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Smallest distance between object and image - lens.

  • Thread starter Kruum
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1. Homework Statement

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

2. 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.

3. 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?
 

Shooting Star

Homework Helper
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3. 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?
You have just arrived at this value as a condition for an extremum (or a stationary value). What do you feel should be the answer? The value is correct, by the way . Can you find values of s and p such that [itex]s+p\neq4f[/itex]?

Not understanding the distinction between a real and a virtual image is not very conducive to learning Physics. Please read up or ask for help here.
 
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Not understanding the distinction between a real and a virtual image is not very conducive to learning Physics. Please read up or ask for help here.
That's the thing, I have read and listened and either misunderstood or then have had contradicting information. First I learned that the magnification for real image is m<0, the image is up-side down, and for virtual image m>0. Now I'm taught that virtual image always appears on the other side of the lens (or mirror). For converging lenses the image appears on the other side of the lens with m<0. :uhh:
 

Shooting Star

Homework Helper
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Last edited by a moderator:
hi there! I'm someone else asking the same question. so is the answer 4f? I don't understand what you mean by extremum and find the values of s and p so that it cannot equal 4f.
 

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