When two light rays intersect after reflecting from a concave mirror they form a real image.
but what happens if, say, a ray from the head of the object A collide with the reflected ray DA'? It might be a basic question but I'm new in optics.
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$\begingroup$ Light rays do not collide with each other. They simply pass through each other with no interaction. $\endgroup$– Solomon SlowCommented Mar 22 at 16:10
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3 Answers
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Let us refer to the figure that you have provided. You can clearly see that the object, which is represented as an arrow, has height. Therefore, the formed image must also have height. However, for the intersection point, which is the F (Focal Point), we cannot define a height. If a screen is placed at the focal point, you will see a point of converged light there.
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The diagram is simplified. You get an image when a whole bundle of rays converge, not just two. Every ray from point A that reflects from the mirror (ideally) reaches point A', not just the two rays shown.
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I do not agree with what SaaN has said.
I do agree with what John has to say (though the answer does not address the main issue of the question).
This simulation might help you to understand how many rays actually meet for an intensified image formation. You can manipulate the object and type of lens to understand why exactly the image is formed there. The place of formation of an image refers to the position where all the refracted rays meet.
Now what happens when a reflected and refracted wave can be judged by superimposing the two waves. Considering the object to be a source, such a meeting point will change the intensity of the waves by an amount which is indistinguishable from the "background" that is formed by the source (which of course, emits light in all directions).
P.S. while using the link, make sure to click the "second point" button (to the bottom right corner) to understand better.
