Our Perception of Light (for an article)
- Thread starter Seawitch
- Start date
- Tagged users None
They are the same brightness ie the same color. Looking forward to the article! I also wonder if the question is formulated correctly as I’m not sure that what I wrote means each area is receiving the same intensity or brightness of light.
Not sure that right if you define light intensity as watts per meter squared. In that case the shadowed area is receiving less light. But I guess it depends on how reads the question
Sarah24!
2500 Club Member
Hello,
That is only if we can assume there are no other variables. The diagram posted shared shows that light is bending, and will depend upon the opening. It will also depend on the angle of refraction. We don’t know if this diagram is in a tank or anything else. We can only assume that the light is provided by the sun. Again light will bend, which if the cylinder is in fact casting a shadow basic physics proves it receives less light.
In addition here is a small sample that was done by the university of Illinois “ Diffracted light can produce fringes of light, dark or colored bands. An optical effect that results from the diffraction of light is the silver lining sometimes found around the edges of clouds or coronas surrounding the sun or moon. The illustration above shows how light (from either the sun or the moon) is bent around small droplets in the cloud.
Optical effects resulting from diffraction are produced through the interference of light waves. To visualize this, imagine light waves as water waves. If water waves were incident upon a float residing on the water surface, the float would bounce up and down in response to the incident waves, producing waves of its own. As these waves spread outward in all directions from the float, they interact with other water waves. If the crests of two waves combine, an amplified wave is produced (constructive interference). However, if a crest of one wave and a trough of another wave combine, they cancel each other out to produce no vertical displacement (destructive interference).
This concept also applies to light waves. When sunlight (or moonlight) encounters a cloud droplet, light waves are altered and interact with one another in a similar manner as the water waves described above. If there is constructive interference, (the crests of two light waves combining), the light will appear brighter. If there is destructive interference, (the trough of one light wave meeting the crest of another), the light will either appear darker or disappear entirely. (Source from University of Illinois 2010 est). Which this will help explain if based on normal nature and sun as the variable.
Even if this was in a tank, we still have to deal with water droplets and light bending from the coral growth. It would also depend on type of light used, and how the light is bending.
That is only if we can assume there are no other variables. The diagram posted shared shows that light is bending, and will depend upon the opening. It will also depend on the angle of refraction. We don’t know if this diagram is in a tank or anything else. We can only assume that the light is provided by the sun. Again light will bend, which if the cylinder is in fact casting a shadow basic physics proves it receives less light.
In addition here is a small sample that was done by the university of Illinois “ Diffracted light can produce fringes of light, dark or colored bands. An optical effect that results from the diffraction of light is the silver lining sometimes found around the edges of clouds or coronas surrounding the sun or moon. The illustration above shows how light (from either the sun or the moon) is bent around small droplets in the cloud.
Optical effects resulting from diffraction are produced through the interference of light waves. To visualize this, imagine light waves as water waves. If water waves were incident upon a float residing on the water surface, the float would bounce up and down in response to the incident waves, producing waves of its own. As these waves spread outward in all directions from the float, they interact with other water waves. If the crests of two waves combine, an amplified wave is produced (constructive interference). However, if a crest of one wave and a trough of another wave combine, they cancel each other out to produce no vertical displacement (destructive interference).
This concept also applies to light waves. When sunlight (or moonlight) encounters a cloud droplet, light waves are altered and interact with one another in a similar manner as the water waves described above. If there is constructive interference, (the crests of two light waves combining), the light will appear brighter. If there is destructive interference, (the trough of one light wave meeting the crest of another), the light will either appear darker or disappear entirely. (Source from University of Illinois 2010 est). Which this will help explain if based on normal nature and sun as the variable.
Even if this was in a tank, we still have to deal with water droplets and light bending from the coral growth. It would also depend on type of light used, and how the light is bending.
lapin
10K Club member
Ultimate Member
Reef Squad
Excellence Award
Reef Squad Emeritus
Hospitality Award
No one said there was a light source casting the shadow. It is a flat picture and the whole thing receives the same light.
I could not hold back, in light of this getting to technical for my old brain
I could not hold back, in light of this getting to technical for my old brain
Hemmdog
10K Club member
Meteorology 456b w/ Sarah24! lol , fountain of knowledge!Hello,
That is only if we can assume there are no other variables. The diagram posted shared shows that light is bending, and will depend upon the opening. It will also depend on the angle of refraction. We don’t know if this diagram is in a tank or anything else. We can only assume that the light is provided by the sun. Again light will bend, which if the cylinder is in fact casting a shadow basic physics proves it receives less light.
In addition here is a small sample that was done by the university of Illinois “ Diffracted light can produce fringes of light, dark or colored bands. An optical effect that results from the diffraction of light is the silver lining sometimes found around the edges of clouds or coronas surrounding the sun or moon. The illustration above shows how light (from either the sun or the moon) is bent around small droplets in the cloud.
Optical effects resulting from diffraction are produced through the interference of light waves. To visualize this, imagine light waves as water waves. If water waves were incident upon a float residing on the water surface, the float would bounce up and down in response to the incident waves, producing waves of its own. As these waves spread outward in all directions from the float, they interact with other water waves. If the crests of two waves combine, an amplified wave is produced (constructive interference). However, if a crest of one wave and a trough of another wave combine, they cancel each other out to produce no vertical displacement (destructive interference).
This concept also applies to light waves. When sunlight (or moonlight) encounters a cloud droplet, light waves are altered and interact with one another in a similar manner as the water waves described above. If there is constructive interference, (the crests of two light waves combining), the light will appear brighter. If there is destructive interference, (the trough of one light wave meeting the crest of another), the light will either appear darker or disappear entirely. (Source from University of Illinois 2010 est). Which this will help explain if based on normal nature and sun as the variable.
Even if this was in a tank, we still have to deal with water droplets and light bending from the coral growth. It would also depend on type of light used, and how the light is bending.
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.
lapin
10K Club member
Ultimate Member
Reef Squad
Excellence Award
Reef Squad Emeritus
Hospitality Award
The question is formulated correctly. Or should I say the hook was baited and we took the bait.
Jase4224
Well-Known Member
I voted A assuming that this picture shows that there is a light source shining onto the board and green pole. It doesn’t matter what colour the squares are, the position of B is clearly being shaded from the light source and A is not.
Imagine the same picture but with just a white board instead of the checkered pattern. A would be in th white area and B would still be shadowed by the green pole.
Imagine the same picture but with just a white board instead of the checkered pattern. A would be in th white area and B would still be shadowed by the green pole.
Sierra_Bravo
Valuable Member
Well, technically the question is flawed. The illusion was designed to demonstrate color perception and has nothing to do light or light intensity. It has to do with how the surrounding darker colored squares around B serve to make it appear a lighter shade of color.
Using light or light intensity, the drawing clearly shows the light source coming from the NE and the cylinder casting a shadow. If we were judging the light intensity based on the picture, A is correct. :)
(Edit - posted before seeing Robin Haselden's response which is the same - sorry Robin!)
Using light or light intensity, the drawing clearly shows the light source coming from the NE and the cylinder casting a shadow. If we were judging the light intensity based on the picture, A is correct. :)
(Edit - posted before seeing Robin Haselden's response which is the same - sorry Robin!)
Sarah24!
2500 Club Member
Hello,
If we all go back to physics for this question it’s was asked if the squares are the same color based on where the clyindar is placed. In the results of the test it stated that
“That the two squares are the same color can be proven using the following methods:
So if we read this test sample it wasn’t about which square has the most light intensity, it’s if the squares have the same color. Based on her question she is asking which has a higher light intensity which would be A because of the refraction point interference with point B lol. So she worded it correctly, but the diagram was used to prove an optical illusion to show that A and B are the same color. It had nothing to do with light intensity:)
Edit sorry didn’t see the other three posts before mine lol
If we all go back to physics for this question it’s was asked if the squares are the same color based on where the clyindar is placed. In the results of the test it stated that
“That the two squares are the same color can be proven using the following methods:
- Open the illusion in an image editing program and using the eyedropper tool to verify that the colors are the same.
- Cut out a cardboard mask. By viewing patches of the squares without the surrounding context, you can remove the effect of the illusion. A piece of cardboard with two circles removed will work as a mask for a computer screen or for a printed piece of paper.
- Connect the squares with a rectangle of the same color, as seen in the figure to the right.” (Source
- Adelson, Edward H. (2005). "Checkershadow Illusion". Retrieved 2007-04-21.
- ^ "Proof of Checkershadow Illusion". Archived from the original on 2004-05-14.
So if we read this test sample it wasn’t about which square has the most light intensity, it’s if the squares have the same color. Based on her question she is asking which has a higher light intensity which would be A because of the refraction point interference with point B lol. So she worded it correctly, but the diagram was used to prove an optical illusion to show that A and B are the same color. It had nothing to do with light intensity:)
Edit sorry didn’t see the other three posts before mine lol
Sod Buster
Valuable Member
The source is constant, so both the same. The intensity is being manipulated by a shadow.
tupes
Valuable Member
I'll have to disagree.
One is light grey and one dark. However this wasnt the question. ;Hilarious:D
Although I think we all know what you meant.
Sarah24!
2500 Club Member
Hello,
In order for the human eye to see and to make colors out we have to have some form of light source to even see. The fact we have green white and charcoal colors proves there is light present, we just don’t know the source. We also know the light source is coming from the ne position which is casting the shadow on B. If no light was present the diagram would be black, because of how the human eye works.
Richard Hercher
Active Member
Here’s how we find out: you will need a closet, a checker board, an opaque cylinder, two light meters (I’ll accept PAR meters since that’s something someone might have on hand in this crazy forum), a flash light, and a mount for the flash light. Place the par meter sensors and cylinder on the checkerboard as indicated, with the reader outside the closet. mount the flashlight inside the closet to model the shadow. Close the door, read the measurements of the sensor. Maybe something for @Bulk Reef Supply to do for a BRS Investigates video?
I’ll wait.
I’ll wait.
PhreeByrd
Active Member
There seems to be a lot of confusion between color and light. If you take the cylinder away, then there are only two colors of squares -- light gray and dark gray. Those colors are not affected by the light. They are simply the colors that they are, as we perceive them under uniform lighting conditions.
Obviously there is a (point) light source because the cylinder casts a shadow onto some of the squares. That shadow changes our perception of the colors of the squares. Squares within the shadow area are darker than their same colored counterparts outside of the shadow area. The trick is that our brains tell us that the shadowed B square is a lighter color than the unshadowed A square. In truth, the shadowed light gray squares (B) are (to our eyes) actually as dark as the unshadowed dark gray (A) squares -- at least, I feel confident that this is the case, although I have not verified this by isolating and comparing those two colors. Our brains tell us that this is not correct, because we know that without the effects of light and shadow, A is darker than B.
In any case, the color of the squares is simply a distraction from the question. The color of the squares (meaning, whether the square is dark gray or light gray) has no bearing on how much light each square receives. The shadow (or lack of it) from the cylinder is the only thing that affects the amount of light received by the two squares. Because B is in the shadow and A is not, then A receives more light than B... regardless of the color the squares are or appear to be.
Obviously there is a (point) light source because the cylinder casts a shadow onto some of the squares. That shadow changes our perception of the colors of the squares. Squares within the shadow area are darker than their same colored counterparts outside of the shadow area. The trick is that our brains tell us that the shadowed B square is a lighter color than the unshadowed A square. In truth, the shadowed light gray squares (B) are (to our eyes) actually as dark as the unshadowed dark gray (A) squares -- at least, I feel confident that this is the case, although I have not verified this by isolating and comparing those two colors. Our brains tell us that this is not correct, because we know that without the effects of light and shadow, A is darker than B.
In any case, the color of the squares is simply a distraction from the question. The color of the squares (meaning, whether the square is dark gray or light gray) has no bearing on how much light each square receives. The shadow (or lack of it) from the cylinder is the only thing that affects the amount of light received by the two squares. Because B is in the shadow and A is not, then A receives more light than B... regardless of the color the squares are or appear to be.
Last edited:
Similar threads
