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## Homework Statement

Find the smallest distance between the object and a real image, when the focal distance of the lens is [tex]f[/tex]

## Homework Equations

[tex] \frac{1}{s}+ \frac{1}{p} = \frac{1}{f}[/tex], where [tex]s[/tex] is the distance of the object from the lens and [tex]p[/tex] is the distance of the image.

## The Attempt at a Solution

I'm not even sure, what I'm trying to do here, since the definition of a real and virtual image is a bit vague to me. But I've got something...

Let [tex]D=s+p[/tex] be the desired distance. From the equation above we get [tex]p = \frac{sf}{s-f}[/tex] so [tex]D= \frac{s^2}{s-f}[/tex]. Then [tex] \frac{ \partial D}{ \partial s}= \frac{2s^2-2sf-s^2}{(s-f)^2}=0 \Rightarrow s=0[/tex] or [tex]s=2f[/tex]. We get the same for [tex]p[/tex], so the distance would be [tex]D=4f[/tex]. Is this correct and does this apply for the minimum distance only or are the real image and the object always equal lenght from the lens?