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I'm working through some past papers and came across the question below and I've shown my ray diagrams as the red and blue lines. For the next question I would think that the image is virtual because light has not actually from the image. However the mark scheme says that it is because the two light rays don't join up in the plane mirror, but they do join up in my diagram. Is my diagram wrong and if so, why would the two rays of light not joining give the indication that the image is virtual? enter image description here

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  • $\begingroup$ It is just a matter of definition. Look in your script/book/notes, your teacher/prof somewhere defined what a virtual image is. The definition will probably contain "not joining" or "diverge", and you have to use this definition in your answer. $\endgroup$ – Solarflare Mar 19 '17 at 12:13
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The actual physical light ray doesn't pass through the image so it is virtual.

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Look for diverging rays which when back produced appear to emanate from a point although the light rays do not actually come from that point.

Another way of deciding is to put a screen where the image is supposed to be.
With a real image you will see the image on the screen but with a virtual image there will be no image visible on the screen.

So looking at your diagram those dashed lines are construction lines to show where the virtual image is located.
The full lines with arrows on them shown the direction in which the light travels.

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