Understanding a Converging Lens and Its Two Positions

[aeroboi]

Homework Statement

The distance between an object and its image is fixed at 40.0 cm. A converging lens of focal length f = 8.96 cm forms a sharp image for two positions of the lens. What is the distance between these two positions?

Relevant Equations

f=(s*s')/(s+s')
s'=(sf)/(s-f)

I am aware that the object would be to the left of the lens and the image would be to the right, but I don't understand what it is mean by " A converging lens of focal length f = 8.96 cm forms a sharp image for two positions of the lens." I don't understand where the two positions would be and why two positions are produced.

I'd appreciate any help visualizing this and a push for an approach. Thanks!

 

[kuruman]

It means that if you fix the distance between the object and the image ($s+s'$) there are two possible choices for the values of $s$ and $s'$.

 

[aeroboi]

It means that if you fix the distance between the object and the image ($s+s'$) there are two possible choices for the values of $s$ and $s'$.

Ok... but isn't that distance just 40cm (as stated in the problem statement)?

 

[kuruman]

Ok... but isn't that distance just 40cm (as stated in the problem statement)?

Yes, that's what it is. By using the thin lens equation you can find the two values of $s$, $s_1$ and $s_2$ that would provide an image to object distance $d=40~\mathrm{cm}$. Assuming that you have found them, how can you use them to find what the problem is asking?