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The
diagram on the left shows an object (red arrow) in front of a
diverging mirror. The ray (solid blue line) is the one that would
have passed through the focus of the mirror and is reflected parallel
to the axis. The shaded blue arrow represents the image. (The other
two rays used to locate the image are not shown for clarity.) Notice
that there are two similar right triangles in the diagram. (The focus
is a vertex of both of them, and the axis is a leg of both of them.
The object forms the vertical leg of the larger one, and the mirror
forms the vertical leg of the smaller one.) Since corresponding sides
of similar triangles are proportional:
solving for Hi gives:
Notice that this is the same equation we used for a converging mirror!
The
diagram at left shows the ray that leaves the object and travels
parallel to the axis before reflecting from the mirror. There are two
similar right triangles in this diagram also, and their corresponding
sides are proportional also:
Solving for DI gives:
Substituting for HI from equation 1 gives:
This is the same as Equation 2 for converging mirrors! Surprise! The equations used to locate images for diverging mirrors are the same ones as for converging mirrors!
