Why are we able to see air bubbles under water?
Title is self explanatory.
I'm assuming both water and air are transparent. (Are these premisses false?)
So, if they are true, how can I clearly distinguish an air bubble under water?
Ps: I don't have a strong scientific background (high school level only), sorry if it's a silly question.
optics water air bubble
New contributor
add a comment |
Title is self explanatory.
I'm assuming both water and air are transparent. (Are these premisses false?)
So, if they are true, how can I clearly distinguish an air bubble under water?
Ps: I don't have a strong scientific background (high school level only), sorry if it's a silly question.
optics water air bubble
New contributor
4
I would use the term transparent instead of invisible in the question
– user1936752
Dec 23 at 20:22
1
Ironically we can see the water in your picture. Perhaps it has to do with light.
– KingDuken
Dec 24 at 21:48
3
Note that it is the same reason as why we can see water drops in air.
– Jan Hudec
Dec 24 at 22:16
@user1936752 You're right, thanks
– ihavenoidea
Dec 25 at 23:01
add a comment |
Title is self explanatory.
I'm assuming both water and air are transparent. (Are these premisses false?)
So, if they are true, how can I clearly distinguish an air bubble under water?
Ps: I don't have a strong scientific background (high school level only), sorry if it's a silly question.
optics water air bubble
New contributor
Title is self explanatory.
I'm assuming both water and air are transparent. (Are these premisses false?)
So, if they are true, how can I clearly distinguish an air bubble under water?
Ps: I don't have a strong scientific background (high school level only), sorry if it's a silly question.
optics water air bubble
optics water air bubble
New contributor
New contributor
edited Dec 25 at 23:01
New contributor
asked Dec 23 at 19:49
ihavenoidea
966
966
New contributor
New contributor
4
I would use the term transparent instead of invisible in the question
– user1936752
Dec 23 at 20:22
1
Ironically we can see the water in your picture. Perhaps it has to do with light.
– KingDuken
Dec 24 at 21:48
3
Note that it is the same reason as why we can see water drops in air.
– Jan Hudec
Dec 24 at 22:16
@user1936752 You're right, thanks
– ihavenoidea
Dec 25 at 23:01
add a comment |
4
I would use the term transparent instead of invisible in the question
– user1936752
Dec 23 at 20:22
1
Ironically we can see the water in your picture. Perhaps it has to do with light.
– KingDuken
Dec 24 at 21:48
3
Note that it is the same reason as why we can see water drops in air.
– Jan Hudec
Dec 24 at 22:16
@user1936752 You're right, thanks
– ihavenoidea
Dec 25 at 23:01
4
4
I would use the term transparent instead of invisible in the question
– user1936752
Dec 23 at 20:22
I would use the term transparent instead of invisible in the question
– user1936752
Dec 23 at 20:22
1
1
Ironically we can see the water in your picture. Perhaps it has to do with light.
– KingDuken
Dec 24 at 21:48
Ironically we can see the water in your picture. Perhaps it has to do with light.
– KingDuken
Dec 24 at 21:48
3
3
Note that it is the same reason as why we can see water drops in air.
– Jan Hudec
Dec 24 at 22:16
Note that it is the same reason as why we can see water drops in air.
– Jan Hudec
Dec 24 at 22:16
@user1936752 You're right, thanks
– ihavenoidea
Dec 25 at 23:01
@user1936752 You're right, thanks
– ihavenoidea
Dec 25 at 23:01
add a comment |
4 Answers
4
active
oldest
votes
Air and water are both transparent to a good enough approximation. However, light travels more slowly in water: the speed of light in air is about 33% faster than in water. As a result, when light passes from one medium to the other, it is partly reflected and partly refracted (bent). For the refracted part, the general rule for determining the bending angle is called Snell's law, which can be expressed like this:
$$
frac{sintheta_text{w}}{sintheta_text{a}}=frac{v_text{w}}{v_text{a}}
approx frac{1}{1.33}
tag{1}
$$
where $v_text{w}$ and $v_text{a}$ are the speed of light in water and air, respectively, and where $theta_text{w}$ and $theta_text{a}$ are the angles of the light ray relative to a line perpendicular to the surface, on the water side and on the air side, respectively.
If the angle on the water side is $theta_text{w} gtrsim 49^circ$, then equation (1) does not have any solution: there is no air-side angle $theta_text{a}$ that satisfies the equation. In this case, as niels nielsen indicated, light propagating inside the water will be completely reflected at the water-air interface. So the rim of the bubble acts like a mirror: if you do a reverse ray-trace from your eye back to near the rim of an air bubble in the water, the angle between the ray and the line perpendicular to the surface of the bubble will be greater than $49^circ$ (this defines what "near the rim" means), so that part of the bubble acts like a mirror for light coming from those angles, as illustrated here:
add a comment |
You can see light reflected off of the surface of a submerged bubble because the index of refraction of the air inside the bubble is different from that of the water that surrounds the bubble.
That difference, if great enough, will turn a bubble surface into a mirror for light rays that approach it from certain directions, thereby making it easy to see.
This condition is easily met for the combination of air and water.
A google search on "refraction" and "total internal reflection" will furnish more examples of this, and explain the math behind it.
1
You said - "...light propagating inside the water will be completely reflected at the water-air interface" Does this mean that it is completely dark inside an air bubble underwater?
– MarkTO
Dec 24 at 17:18
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
– niels nielsen
Dec 24 at 20:19
add a comment |
The inside of the bubble is not dark because only the light making a more glancing intersection than 49 degrees is completely reflected. Light that hits near the center of the bubble with respect to the direction it is traveling will be mostly transmitted into the bubble, illuminating the interior.
New contributor
add a comment |
For the exact same reason why you can see rain. Light waves hit it and the direction of propagation(travel) of the wave changes because it goes from one medium to another. Its movement causes the refraction to happen differently making it different from it's surroundings and visible to you.
New contributor
add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
votes
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active
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Air and water are both transparent to a good enough approximation. However, light travels more slowly in water: the speed of light in air is about 33% faster than in water. As a result, when light passes from one medium to the other, it is partly reflected and partly refracted (bent). For the refracted part, the general rule for determining the bending angle is called Snell's law, which can be expressed like this:
$$
frac{sintheta_text{w}}{sintheta_text{a}}=frac{v_text{w}}{v_text{a}}
approx frac{1}{1.33}
tag{1}
$$
where $v_text{w}$ and $v_text{a}$ are the speed of light in water and air, respectively, and where $theta_text{w}$ and $theta_text{a}$ are the angles of the light ray relative to a line perpendicular to the surface, on the water side and on the air side, respectively.
If the angle on the water side is $theta_text{w} gtrsim 49^circ$, then equation (1) does not have any solution: there is no air-side angle $theta_text{a}$ that satisfies the equation. In this case, as niels nielsen indicated, light propagating inside the water will be completely reflected at the water-air interface. So the rim of the bubble acts like a mirror: if you do a reverse ray-trace from your eye back to near the rim of an air bubble in the water, the angle between the ray and the line perpendicular to the surface of the bubble will be greater than $49^circ$ (this defines what "near the rim" means), so that part of the bubble acts like a mirror for light coming from those angles, as illustrated here:
add a comment |
Air and water are both transparent to a good enough approximation. However, light travels more slowly in water: the speed of light in air is about 33% faster than in water. As a result, when light passes from one medium to the other, it is partly reflected and partly refracted (bent). For the refracted part, the general rule for determining the bending angle is called Snell's law, which can be expressed like this:
$$
frac{sintheta_text{w}}{sintheta_text{a}}=frac{v_text{w}}{v_text{a}}
approx frac{1}{1.33}
tag{1}
$$
where $v_text{w}$ and $v_text{a}$ are the speed of light in water and air, respectively, and where $theta_text{w}$ and $theta_text{a}$ are the angles of the light ray relative to a line perpendicular to the surface, on the water side and on the air side, respectively.
If the angle on the water side is $theta_text{w} gtrsim 49^circ$, then equation (1) does not have any solution: there is no air-side angle $theta_text{a}$ that satisfies the equation. In this case, as niels nielsen indicated, light propagating inside the water will be completely reflected at the water-air interface. So the rim of the bubble acts like a mirror: if you do a reverse ray-trace from your eye back to near the rim of an air bubble in the water, the angle between the ray and the line perpendicular to the surface of the bubble will be greater than $49^circ$ (this defines what "near the rim" means), so that part of the bubble acts like a mirror for light coming from those angles, as illustrated here:
add a comment |
Air and water are both transparent to a good enough approximation. However, light travels more slowly in water: the speed of light in air is about 33% faster than in water. As a result, when light passes from one medium to the other, it is partly reflected and partly refracted (bent). For the refracted part, the general rule for determining the bending angle is called Snell's law, which can be expressed like this:
$$
frac{sintheta_text{w}}{sintheta_text{a}}=frac{v_text{w}}{v_text{a}}
approx frac{1}{1.33}
tag{1}
$$
where $v_text{w}$ and $v_text{a}$ are the speed of light in water and air, respectively, and where $theta_text{w}$ and $theta_text{a}$ are the angles of the light ray relative to a line perpendicular to the surface, on the water side and on the air side, respectively.
If the angle on the water side is $theta_text{w} gtrsim 49^circ$, then equation (1) does not have any solution: there is no air-side angle $theta_text{a}$ that satisfies the equation. In this case, as niels nielsen indicated, light propagating inside the water will be completely reflected at the water-air interface. So the rim of the bubble acts like a mirror: if you do a reverse ray-trace from your eye back to near the rim of an air bubble in the water, the angle between the ray and the line perpendicular to the surface of the bubble will be greater than $49^circ$ (this defines what "near the rim" means), so that part of the bubble acts like a mirror for light coming from those angles, as illustrated here:
Air and water are both transparent to a good enough approximation. However, light travels more slowly in water: the speed of light in air is about 33% faster than in water. As a result, when light passes from one medium to the other, it is partly reflected and partly refracted (bent). For the refracted part, the general rule for determining the bending angle is called Snell's law, which can be expressed like this:
$$
frac{sintheta_text{w}}{sintheta_text{a}}=frac{v_text{w}}{v_text{a}}
approx frac{1}{1.33}
tag{1}
$$
where $v_text{w}$ and $v_text{a}$ are the speed of light in water and air, respectively, and where $theta_text{w}$ and $theta_text{a}$ are the angles of the light ray relative to a line perpendicular to the surface, on the water side and on the air side, respectively.
If the angle on the water side is $theta_text{w} gtrsim 49^circ$, then equation (1) does not have any solution: there is no air-side angle $theta_text{a}$ that satisfies the equation. In this case, as niels nielsen indicated, light propagating inside the water will be completely reflected at the water-air interface. So the rim of the bubble acts like a mirror: if you do a reverse ray-trace from your eye back to near the rim of an air bubble in the water, the angle between the ray and the line perpendicular to the surface of the bubble will be greater than $49^circ$ (this defines what "near the rim" means), so that part of the bubble acts like a mirror for light coming from those angles, as illustrated here:
edited Dec 23 at 22:34
answered Dec 23 at 21:50
Dan Yand
6,7271830
6,7271830
add a comment |
add a comment |
You can see light reflected off of the surface of a submerged bubble because the index of refraction of the air inside the bubble is different from that of the water that surrounds the bubble.
That difference, if great enough, will turn a bubble surface into a mirror for light rays that approach it from certain directions, thereby making it easy to see.
This condition is easily met for the combination of air and water.
A google search on "refraction" and "total internal reflection" will furnish more examples of this, and explain the math behind it.
1
You said - "...light propagating inside the water will be completely reflected at the water-air interface" Does this mean that it is completely dark inside an air bubble underwater?
– MarkTO
Dec 24 at 17:18
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
– niels nielsen
Dec 24 at 20:19
add a comment |
You can see light reflected off of the surface of a submerged bubble because the index of refraction of the air inside the bubble is different from that of the water that surrounds the bubble.
That difference, if great enough, will turn a bubble surface into a mirror for light rays that approach it from certain directions, thereby making it easy to see.
This condition is easily met for the combination of air and water.
A google search on "refraction" and "total internal reflection" will furnish more examples of this, and explain the math behind it.
1
You said - "...light propagating inside the water will be completely reflected at the water-air interface" Does this mean that it is completely dark inside an air bubble underwater?
– MarkTO
Dec 24 at 17:18
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
– niels nielsen
Dec 24 at 20:19
add a comment |
You can see light reflected off of the surface of a submerged bubble because the index of refraction of the air inside the bubble is different from that of the water that surrounds the bubble.
That difference, if great enough, will turn a bubble surface into a mirror for light rays that approach it from certain directions, thereby making it easy to see.
This condition is easily met for the combination of air and water.
A google search on "refraction" and "total internal reflection" will furnish more examples of this, and explain the math behind it.
You can see light reflected off of the surface of a submerged bubble because the index of refraction of the air inside the bubble is different from that of the water that surrounds the bubble.
That difference, if great enough, will turn a bubble surface into a mirror for light rays that approach it from certain directions, thereby making it easy to see.
This condition is easily met for the combination of air and water.
A google search on "refraction" and "total internal reflection" will furnish more examples of this, and explain the math behind it.
answered Dec 23 at 20:21
niels nielsen
16.1k42653
16.1k42653
1
You said - "...light propagating inside the water will be completely reflected at the water-air interface" Does this mean that it is completely dark inside an air bubble underwater?
– MarkTO
Dec 24 at 17:18
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
– niels nielsen
Dec 24 at 20:19
add a comment |
1
You said - "...light propagating inside the water will be completely reflected at the water-air interface" Does this mean that it is completely dark inside an air bubble underwater?
– MarkTO
Dec 24 at 17:18
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
– niels nielsen
Dec 24 at 20:19
1
1
You said - "...light propagating inside the water will be completely reflected at the water-air interface" Does this mean that it is completely dark inside an air bubble underwater?
– MarkTO
Dec 24 at 17:18
You said - "...light propagating inside the water will be completely reflected at the water-air interface" Does this mean that it is completely dark inside an air bubble underwater?
– MarkTO
Dec 24 at 17:18
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
– niels nielsen
Dec 24 at 20:19
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
– niels nielsen
Dec 24 at 20:19
add a comment |
The inside of the bubble is not dark because only the light making a more glancing intersection than 49 degrees is completely reflected. Light that hits near the center of the bubble with respect to the direction it is traveling will be mostly transmitted into the bubble, illuminating the interior.
New contributor
add a comment |
The inside of the bubble is not dark because only the light making a more glancing intersection than 49 degrees is completely reflected. Light that hits near the center of the bubble with respect to the direction it is traveling will be mostly transmitted into the bubble, illuminating the interior.
New contributor
add a comment |
The inside of the bubble is not dark because only the light making a more glancing intersection than 49 degrees is completely reflected. Light that hits near the center of the bubble with respect to the direction it is traveling will be mostly transmitted into the bubble, illuminating the interior.
New contributor
The inside of the bubble is not dark because only the light making a more glancing intersection than 49 degrees is completely reflected. Light that hits near the center of the bubble with respect to the direction it is traveling will be mostly transmitted into the bubble, illuminating the interior.
New contributor
edited Dec 25 at 1:25
New contributor
answered Dec 25 at 0:08
Gus Michel
563
563
New contributor
New contributor
add a comment |
add a comment |
For the exact same reason why you can see rain. Light waves hit it and the direction of propagation(travel) of the wave changes because it goes from one medium to another. Its movement causes the refraction to happen differently making it different from it's surroundings and visible to you.
New contributor
add a comment |
For the exact same reason why you can see rain. Light waves hit it and the direction of propagation(travel) of the wave changes because it goes from one medium to another. Its movement causes the refraction to happen differently making it different from it's surroundings and visible to you.
New contributor
add a comment |
For the exact same reason why you can see rain. Light waves hit it and the direction of propagation(travel) of the wave changes because it goes from one medium to another. Its movement causes the refraction to happen differently making it different from it's surroundings and visible to you.
New contributor
For the exact same reason why you can see rain. Light waves hit it and the direction of propagation(travel) of the wave changes because it goes from one medium to another. Its movement causes the refraction to happen differently making it different from it's surroundings and visible to you.
New contributor
New contributor
answered Dec 25 at 1:16
Es Carter
1
1
New contributor
New contributor
add a comment |
add a comment |
ihavenoidea is a new contributor. Be nice, and check out our Code of Conduct.
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4
I would use the term transparent instead of invisible in the question
– user1936752
Dec 23 at 20:22
1
Ironically we can see the water in your picture. Perhaps it has to do with light.
– KingDuken
Dec 24 at 21:48
3
Note that it is the same reason as why we can see water drops in air.
– Jan Hudec
Dec 24 at 22:16
@user1936752 You're right, thanks
– ihavenoidea
Dec 25 at 23:01