :hihi: :ebomb:

 

It's a bathroom mirror, whats front-surfaced silvered and back surface silvered.