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physucsc11
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Homework Statement
A short thin object (like a short length of wire) of length L is placed along the axis of a spherical mirror (perpendicular to the glass surface). Show that its image has length L' = (m^2)*L
so the longitudinal magnification is equal to -m^2 where m is the normal "lateral" magnification. Why the minus sign? (Hint: Find the image positions of both ends of the wire, and assume L is very small.)
Homework Equations
[tex]\frac{1}{d_{i}}[/tex] = [tex]\frac{1}{f}[/tex] - [tex]\frac{1}{d_{0}}[/tex]
m = - [tex]\frac{d_{i}}{d_{0}}[/tex]
The Attempt at a Solution
I'm kind of stuck. I was trying to draw ray diagrams but it doesn't make sense how to draw them for an object that is a wire, that is, is very thin. It would be nice if somebody could just give me helpful advice how to approach the problem.