If light rays are parallel to each other , if they are from infinity, they are not gonna make an image because they won't intersect with each other. So we are not gonna see the image. Does that means we are not gonna see the light rays from infinity? Would it be just blank? Why do we assume that light rays are parallel in the first place?
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asked Jan 1, 2015 at 7:06
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$\begingroup$ probably you are talking about the light rays from the sun, which can be treated as a point source in most cases since it's far away from us. in this case the light rays can be treated as parallel for most practical purposes again because it's far away from us. $\endgroup$– M. ZengCommented Jan 1, 2015 at 7:25
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4$\begingroup$ This might help you: physics.stackexchange.com/q/155075 $\endgroup$– SensebeCommented Jan 1, 2015 at 11:05
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$\begingroup$ Two distinct lines meet at exactly one point in projective space. We define two lines to be parallel if that point is at infinity. This is a definition, not an assumption. $\endgroup$– WillOCommented Jan 1, 2015 at 15:34
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$\begingroup$ Possible duplicate of physics.stackexchange.com/q/155075 $\endgroup$– LeoCommented Nov 18, 2015 at 20:40
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4 Answers
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We consider light rays from infinity parallel because they are parallel. This can be shown using some high school level trignometry. Imagine any two lines that meet at a given point. As the intersection point is pushed further away, the lines become more parallel. The more mathematical proof follows from trigonometry. Consider an triangle who's vertices lie on the points A,B,C. It has angles A, B and C at the corresponding vertices. Let angles A and B be 90°. This implies that AC is parallel to BC. Also tan(B) by definition is CA/AB. Since tan(90°) is known to be infinity, it follows that CA/AB = infinity. Since AB is not zero, CA is infinity. Remember that because we considered ABC to form a triangle, C was the intersection of AC and BC. Since AC is infinity, the two parallel lines may said to meet at infinity.
As for your original question regarding image formation, the image of an object at infinity can be produced using a curved lens( such as the eyes) or curved mirror, as these have the property to focus light. They can refract parallel rays of light to pass through a single point. But on a plane mirror, which does not bend light, no virtual image would be formed. Keep in mind that there are many reasons why you can't see an actual object at infinity, but optics is more about approximation. For example, we can treat very far objects as infinitely away.
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3$\begingroup$ I think this answer can be made a lot better by including a picture of the mentioned triangle $\endgroup$ Commented Jan 1, 2015 at 14:43
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In reality, light rays are never perfectly parallel because the distance to their source point is always a finite distance.
However, if that distance is vast ( across the Universe, or just across a Galaxy ) then the distance traveled is vastly greater than the distance of separation of the two rays, so the rays are very nearly parallel.
So, in theory, if the distance traveled is approximately infinite, and the separation of the rays is close to zero, then we can call them parallel.
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Well, consider some wave fronts emerging from the light source. Now, as the distance from the source increases, so will the curve of the wave fronts increase. As we go farther and farther from the source, the curve of the wave fronts will become nearly straight. Light rays are considered to be falling normally on these curves and as with increased distance the curves become nearly straight, the light rays are assumed to be parallel to each other.
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It can be said with the help of this example given below.
See whenever we are standing on railway tracks and see long distance from same place, we observe that the railway tracks instead of being parallel meet together. We fell it but it's not true as parallel lines do not meet. Same is with it too. Actually we consider that light rays meet together at infinity
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$\begingroup$ This doesn't really seem to answer the question of why it's there case, only that it is the case. $\endgroup$ Commented Nov 27, 2018 at 16:43