Finding the focal length of a converging lens with VERY limited info

[Overdose_]

Homework Statement

A candle, a converging lens and a white screen are placed in a line with the lens between the candle and the screen. A distance of 72cm separates the candle and screen. As the lens is moved to all points between the candle and the screen, only one focused image of the candle can be made on the screen. What is the focal length of the converging lens?

Relevant Equations

The thin lens equation

This is the solution but I do not understand??? How could they just assume the object distance equals 2f? No magnification is given here.
We know the candle produces a real image and the distance is limited, so that means it must be somewhere between the principal focus, the center of curvature, and possibly a little beyond that. Doesn't that mean, then, that the object distance can be anywhere between 2 point something f to 1f? Only one image is produced, what if that ONE image isn't produced at 2f? Then wouldn't the assumption be invalid? What? Please help

 

[haruspex]

How could they just assume the object distance equals 2f?

In the equation relating $f, d_i, d_o$, think about the symmetry.

 

[Nik_2213]

Provided you have a 'simple' convex lens, be it Convex, Biconvex, Plano-convex, Positive meniscus or distributed 'Fresnel', there's a single focal length for each wave-length. ( Hence colour aberration !!)
A 'meta-material' may have dual or multiple focal lengths for same wave-length...

 

[kuruman]

If you don't see the symmetry argument, consider this. Since $d_i+d_o=L=72~$cm, you can write the thin lens equation as $$\frac{1}{d_o}+\frac{1}{L-d_o}=\frac{1}{f}.$$ This gives you a quadratic in $d_o$ which, generally, has two solutions. This means that, in general, there are two values of $d_o$ for a given focal length. What must be true for the quadratic to have only one solution?

 

[BvU]

How could they just assume the object distance equals 2f?

The exercise becomes easier when you reason from the other end: If the 72 cm is greater than 4f, there will be two positions that give a sharp image (with the symmetry @haruspex mentioned). If it's less, there will be no such postion at all.

$\$

 

[Overdose_]

If you don't see the symmetry argument, consider this. Since $d_i+d_o=L=72~$cm, you can write the thin lens equation as $$\frac{1}{d_o}+\frac{1}{L-d_o}=\frac{1}{f}.$$ This gives you a quadratic in $d_o$ which, generally, has two solutions. This means that, in general, there are two values of $d_o$ for a given focal length. What must be true for the quadratic to have only one solution?

Thank you so much I understand now!! The two solutions are the same so object distance and image distance must be the same!!!

 

[Overdose_]

In the equation relating $f, d_i, d_o$, think about the symmetry.

Thank you! I finally found a video on the conjugate foci method and now I completely understand!

 

[haruspex]

Provided you have a 'simple' convex lens, be it Convex, Biconvex, Plano-convex, Positive meniscus or distributed 'Fresnel', there's a single focal length for each wave-length. ( Hence colour aberration !!)
A 'meta-material' may have dual or multiple focal lengths for same wave-length...

All true, but, sadly, nothing to do with the question at hand.