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How photons get into the eyes?
How do virtual-photons curve in a magetic field?
$begingroup$
I hope you will understand me correctly because some things I translate.
It is known that we see the world around us thanks to photons that
are reflected from the surfaces of objects,so i have the following question:
If you imagine...
I do not know...a huge gray column of 200 meters from the eyes...
why are the photons reflected off this pole flying straight into your eyes
all the time while you're looking?I mean.....is it some huge stream flying in all directions,
particles from which will necessarily fall into the eyes?How does this stream not mix with others?
What does that even look like?An infinite number of randomly intersecting and moving points?
How do we distinguish which photons are reflected from what?
photons vision
edited 8 hours ago
Qmechanic♦
109k122081281
asked 8 hours ago
LyyLyy
182
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I hope you will understand me correctly because some things I translate.
It is known that we see the world around us thanks to photons that
are reflected from the surfaces of objects,so i have the following question:
If you imagine...
I do not know...a huge gray column of 200 meters from the eyes...
why are the photons reflected off this pole flying straight into your eyes
all the time while you're looking?I mean.....is it some huge stream flying in all directions,
particles from which will necessarily fall into the eyes?How does this stream not mix with others?
What does that even look like?An infinite number of randomly intersecting and moving points?
How do we distinguish which photons are reflected from what?
photons vision
edited 8 hours ago
Qmechanic♦
109k122081281
asked 8 hours ago
LyyLyy
182
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I hope you will understand me correctly because some things I translate.
It is known that we see the world around us thanks to photons that
are reflected from the surfaces of objects,so i have the following question:
If you imagine...
I do not know...a huge gray column of 200 meters from the eyes...
why are the photons reflected off this pole flying straight into your eyes
all the time while you're looking?I mean.....is it some huge stream flying in all directions,
particles from which will necessarily fall into the eyes?How does this stream not mix with others?
What does that even look like?An infinite number of randomly intersecting and moving points?
How do we distinguish which photons are reflected from what?
photons vision
edited 8 hours ago
Qmechanic♦
109k122081281
asked 8 hours ago
LyyLyy
182
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I hope you will understand me correctly because some things I translate.
It is known that we see the world around us thanks to photons that
are reflected from the surfaces of objects,so i have the following question:
If you imagine...
I do not know...a huge gray column of 200 meters from the eyes...
why are the photons reflected off this pole flying straight into your eyes
all the time while you're looking?I mean.....is it some huge stream flying in all directions,
particles from which will necessarily fall into the eyes?How does this stream not mix with others?
What does that even look like?An infinite number of randomly intersecting and moving points?
How do we distinguish which photons are reflected from what?
photons vision
photons vision
edited 8 hours ago
Qmechanic♦
109k122081281
asked 8 hours ago
LyyLyy
182
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Qmechanic♦
109k122081281
asked 8 hours ago
LyyLyy
182
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Qmechanic♦
109k122081281
edited 8 hours ago
Qmechanic♦
109k122081281
edited 8 hours ago
Qmechanic♦
109k122081281
109k122081281
asked 8 hours ago
LyyLyy
182
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
LyyLyy
182
asked 8 hours ago
LyyLyy
182
182
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lyy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Yes - we are surrounded by a "sea of photons".
An individual object that reflects light (let's assume a Lambertian reflector - something that reflects incident photons in all directions) sends some fraction of the incident photons in all directions. "Some fraction" because the surface will absorb some light (there is no such thing as 100% white).
The propagation of photons follows linear laws (at normal light intensities) so that two photons, like waves, can travel on intersecting paths and continue along their way without disturbing each other.
Finally it is worth calculating how many photons hit a unit area per unit time. If we assume sunlight, we know that the intensity of the light is about 1 kW / m$^2$. For the purpose of approximation, if we assume every photon had a wavelength of 500 nm, it would have an energy of $E = frachlambda = 3.97 cdot 10^-19 J$. So one square meter is hit with approximately $2.5cdot 10^21$ photons. Let's assume your grey column reflects just 20% of these and that the visible component of light is about 1/10th of the total light (for the sake of this argument I can be off by an order of magnitude... this is for illustration only).
At a distance of 200 m, these photons would have spread over a sphere with a surface of $4pi R^2 approx 500,000 m^2$, or $10^14$ photons per square meter per second.
If your pupil has a diameter of 4 mm, an area of $12 mm^2$, it will be hit by about $12cdot 10^8$ photons per second from one square meter of grey surface illuminated by the sun from 200 m away.
At that distance, the angular size of that object is about 1/200 th of a radian. "Normal" vision is defined as the ability to resolve objects that are about 5 minutes of arc (there are 60 minutes to a degree and about 57 degrees to a radian). in other words, you should be able to resolve 1/(57*(60/5)) or about 1/600 of a radian. That's still lots of photons...
Finally you ask "how do we distinguish what photons are reflected from what"? For this we have to thank the lens in our eye. A photon has a particular direction, and thanks to the lens its energy ends up on a particular part of the retina (this is what we call "focusing"). Photons from different directions end up in a different place. Nerves on the back of the retina tell us where the photons landed - and even what color they were. The visual cortex (part of the brain) uses that information to make a picture of the surrounding world in our mind.
It's nothing short of miraculous.
answered 8 hours ago
FlorisFloris
107k11189326
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes - we are surrounded by a "sea of photons".
An individual object that reflects light (let's assume a Lambertian reflector - something that reflects incident photons in all directions) sends some fraction of the incident photons in all directions. "Some fraction" because the surface will absorb some light (there is no such thing as 100% white).
The propagation of photons follows linear laws (at normal light intensities) so that two photons, like waves, can travel on intersecting paths and continue along their way without disturbing each other.
Finally it is worth calculating how many photons hit a unit area per unit time. If we assume sunlight, we know that the intensity of the light is about 1 kW / m$^2$. For the purpose of approximation, if we assume every photon had a wavelength of 500 nm, it would have an energy of $E = frachlambda = 3.97 cdot 10^-19 J$. So one square meter is hit with approximately $2.5cdot 10^21$ photons. Let's assume your grey column reflects just 20% of these and that the visible component of light is about 1/10th of the total light (for the sake of this argument I can be off by an order of magnitude... this is for illustration only).
At a distance of 200 m, these photons would have spread over a sphere with a surface of $4pi R^2 approx 500,000 m^2$, or $10^14$ photons per square meter per second.
If your pupil has a diameter of 4 mm, an area of $12 mm^2$, it will be hit by about $12cdot 10^8$ photons per second from one square meter of grey surface illuminated by the sun from 200 m away.
At that distance, the angular size of that object is about 1/200 th of a radian. "Normal" vision is defined as the ability to resolve objects that are about 5 minutes of arc (there are 60 minutes to a degree and about 57 degrees to a radian). in other words, you should be able to resolve 1/(57*(60/5)) or about 1/600 of a radian. That's still lots of photons...
Finally you ask "how do we distinguish what photons are reflected from what"? For this we have to thank the lens in our eye. A photon has a particular direction, and thanks to the lens its energy ends up on a particular part of the retina (this is what we call "focusing"). Photons from different directions end up in a different place. Nerves on the back of the retina tell us where the photons landed - and even what color they were. The visual cortex (part of the brain) uses that information to make a picture of the surrounding world in our mind.
It's nothing short of miraculous.
answered 8 hours ago
FlorisFloris
107k11189326
$endgroup$
add a comment |
$begingroup$
Yes - we are surrounded by a "sea of photons".
An individual object that reflects light (let's assume a Lambertian reflector - something that reflects incident photons in all directions) sends some fraction of the incident photons in all directions. "Some fraction" because the surface will absorb some light (there is no such thing as 100% white).
The propagation of photons follows linear laws (at normal light intensities) so that two photons, like waves, can travel on intersecting paths and continue along their way without disturbing each other.
Finally it is worth calculating how many photons hit a unit area per unit time. If we assume sunlight, we know that the intensity of the light is about 1 kW / m$^2$. For the purpose of approximation, if we assume every photon had a wavelength of 500 nm, it would have an energy of $E = frachlambda = 3.97 cdot 10^-19 J$. So one square meter is hit with approximately $2.5cdot 10^21$ photons. Let's assume your grey column reflects just 20% of these and that the visible component of light is about 1/10th of the total light (for the sake of this argument I can be off by an order of magnitude... this is for illustration only).
At a distance of 200 m, these photons would have spread over a sphere with a surface of $4pi R^2 approx 500,000 m^2$, or $10^14$ photons per square meter per second.
If your pupil has a diameter of 4 mm, an area of $12 mm^2$, it will be hit by about $12cdot 10^8$ photons per second from one square meter of grey surface illuminated by the sun from 200 m away.
At that distance, the angular size of that object is about 1/200 th of a radian. "Normal" vision is defined as the ability to resolve objects that are about 5 minutes of arc (there are 60 minutes to a degree and about 57 degrees to a radian). in other words, you should be able to resolve 1/(57*(60/5)) or about 1/600 of a radian. That's still lots of photons...
Finally you ask "how do we distinguish what photons are reflected from what"? For this we have to thank the lens in our eye. A photon has a particular direction, and thanks to the lens its energy ends up on a particular part of the retina (this is what we call "focusing"). Photons from different directions end up in a different place. Nerves on the back of the retina tell us where the photons landed - and even what color they were. The visual cortex (part of the brain) uses that information to make a picture of the surrounding world in our mind.
It's nothing short of miraculous.
answered 8 hours ago
FlorisFloris
107k11189326
$endgroup$
add a comment |
$begingroup$
Yes - we are surrounded by a "sea of photons".
An individual object that reflects light (let's assume a Lambertian reflector - something that reflects incident photons in all directions) sends some fraction of the incident photons in all directions. "Some fraction" because the surface will absorb some light (there is no such thing as 100% white).
The propagation of photons follows linear laws (at normal light intensities) so that two photons, like waves, can travel on intersecting paths and continue along their way without disturbing each other.
Finally it is worth calculating how many photons hit a unit area per unit time. If we assume sunlight, we know that the intensity of the light is about 1 kW / m$^2$. For the purpose of approximation, if we assume every photon had a wavelength of 500 nm, it would have an energy of $E = frachlambda = 3.97 cdot 10^-19 J$. So one square meter is hit with approximately $2.5cdot 10^21$ photons. Let's assume your grey column reflects just 20% of these and that the visible component of light is about 1/10th of the total light (for the sake of this argument I can be off by an order of magnitude... this is for illustration only).
At a distance of 200 m, these photons would have spread over a sphere with a surface of $4pi R^2 approx 500,000 m^2$, or $10^14$ photons per square meter per second.
If your pupil has a diameter of 4 mm, an area of $12 mm^2$, it will be hit by about $12cdot 10^8$ photons per second from one square meter of grey surface illuminated by the sun from 200 m away.
At that distance, the angular size of that object is about 1/200 th of a radian. "Normal" vision is defined as the ability to resolve objects that are about 5 minutes of arc (there are 60 minutes to a degree and about 57 degrees to a radian). in other words, you should be able to resolve 1/(57*(60/5)) or about 1/600 of a radian. That's still lots of photons...
Finally you ask "how do we distinguish what photons are reflected from what"? For this we have to thank the lens in our eye. A photon has a particular direction, and thanks to the lens its energy ends up on a particular part of the retina (this is what we call "focusing"). Photons from different directions end up in a different place. Nerves on the back of the retina tell us where the photons landed - and even what color they were. The visual cortex (part of the brain) uses that information to make a picture of the surrounding world in our mind.
It's nothing short of miraculous.
answered 8 hours ago
FlorisFloris
107k11189326
$endgroup$
Yes - we are surrounded by a "sea of photons".
An individual object that reflects light (let's assume a Lambertian reflector - something that reflects incident photons in all directions) sends some fraction of the incident photons in all directions. "Some fraction" because the surface will absorb some light (there is no such thing as 100% white).
The propagation of photons follows linear laws (at normal light intensities) so that two photons, like waves, can travel on intersecting paths and continue along their way without disturbing each other.
Finally it is worth calculating how many photons hit a unit area per unit time. If we assume sunlight, we know that the intensity of the light is about 1 kW / m$^2$. For the purpose of approximation, if we assume every photon had a wavelength of 500 nm, it would have an energy of $E = frachlambda = 3.97 cdot 10^-19 J$. So one square meter is hit with approximately $2.5cdot 10^21$ photons. Let's assume your grey column reflects just 20% of these and that the visible component of light is about 1/10th of the total light (for the sake of this argument I can be off by an order of magnitude... this is for illustration only).
At a distance of 200 m, these photons would have spread over a sphere with a surface of $4pi R^2 approx 500,000 m^2$, or $10^14$ photons per square meter per second.
If your pupil has a diameter of 4 mm, an area of $12 mm^2$, it will be hit by about $12cdot 10^8$ photons per second from one square meter of grey surface illuminated by the sun from 200 m away.
At that distance, the angular size of that object is about 1/200 th of a radian. "Normal" vision is defined as the ability to resolve objects that are about 5 minutes of arc (there are 60 minutes to a degree and about 57 degrees to a radian). in other words, you should be able to resolve 1/(57*(60/5)) or about 1/600 of a radian. That's still lots of photons...
Finally you ask "how do we distinguish what photons are reflected from what"? For this we have to thank the lens in our eye. A photon has a particular direction, and thanks to the lens its energy ends up on a particular part of the retina (this is what we call "focusing"). Photons from different directions end up in a different place. Nerves on the back of the retina tell us where the photons landed - and even what color they were. The visual cortex (part of the brain) uses that information to make a picture of the surrounding world in our mind.
It's nothing short of miraculous.
answered 8 hours ago
FlorisFloris
107k11189326
answered 8 hours ago
FlorisFloris
107k11189326
answered 8 hours ago
FlorisFloris
107k11189326
answered 8 hours ago
FlorisFloris
107k11189326
107k11189326
add a comment |
add a comment |
Lyy is a new contributor. Be nice, and check out our Code of Conduct.
Lyy is a new contributor. Be nice, and check out our Code of Conduct.
Lyy is a new contributor. Be nice, and check out our Code of Conduct.
Lyy is a new contributor. Be nice, and check out our Code of Conduct.
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