Why are we able to see air bubbles under water?
$begingroup$
Title is self explanatory.
I'm assuming both water and air are transparent.
So, if they are true, how can I clearly distinguish an air bubble under water?
optics water air bubble
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
$endgroup$
add a comment |
$begingroup$
Title is self explanatory.
I'm assuming both water and air are transparent.
So, if they are true, how can I clearly distinguish an air bubble under water?
optics water air bubble
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
$endgroup$
4
$begingroup$
I would use the term transparent instead of invisible in the question
$endgroup$
– user1936752
Dec 23 '18 at 20:22
1
$begingroup$
Ironically we can see the water in your picture. Perhaps it has to do with light.
$endgroup$
– KingDuken
Dec 24 '18 at 21:48
3
$begingroup$
Note that it is the same reason as why we can see water drops in air.
$endgroup$
– Jan Hudec
Dec 24 '18 at 22:16
add a comment |
$begingroup$
Title is self explanatory.
I'm assuming both water and air are transparent.
So, if they are true, how can I clearly distinguish an air bubble under water?
optics water air bubble
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
$endgroup$
Title is self explanatory.
I'm assuming both water and air are transparent.
So, if they are true, how can I clearly distinguish an air bubble under water?
optics water air bubble
optics water air bubble
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
edited Jan 7 at 19:45
ihavenoidea
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
asked Dec 23 '18 at 19:49
ihavenoideaihavenoidea
2237
2237
4
$begingroup$
I would use the term transparent instead of invisible in the question
$endgroup$
– user1936752
Dec 23 '18 at 20:22
1
$begingroup$
Ironically we can see the water in your picture. Perhaps it has to do with light.
$endgroup$
– KingDuken
Dec 24 '18 at 21:48
3
$begingroup$
Note that it is the same reason as why we can see water drops in air.
$endgroup$
– Jan Hudec
Dec 24 '18 at 22:16
add a comment |
4
$begingroup$
I would use the term transparent instead of invisible in the question
$endgroup$
– user1936752
Dec 23 '18 at 20:22
1
$begingroup$
Ironically we can see the water in your picture. Perhaps it has to do with light.
$endgroup$
– KingDuken
Dec 24 '18 at 21:48
3
$begingroup$
Note that it is the same reason as why we can see water drops in air.
$endgroup$
– Jan Hudec
Dec 24 '18 at 22:16
4
4
$begingroup$
I would use the term transparent instead of invisible in the question
$endgroup$
– user1936752
Dec 23 '18 at 20:22
$begingroup$
I would use the term transparent instead of invisible in the question
$endgroup$
– user1936752
Dec 23 '18 at 20:22
1
1
$begingroup$
Ironically we can see the water in your picture. Perhaps it has to do with light.
$endgroup$
– KingDuken
Dec 24 '18 at 21:48
$begingroup$
Ironically we can see the water in your picture. Perhaps it has to do with light.
$endgroup$
– KingDuken
Dec 24 '18 at 21:48
3
3
$begingroup$
Note that it is the same reason as why we can see water drops in air.
$endgroup$
– Jan Hudec
Dec 24 '18 at 22:16
$begingroup$
Note that it is the same reason as why we can see water drops in air.
$endgroup$
– Jan Hudec
Dec 24 '18 at 22:16
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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:
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
$endgroup$
1
$begingroup$
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?
$endgroup$
– MarkTO
Dec 24 '18 at 17:18
$begingroup$
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
$endgroup$
– niels nielsen
Dec 24 '18 at 20:19
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
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:
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
$endgroup$
add a comment |
$begingroup$
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:
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
$endgroup$
add a comment |
$begingroup$
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:
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
$endgroup$
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:
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
edited Dec 23 '18 at 22:34
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
answered Dec 23 '18 at 21:50
Chiral AnomalyChiral Anomaly
13.1k21744
13.1k21744
add a comment |
add a comment |
$begingroup$
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 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
$endgroup$
1
$begingroup$
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?
$endgroup$
– MarkTO
Dec 24 '18 at 17:18
$begingroup$
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
$endgroup$
– niels nielsen
Dec 24 '18 at 20:19
add a comment |
$begingroup$
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 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
$endgroup$
1
$begingroup$
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?
$endgroup$
– MarkTO
Dec 24 '18 at 17:18
$begingroup$
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
$endgroup$
– niels nielsen
Dec 24 '18 at 20:19
add a comment |
$begingroup$
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 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
$endgroup$
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 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
answered Dec 23 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
answered Dec 23 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
answered Dec 23 '18 at 20:21
niels nielsenniels nielsen
21.1k53062
21.1k53062
1
$begingroup$
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?
$endgroup$
– MarkTO
Dec 24 '18 at 17:18
$begingroup$
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
$endgroup$
– niels nielsen
Dec 24 '18 at 20:19
add a comment |
1
$begingroup$
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?
$endgroup$
– MarkTO
Dec 24 '18 at 17:18
$begingroup$
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
$endgroup$
– niels nielsen
Dec 24 '18 at 20:19
1
1
$begingroup$
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?
$endgroup$
– MarkTO
Dec 24 '18 at 17:18
$begingroup$
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?
$endgroup$
– MarkTO
Dec 24 '18 at 17:18
$begingroup$
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
$endgroup$
– niels nielsen
Dec 24 '18 at 20:19
$begingroup$
No, it means that the bubble will behave like a bent mirror- which is exactly what the bubbles in the picture above look like.
$endgroup$
– niels nielsen
Dec 24 '18 at 20:19
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
$endgroup$
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.
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
edited Dec 25 '18 at 1:25
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
answered Dec 25 '18 at 0:08
Gus MichelGus Michel
563
563
add a comment |
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
$endgroup$
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.
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
answered Dec 25 '18 at 1:16
Es CarterEs Carter
1
1
add a comment |
add a comment |
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$begingroup$
I would use the term transparent instead of invisible in the question
$endgroup$
– user1936752
Dec 23 '18 at 20:22
1
$begingroup$
Ironically we can see the water in your picture. Perhaps it has to do with light.
$endgroup$
– KingDuken
Dec 24 '18 at 21:48
3
$begingroup$
Note that it is the same reason as why we can see water drops in air.
$endgroup$
– Jan Hudec
Dec 24 '18 at 22:16