1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric optics

  1. Jan 27, 2014 #1
    Why is the position of an image the intersection of 2+ rays?
  2. jcsd
  3. Jan 27, 2014 #2
    This one of those simple questions that require a lot of knowledge in order to provide a reasonably simple and correct answer. To be short its like this.

    Consider a simplest optical system: a flat mirror.
    Instead of an arrow (standard object for optical systems), we use small lamp that emits light in all directions uniformly. Some rays from the lamp reach the mirror and are reflected by it so they intersect behind the mirror in an imaginary point where the mirror image of the real lamp appears.

    But how do we see the image? We see the image the same way we see the object.
    Observer (a human or optical devices with image sensor) analyzes all the rays that reach its sensor and by "analyzing" their propagation direction it detects two points that emit light.
    Because some rays come directly from the real lamp and some rays are reflected by the mirror and appear like they are coming from another imaginary point -- image of the lamp.
    Thus mirror "fools" our observer to detect two points in space which emit light.
    If the mirror is of perfect quality (and no other object are reflected in it) it will be nearly impossible for observer to distinguish the lamp with its image from a pair of two small identical lamps but without a mirror between them.

    Generally, in order to make image of an object in some point in space, optical system should collect as many rays emitted by the object as possible and direct them into this point. But these rays are not stopped there, they continue to propagate. But now they propagate like they were emitted by the imaginary copy of the object.
    When observer (eye or camera) interacts with these rays, it can detect only their current propagation direction. Past "history" (all the refractions or reflections) of these rays is not detectable.
    Therefore observer is "fooled" to believe that these rays come from this imaginary point and believes that the object is right there.
    But in order to "fool" the observer to perceive the image of the object, optical system should first bring the rays in some point. Without this condition there will be no image to detect.

    If you try to build image of an object after, say, convex lens, you'll see that not just two, but all the rays that come from a single point of the object and refracted by the lens are collected in corresponding point of the image. In real world the more rays are collected by the lens the better quality of the image can be achieved. This is why large-diameter camera lenses are capable of capturing really nice photos. Such quality is way much harder to achieve using simple small-lens cameras (for example, in mobile phones). This is especially noticeable when object is very close to the lens (so called macro photography): the amount of light collected by big lens is many times larger comparing to small lens. But at the same time amount of light captured by big and small lenses and thus the final image quality is much closer when distant objects are imaged.

    P.S. Sorry for such a long story. You just caught me in "writing" mood
    Last edited: Jan 27, 2014
  4. Jan 27, 2014 #3
    Thanks a lot! Also, is the principal axis, since it's a radius, 90 degrees to the reflective side of a curved mirror? Why? And also, is any ray through centre of curvature 90 degrees to the reflective side? Why? And finally, do laws of reflection apply to curved (spherical) mirrors?
  5. Jan 27, 2014 #4
    See nice explanation here: http://en.wikipedia.org/wiki/Curved_mirror

    Of course they do apply. But locally :eek:)
    If you want to draw a ray reflected by a curved surface, first find the point where incident ray intersects with that surface. Then draw tangent plane to this surface at the intersection point.
    Now in the close vicinity of this intersection point curved mirror may be replaced with imaginary flat one which is given by the tangent plane. And you already know how to draw a ray reflected by a flat mirror ;o)
    One example is presented in the mentioned above wikipedia article on curved mirrors in "Ray transfer matrix of spherical mirrors" chapter (incident and refleted rays are in black color).

    Because if you check carefully how the reflection occurs (see my answer to the previous question) you'll see that any ray that passes through center of curvature hits the curved mirror exactly along the "local" perpendicular to it.
    Due to this such rays bounce exactly back.
  6. Jan 29, 2014 #5


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    Why would you ever think that something described as a "Law" would not be pretty universal? If you carefully read the (two basic) lawn there is no mention of a Plane surface - just the Normal to the surface. This makes the law apply to any surface. But the important thing is the first law which says the incident ray, normal and reflected ray are all in the same plane. Interpreting this for what it really means and applying it to any arbitrary curved surface can be very instructive.

    The principle axis is the axis of symmetry of the curved face (mirror or lens) which could be a sphere or a parabola (or any shape with basic rotational symmetry) - not just an arbitrarily chosen line.

    I think you will get a lot of help by looking at as many diagrams as you can. Verbal descriptions can be a bit limiting and the web is full of them.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook