# Optics - distance between object and image

1. Jul 21, 2009

### Smile101

hello,

1. The problem statement, all variables and given/known data
A lens has a focal length of 20 cm and a magnification of 4. How far apart are the object and the image?

2. Relevant equations
m=di(image)/do(object)
1/di + 1/do = 1/f

3. The attempt at a solution
Well, I don't really know what to do, I feel like I don't have enough information. Your help will be very appreciated!

2. Jul 21, 2009

### Redbelly98

Staff Emeritus
Re: Optics-Lens

Since you have 2 equations in 2 unknowns ... solve 1 equation for 1 of the unknowns, then substitute that into the 2nd equation.

3. Jul 21, 2009

### Smile101

Re: Optics-Lens

Yes, I was thinking about that but how do I do it? Seeing as if i did m=di/do 4=di/do --but it leads me to a dead end

or if i used 1/di+1/do=1/f i can only go as far as 1/di+1/do=1/20 then once again i get to a dead end!

4. Jul 21, 2009

### Redbelly98

Staff Emeritus
Re: Optics-Lens

Okay, so far so good. Now solve that equation for di:

di = _____ ?​

5. Jul 21, 2009

### Smile101

Re: Optics-Lens

umm.. di=4(do) but since we dont have do we can't solve for di

6. Jul 21, 2009

### Redbelly98

Staff Emeritus
Re: Optics-Lens

You can now substitute 4do for di in the other equation ... that eliminates di, leaving do as the only unknown.

7. Jul 21, 2009

### Smile101

Re: Optics-Lens

so the equation is : 1/di +1/do =1/f

1/4do + 1/do = 1/20

Now I'm guessing we want to make do single... but i have to get rid of the 4 so since i'm dividing....

1/do^2 = 1/20 - 1/4

8. Jul 21, 2009

### rl.bhat

Re: Optics-Lens

Multiply do to both sides and solve for do.

9. Jul 22, 2009

### Redbelly98

Staff Emeritus
Re: Optics-Lens

Yes, exactly. It's just algebra at this point.

10. Jul 25, 2009

### Smile101

Re: Optics-Lens

sorry for the late reply but, thanks a lot for your help, I really appreciated it! :)