# Converging lens

1. Mar 30, 2007

### Aikenfan

1. The problem statement, all variables and given/known data

If anyone can help me, i would really appreciate it! thanks

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 30, 2007

### Aikenfan

would i do:
15.5^-1 - 2.96^-1
Ans^-1
-3.658

3. Mar 30, 2007

### hage567

i does not equal 2.96. You've just dropped a variable, you can't do that. If M = -i/o, then i = -M*o, right? Remember there should be a negative sign in the magnification equation when you are dealing with object and image distances. Now you know what i is in terms of o, so using the lens equation and a little algebra you can solve for o.

4. Mar 30, 2007

### Aikenfan

1/15.5 = 1/2.96*o + 1/o

Im sorry i believe i typed that wrong the first time, i forgot the "o" after 2.96
do i find the common denominator, i am not sure what to do

Last edited: Mar 30, 2007
5. Mar 30, 2007

### hage567

You are still forgetting the negative sign in the M equation. To find "o", you can just take it out so you will have (1/o)*(1-(1/2.96)) on the right side of the lens equation. You can then cross multiply to solve for "o".

Once you get your answer for "o", find "i". You can put them both back into the lens equation and check if they are right, since you were given f.

6. Mar 30, 2007

### Aikenfan

so o is equal to 0.663?

Last edited: Mar 30, 2007
7. Mar 30, 2007

### hage567

No, you're not quite finished yet. You should be at (1/15.5) = (1/o)*(0.663). Now you must isolate "o" and solve for it. You can just cross multiply, since what you have is the same as (1/15.5)=(0.663/o). Do you see what I mean?

8. Mar 30, 2007

### Aikenfan

when i cross multiply, i get:
15.5 (0.663)
=
10.2765
does that look correct?

Last edited: Mar 30, 2007
9. Mar 30, 2007

### hage567

Yes, I think that looks OK.