# Optics Problem -- possible values for the object distance?

1. Aug 31, 2015

### TomInPhysics

1. The problem statement, all variables and given/known data
An object and its lens-produced real image are 2.4 m apart. If the lens has 55-cm focal length, what are the possible values for the object distance?

2. Relevant equations
1/s+1/s'=1/f

3. The attempt at a solution
The object and the image are 2.4m apart so I thought "Hey, that is the same as 240cm apart, cool."

So now I have an image and an object that are 240cm apart and I thought, "well... that is basically the same as saying the image is either 240cm to the left of the object or 240cm to the right of the object. Since we dunno for sure we'll have to solve it both ways."

At this point I'm thinking: s'=s-240 OR s'=s+240

So now I'm like "Sweet, lets plug that baby into the lens equation!"

1/s+(1/s-240)=1/55 OR 1/s+(1/s+240)=1/55

Then I flipped everything since it was all over one to make it easier on my eyeballs to see...

s+2-240=55 OR s+s+240=55

Then I was all like "wham blam math mojo stuff" and finished with

295=s^2 OR -185=s^2

And then from their I thought I was wrong.

So any help would be great, thanks!

2. Aug 31, 2015

### SammyS

Staff Emeritus
It might surprise you to know that $\displaystyle \ \frac{1}{\displaystyle\ \frac{1}{a}+\frac{1}{b}\ }\ne a+b \ .$

Try it with some numbers.

You need a common denominator to simplify the expressions on the left hand side of your equations.