Our Perception of Light (for an article)

[Kyle Rinker]

I believe there is a vast difference between whether the two squares are the same color in this exact moment in time and whether one square or the other is receiving more light in this exact moment of time.

 

[Snuggs]

The original illusion refers to tha actual color of the squares. Which are the same color. But it appears that A gets more light

 

[WallyB]

I know this picture... A looks to be recieving more light, but if you cut the two squares and remove the surrounding picture you will see that both A and B are the same color.

You are Dead on!! Both squares are exact same Shades. They only look different with the illusion/comparison to the other objects. It's happening in your head.

It's the same when the Moon Looks Huge, when Low on Horizon Compared to Buildings and Trees close to it. I show that to my kids all the time.
Cover the Horizon with your hand when looking at the Large moon. It's shrinks instantly. (After All the moon doesn't get closer or farther when it rises up into the Sky, but a little while later in the Empty Sky, it's looks way smaller).

 

[vetteguy53081]

I voted. B represents shading which is the enemy of Many coral.

 

[Mical]

Try this - https://en.wikipedia.org/wiki/Checker_shadow_illusion

 

[siggy]

Love this kind of stuff! .....Note the question did not state that the grid was indicative of photons revived

 

[MnFish1]

The question is formulated correctly. Or should I say the hook was baited and we took the bait.

Well - Actually - this is a famous optical illusion - with the point being that A and B (though they look different) are the same color (ie the same brightness). So if you read the question 'which is brighter (or lighter) A or B' - the answer is they are the same. If you read the question - on a more physics basis - The light intensity is less in the shadowed area.

 

[MnFish1]

The question is not correctly formulated IMO. The correct question should be, "Which of these squares is darker?" A is in fact recieving more light because clearly the light source is from the top right of the drawing. Them being the same color is an optical illusion made to trick your brain. If that were a real photograph, A and B would in fact be different colors.

I agree about the question - it depends on what the question was designed to 'prove'. (hopefully the article is coming).

As to the illusion part - I think you're incorrect - The optical illusion is that they appear to be different colors when in fact they are the same color - and with the right light, etc - you could produce the same 'illusion' in real life. The brain sees a pattern and keeps up the pattern even though in reality, the shadow makes the (shadowed) white square the same color as the (unshadowed) black square the brain sees 2 different colors

 

[WallyB]

I took sample of the Squares (using computer) NOT MY BRAIN's Perception. And placed them to compare on the RIGHT SIDE.

Analyze the results yourself (but to me the Image is By Law of Physics impossible, if Each Black and White squares are the same Shaded MATERIAL )

Thus is image is Altered (shaded) to make a interesting Point (on Brain Perception). Fun stuff. Great Brain teaser!!

So to me this Image is Physically not possible. To me the White and Black squares cannot be all the same in a real physical world.

I wonder what the Article will try to convey.

 

[MnFish1]

I took sample of the Squares (using computer) NOT MY BRAIN's Perception. And placed them to compare on the RIGHT SIDE.

Analyze the results yourself (but to me the Image is By Law of Physics impossible, if Each Black and White squares are the same Shaded MATERIAL )

Thus is image is Altered (shaded) to make a interesting Point (on Brain Perception). Fun stuff. Great Brain teaser!!

So to me this Image is Physically not possible. To me the White and Black squares cannot be all the same in a real physical world.

I wonder what the Article will try to convey.

I dont see why the image is physically 'not possible'. Its the whole principle of a shadow. Take a white wall with high light - it will look white. Shade it it will look gray. Shade it more it will look darker. Turn the light off - it will look black......

 

[WallyB]

I dont see why the image is physically 'not possible'. Its the whole principle of a shadow. Take a white wall with high light - it will look white. Shade it it will look gray. Shade it more it will look darker. Turn the light off - it will look black......

I agree that shading a white square will make it darker.
However look at the shadow cast. It appears the light source is coming from Top Right.
Wouldn't the Square with the RED Arrow be brighter then the Square with the ORANGE Arrow (since closer to light, farther from viewer and also getting reflection from the Column).
Yet the Orange Arrow Square is brighter. (That could be just the angle we are observing the objects in the image).

This is all fun, and great analysis/discussion.....and maybe I'm over analyzing (like others have tried to explain diffraction of light which was interesting too).

I'm still curious what the Article will try to convey.

 

[WallyB]

I dont see why the image is physically 'not possible'. Its the whole principle of a shadow. Take a white wall with high light - it will look white. Shade it it will look gray. Shade it more it will look darker. Turn the light off - it will look black......

More Fun, since I like brain teasers, and analysis.

I made a Real Physical Model (and took a Photo), using Checkerboard Printout on Paper, A Light, and a MaxiJet Funnel attachment to cast shadow..

The Brain Illusion Holds (B) Looks lighter than (A) but is Actually Darker.

But (D) is Lighter then (C) since closer to Light, contrary to the Famous Image.

.......Onward to upcoming Article.

 

[MnFish1]

More Fun, since I like brain teasers, and analysis.

I made a Real Physical Model (and took a Photo), using Checkerboard Printout on Paper, A Light, and a MaxiJet Funnel attachment to cast shadow..

The Brain Illusion Holds (B) Looks lighter than (A) but is Actually Darker.

But (D) is Lighter then (C) since closer to Light, contrary to the Famous Image.

.......Onward to upcoming Article.

Yep - I see what you mean - I wasn't comparing the 2 white areas. I just meant that its possible to 'create' a scenario where the black and the white area looked similar. Hopefully they will use your model in the article. Nice job.

 

[Leyth]

I don't think you can answer the question without knowing if the light source is LED or T5. If it is LED, I will also need the schedule posted please. Thanks.

 

[PhreeByrd]

I don't think you can answer the question without knowing if the light source is LED or T5. If it is LED, I will also need the schedule posted please. Thanks.

I think it's the sun...

 

[Paul B]

I voted B but I was shinning an LED flashlight on it as I voted.

(I also know they get the same light. I think anyway. OK, I got nothing. )

 

[MabuyaQ]

Eyes can't see the intensity of light cast on an object as all objects (partly) absorb light (of a certain spectum), not to mention that we don't have full spectrum vision. What we truly see is the intensity of the light reflected by an object within the spectral range of the human eye, nothing more, nothing less. Color perception is therefore created by this spectral range and the intensity of the light casted on that object within that spectral range. So in this image all light squares are the same color and all the dark squares are the same color. The 'darker' squares (both light and dark in color) in the shadow casted by the green object receive less intense light and therefore reflect less intens light which is perceived as a different/darker color by the human eye. Now the real question is what if there was a second lightsource but in a spectrum we can't see and 100x more intense than the 'visible' light source that is also casting a shadow (invisible to the human eye) but not on square B but only on square A?
SO the real answer is we simple can't tell because the human eye is not capable to do so.

 

[ReefGeezer]

I'm not familiar with the image or what it was intended to illustrate, and I've never been accused of being too smart. I'll give it a shot anyway and prove I'm not too smart!

The squares do appear to be the same tone in the image If a physical representation of the image was set up and A was a darker color than B, which would receive more light? Let's assume bending/refraction/secondary light source/etc., or some other phenomenon that I'm too dumb to understand, allowed as many photons inside the shadowed area around Square B as that of the fully exposed Square A. While both would be exposed to the same # of photons, Square A would "receive" more light because it would absorb more photons.

 

[MnFish1]

I'm not familiar with the image or what it was intended to illustrate, and I've never been accused of being too smart. I'll give it a shot anyway and prove I'm not too smart!

The squares do appear to be the same tone in the image If a physical representation of the image was set up and A was a darker color than B, which would receive more light? Let's assume bending/refraction/secondary light source/etc., or some other phenomenon that I'm too dumb to understand, allowed as many photons inside the shadowed area around Square B as that of the fully exposed Square A. While both would be exposed to the same # of photons, Square A would "receive" more light because it would absorb more photons.

Where is the article

 

[MnFish1]

Eyes can't see the intensity of light cast on an object as all objects (partly) absorb light (of a certain spectum), not to mention that we don't have full spectrum vision. What we truly see is the intensity of the light reflected by an object within the spectral range of the human eye, nothing more, nothing less. Color perception is therefore created by this spectral range and the intensity of the light casted on that object within that spectral range. So in this image all light squares are the same color and all the dark squares are the same color. The 'darker' squares (both light and dark in color) in the shadow casted by the green object receive less intense light and therefore reflect less intens light which is perceived as a different/darker color by the human eye. Now the real question is what if there was a second lightsource but in a spectrum we can't see and 100x more intense than the 'visible' light source that is also casting a shadow (invisible to the human eye) but not on square B but only on square A?
SO the real answer is we simple can't tell because the human eye is not capable to do so.

Actually that is not the question - but it is interesting that you thought of a second question