Geometric optics: Thin lense equation

[Beth N]

Homework Statement

A 2.0-cm-tall candle flame is 2.0 m from a wall. You happen to have a lens with a focal length of 32 cm. How many places can you put the lens to form a well-focused image of the candle flame on the wall? For each location, what are the height and orientation of the image?

Homework Equations

$\frac {1} {f} = \frac {1} {s'} + \frac {1} {s}$ Where s' is image distance, s is object distance, and f is focal length of the lens. The lens is converging because its focal length is positive.

The Attempt at a Solution

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I just have a hard time approach in the question. I came up with a quadratic formula but still got wrong answer. Apologize for not writing this in Latex I am in quite a rush.

s'+s=2
s's=2/f=2/0.32=6.25

The answer key provided is s′=160 cm, h′=0.50 cm; s′=40 cm, h′=8.0 cm

Thank you so much!

 

[TSny]

s's=2/f

This equation is incorrect. Check your derivation of it. (Note that the left side of your equation has units of m² while the right side is dimensionless since you have 2m divided by the focal length in meters.)

 

[Beth N]

This equation is incorrect. Check your derivation of it. (Note that the left side of your equation has units of m² while the right side is dimensionless since you have 2m divided by the focal length in meters.)

Is this the correct approach though? I'm not sure what other equation to use

 

[TSny]

Yes, your approach is correct. Can you write out the steps that led you to s's=2/f? Then we can identify the specific mistake that you must have made.