What is the solution for object distance in a diverging lens?

[ghostops]

Homework Statement

Now consider a diverging lens with focal length f=−15cm, producing an upright image that is 5/9 as tall as the object.
What is the object distance? You will need to use the magnification equation to find a relationship between s and s′s′. Then substitute into the thin lens equation to solve for s.

Homework Equations

1) M = -s^'/s

2) 1/s+1/s^'=1/f

The Attempt at a Solution

using equation 2 I solved for s^' then subsituted that into equation 1 and solved for s and from that put it back into equation s and solved for s and ended up with the answer of -65/3.

However this was the wrong answer and do not know what I did wrong.

 

[jtbell]

using equation 2 I solved for s^'

And what did you get? Show exactly what you substituted into the equation.

then subsituted that into equation 1 and solved for s

Ditto.

and from that put it back into equation s and solved for s and ended up with the answer of -65/3.

Double ditto. (Well, you've told us the answer you got, but we still need to see exactly how you got it.)

However this was the wrong answer and do not know what I did wrong.

It's hard for us to tell you what you did wrong, if you don't tell us exactly what you did. If you show us all the steps explicitly, we can tell you which step went wrong.

 

[ghostops]

step 1: s = (1/f-1/s')^-1

step 2: m=-s'/((1/f-1/s)^-1)
m=-s'(1/f-1/s)
m=-(s'/f-1)
(f/s')m=1
f/s'=1/m
f=s'/m
s' = f(m)

step 3: 1/s+1/fm=1/f
1/s=1/f-1/fm
s=(f-fm)^-1

 

[dauto]

Double check your algebra. There is at least one fairly serious mistakes there. In step 2, how did you get from the third line to the fourth line?

 

[jtbell]

step 1: s = (1/f-1/s')^-1

step 2: m=-s'/s=-s'/((1/f-1/s)^-1)

Check this step carefully. I added an intermediate step in red to (hopefully) make the error more obvious.

 

[ghostops]

yeah I see it now

so starting my from my mistake
m=-s'/s=-s'/(f-s') = (-s'/f)-(s'/-s)'
m = -s'/f+1

m = -s'/f+1
m-1=-s'/f
fm+-f=-s'
-fm+f=s'

1/s-1/fm+f=1/f
1/s=1/f+1/fm-f
1/s = 1/f + 1/fm -1/f
s=(fm)^-1

 

[jtbell]

m=-s'/s=-s'/(f-s') = (-s'/f)-(s'/-s)'

No, the part that I put in red isn't true. What you're trying to do here is similar to 1/(a-b) = 1/a - 1/b, which doesn't work. (Try it with a numerical example if you need to convince yourself.)

 

[dauto]

yeah I see it now

so starting my from my mistake
m=-s'/s=-s'/(f-s') = (-s'/f)-(s'/-s)'
m = -s'/f+1

m = -s'/f+1
m-1=-s'/f
fm+-f=-s'
-fm+f=s'

1/s-1/fm+f=1/f
1/s=1/f+1/fm-f
1/s = 1/f + 1/fm -1/f
s=(fm)^-1

how did you go from -fm+f=s'
to 1/s-1/fm+f=1/f ?

You 're making too many silly algebra mistakes. You need more practice.