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Why do reflections stretch like that?
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When light reflects off a "flat ground" surface, the reflection stretches vertically but not horizontally. When light reflects off a "vertical wall" surface, the reflection stretches horizontally but not vertically. Reflections seem to stretch in the direction that the reflector plane "goes towards" you, if you know what I mean.
Why do reflections "stretch" like that?
visible-light
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When light reflects off a "flat ground" surface, the reflection stretches vertically but not horizontally. When light reflects off a "vertical wall" surface, the reflection stretches horizontally but not vertically. Reflections seem to stretch in the direction that the reflector plane "goes towards" you, if you know what I mean.
Why do reflections "stretch" like that?
visible-light
asked 2 hours ago
clickbait
162210
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When light reflects off a "flat ground" surface, the reflection stretches vertically but not horizontally. When light reflects off a "vertical wall" surface, the reflection stretches horizontally but not vertically. Reflections seem to stretch in the direction that the reflector plane "goes towards" you, if you know what I mean.
Why do reflections "stretch" like that?
visible-light
asked 2 hours ago
clickbait
162210
When light reflects off a "flat ground" surface, the reflection stretches vertically but not horizontally. When light reflects off a "vertical wall" surface, the reflection stretches horizontally but not vertically. Reflections seem to stretch in the direction that the reflector plane "goes towards" you, if you know what I mean.
Why do reflections "stretch" like that?
visible-light
visible-light
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clickbait
162210
asked 2 hours ago
clickbait
162210
edited 1 hour ago
asked 2 hours ago
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asked 2 hours ago
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It is because light reflects off of the whole distance on the ground (or water) along to you where the angle is working so that light bounces into your eyes (or the camera).
It is not stretched out, it is that light hits everything from the Sun (in this picture it is Sunlight), and you only see along the line to be reflected that has an angle of reflection that bounces off the photons in the camera.
answered 1 hour ago
Ãrpád Szendrei
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Because the surface isn't completely smooth the light is bounced back to you at different points. If the surface was totally smooth (like a mirror) you would see a reflection that is not stretched out!
answered 2 hours ago
Mio
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This hits on the single biggest favcor in the images in the question. However, you get geometric distortions of extended source even with smooth mirrors (though they tend to be much smaller than those images),
– dmckee♦
33 mins ago
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There are, generally, two types of reflections, diffuse (a wall or anything matte) and specular (a mirror or anything shiny). Most surfaces have a little of both.
Diffusely reflected light spreads out in all directions, therefore most objects could be seen from any direction.
Specularly reflected light follows the law of reflection: the angle of reflection is equal to the angle of incidence and both incident and reflected rays lie in the plane normal to the reflecting surface.
If a light source produced a narrow beam (like a laser), its specular reflection could be visible only from a particular observation angle, i.e., the eye of the observer would have to be located somewhere along the reflected beam.
If the light source emitted a wide beam, its specular reflection would be visible from various angles within that beam, each incident and reflected ray still confined to a plane normal to the reflecting surface. The overall reflection, would be a mirror image of the source.
On both pictures in the post, the reflecting surface is pretty specular on a small scale, but, due to unevenness and imperfections of the road surface or ripples on the surface of the water, it acts as diffuse on a larger scale: that is the reason we can see the road and the water on both sides of the bright vertical stretches.
Despite the imperfections and ripples, though, we still see bright vertical stretches, limited, again, to the normal planes formed by pairs of incident and reflected rays. This is because, at any distance (within a wide range of distances), each ripple (or other kind of imperfection) will have a small patch on its surface, which will be oriented exactly to support a specular reflection between a particular point of the light source and the eye of an observer.
When such multitude of tiny reflections are viewed from a distance, they form an almost continuous bright stretch evident on both pictures. Again, the reason those stretches appear to be vertical is because such is a projection of the intersection line between the planes containing incident and reflected rays and the horizontal surface of the road or the water.
answered 5 mins ago
V.F.
8,8122923
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It is because light reflects off of the whole distance on the ground (or water) along to you where the angle is working so that light bounces into your eyes (or the camera).
It is not stretched out, it is that light hits everything from the Sun (in this picture it is Sunlight), and you only see along the line to be reflected that has an angle of reflection that bounces off the photons in the camera.
answered 1 hour ago
Ãrpád Szendrei
3,1561421
add a comment |Â
up vote
2
down vote
It is because light reflects off of the whole distance on the ground (or water) along to you where the angle is working so that light bounces into your eyes (or the camera).
It is not stretched out, it is that light hits everything from the Sun (in this picture it is Sunlight), and you only see along the line to be reflected that has an angle of reflection that bounces off the photons in the camera.
answered 1 hour ago
Ãrpád Szendrei
3,1561421
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is because light reflects off of the whole distance on the ground (or water) along to you where the angle is working so that light bounces into your eyes (or the camera).
It is not stretched out, it is that light hits everything from the Sun (in this picture it is Sunlight), and you only see along the line to be reflected that has an angle of reflection that bounces off the photons in the camera.
answered 1 hour ago
Ãrpád Szendrei
3,1561421
It is because light reflects off of the whole distance on the ground (or water) along to you where the angle is working so that light bounces into your eyes (or the camera).
It is not stretched out, it is that light hits everything from the Sun (in this picture it is Sunlight), and you only see along the line to be reflected that has an angle of reflection that bounces off the photons in the camera.
answered 1 hour ago
Ãrpád Szendrei
3,1561421
answered 1 hour ago
Ãrpád Szendrei
3,1561421
answered 1 hour ago
Ãrpád Szendrei
3,1561421
answered 1 hour ago
Ãrpád Szendrei
3,1561421
3,1561421
add a comment |Â
add a comment |Â
up vote
1
down vote
Because the surface isn't completely smooth the light is bounced back to you at different points. If the surface was totally smooth (like a mirror) you would see a reflection that is not stretched out!
answered 2 hours ago
Mio
342
New contributor
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This hits on the single biggest favcor in the images in the question. However, you get geometric distortions of extended source even with smooth mirrors (though they tend to be much smaller than those images),
– dmckee♦
33 mins ago
add a comment |Â
up vote
1
down vote
Because the surface isn't completely smooth the light is bounced back to you at different points. If the surface was totally smooth (like a mirror) you would see a reflection that is not stretched out!
answered 2 hours ago
Mio
342
New contributor
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This hits on the single biggest favcor in the images in the question. However, you get geometric distortions of extended source even with smooth mirrors (though they tend to be much smaller than those images),
– dmckee♦
33 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Because the surface isn't completely smooth the light is bounced back to you at different points. If the surface was totally smooth (like a mirror) you would see a reflection that is not stretched out!
answered 2 hours ago
Mio
342
New contributor
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Because the surface isn't completely smooth the light is bounced back to you at different points. If the surface was totally smooth (like a mirror) you would see a reflection that is not stretched out!
answered 2 hours ago
Mio
342
New contributor
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Mio
342
New contributor
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Mio
342
answered 2 hours ago
Mio
342
342
New contributor
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This hits on the single biggest favcor in the images in the question. However, you get geometric distortions of extended source even with smooth mirrors (though they tend to be much smaller than those images),
– dmckee♦
33 mins ago
add a comment |Â
This hits on the single biggest favcor in the images in the question. However, you get geometric distortions of extended source even with smooth mirrors (though they tend to be much smaller than those images),
– dmckee♦
33 mins ago
This hits on the single biggest favcor in the images in the question. However, you get geometric distortions of extended source even with smooth mirrors (though they tend to be much smaller than those images),
– dmckee♦
33 mins ago
This hits on the single biggest favcor in the images in the question. However, you get geometric distortions of extended source even with smooth mirrors (though they tend to be much smaller than those images),
– dmckee♦
33 mins ago
add a comment |Â
up vote
0
down vote
There are, generally, two types of reflections, diffuse (a wall or anything matte) and specular (a mirror or anything shiny). Most surfaces have a little of both.
Diffusely reflected light spreads out in all directions, therefore most objects could be seen from any direction.
Specularly reflected light follows the law of reflection: the angle of reflection is equal to the angle of incidence and both incident and reflected rays lie in the plane normal to the reflecting surface.
If a light source produced a narrow beam (like a laser), its specular reflection could be visible only from a particular observation angle, i.e., the eye of the observer would have to be located somewhere along the reflected beam.
If the light source emitted a wide beam, its specular reflection would be visible from various angles within that beam, each incident and reflected ray still confined to a plane normal to the reflecting surface. The overall reflection, would be a mirror image of the source.
On both pictures in the post, the reflecting surface is pretty specular on a small scale, but, due to unevenness and imperfections of the road surface or ripples on the surface of the water, it acts as diffuse on a larger scale: that is the reason we can see the road and the water on both sides of the bright vertical stretches.
Despite the imperfections and ripples, though, we still see bright vertical stretches, limited, again, to the normal planes formed by pairs of incident and reflected rays. This is because, at any distance (within a wide range of distances), each ripple (or other kind of imperfection) will have a small patch on its surface, which will be oriented exactly to support a specular reflection between a particular point of the light source and the eye of an observer.
When such multitude of tiny reflections are viewed from a distance, they form an almost continuous bright stretch evident on both pictures. Again, the reason those stretches appear to be vertical is because such is a projection of the intersection line between the planes containing incident and reflected rays and the horizontal surface of the road or the water.
answered 5 mins ago
V.F.
8,8122923
add a comment |Â
up vote
0
down vote
There are, generally, two types of reflections, diffuse (a wall or anything matte) and specular (a mirror or anything shiny). Most surfaces have a little of both.
Diffusely reflected light spreads out in all directions, therefore most objects could be seen from any direction.
Specularly reflected light follows the law of reflection: the angle of reflection is equal to the angle of incidence and both incident and reflected rays lie in the plane normal to the reflecting surface.
If a light source produced a narrow beam (like a laser), its specular reflection could be visible only from a particular observation angle, i.e., the eye of the observer would have to be located somewhere along the reflected beam.
If the light source emitted a wide beam, its specular reflection would be visible from various angles within that beam, each incident and reflected ray still confined to a plane normal to the reflecting surface. The overall reflection, would be a mirror image of the source.
On both pictures in the post, the reflecting surface is pretty specular on a small scale, but, due to unevenness and imperfections of the road surface or ripples on the surface of the water, it acts as diffuse on a larger scale: that is the reason we can see the road and the water on both sides of the bright vertical stretches.
Despite the imperfections and ripples, though, we still see bright vertical stretches, limited, again, to the normal planes formed by pairs of incident and reflected rays. This is because, at any distance (within a wide range of distances), each ripple (or other kind of imperfection) will have a small patch on its surface, which will be oriented exactly to support a specular reflection between a particular point of the light source and the eye of an observer.
When such multitude of tiny reflections are viewed from a distance, they form an almost continuous bright stretch evident on both pictures. Again, the reason those stretches appear to be vertical is because such is a projection of the intersection line between the planes containing incident and reflected rays and the horizontal surface of the road or the water.
answered 5 mins ago
V.F.
8,8122923
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There are, generally, two types of reflections, diffuse (a wall or anything matte) and specular (a mirror or anything shiny). Most surfaces have a little of both.
Diffusely reflected light spreads out in all directions, therefore most objects could be seen from any direction.
Specularly reflected light follows the law of reflection: the angle of reflection is equal to the angle of incidence and both incident and reflected rays lie in the plane normal to the reflecting surface.
If a light source produced a narrow beam (like a laser), its specular reflection could be visible only from a particular observation angle, i.e., the eye of the observer would have to be located somewhere along the reflected beam.
If the light source emitted a wide beam, its specular reflection would be visible from various angles within that beam, each incident and reflected ray still confined to a plane normal to the reflecting surface. The overall reflection, would be a mirror image of the source.
On both pictures in the post, the reflecting surface is pretty specular on a small scale, but, due to unevenness and imperfections of the road surface or ripples on the surface of the water, it acts as diffuse on a larger scale: that is the reason we can see the road and the water on both sides of the bright vertical stretches.
Despite the imperfections and ripples, though, we still see bright vertical stretches, limited, again, to the normal planes formed by pairs of incident and reflected rays. This is because, at any distance (within a wide range of distances), each ripple (or other kind of imperfection) will have a small patch on its surface, which will be oriented exactly to support a specular reflection between a particular point of the light source and the eye of an observer.
When such multitude of tiny reflections are viewed from a distance, they form an almost continuous bright stretch evident on both pictures. Again, the reason those stretches appear to be vertical is because such is a projection of the intersection line between the planes containing incident and reflected rays and the horizontal surface of the road or the water.
answered 5 mins ago
V.F.
8,8122923
There are, generally, two types of reflections, diffuse (a wall or anything matte) and specular (a mirror or anything shiny). Most surfaces have a little of both.
Diffusely reflected light spreads out in all directions, therefore most objects could be seen from any direction.
Specularly reflected light follows the law of reflection: the angle of reflection is equal to the angle of incidence and both incident and reflected rays lie in the plane normal to the reflecting surface.
If a light source produced a narrow beam (like a laser), its specular reflection could be visible only from a particular observation angle, i.e., the eye of the observer would have to be located somewhere along the reflected beam.
If the light source emitted a wide beam, its specular reflection would be visible from various angles within that beam, each incident and reflected ray still confined to a plane normal to the reflecting surface. The overall reflection, would be a mirror image of the source.
On both pictures in the post, the reflecting surface is pretty specular on a small scale, but, due to unevenness and imperfections of the road surface or ripples on the surface of the water, it acts as diffuse on a larger scale: that is the reason we can see the road and the water on both sides of the bright vertical stretches.
Despite the imperfections and ripples, though, we still see bright vertical stretches, limited, again, to the normal planes formed by pairs of incident and reflected rays. This is because, at any distance (within a wide range of distances), each ripple (or other kind of imperfection) will have a small patch on its surface, which will be oriented exactly to support a specular reflection between a particular point of the light source and the eye of an observer.
When such multitude of tiny reflections are viewed from a distance, they form an almost continuous bright stretch evident on both pictures. Again, the reason those stretches appear to be vertical is because such is a projection of the intersection line between the planes containing incident and reflected rays and the horizontal surface of the road or the water.
answered 5 mins ago
V.F.
8,8122923
answered 5 mins ago
V.F.
8,8122923
answered 5 mins ago
V.F.
8,8122923
answered 5 mins ago
V.F.
8,8122923
8,8122923
add a comment |Â
add a comment |Â
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