# Formation of Images by lenses!

1. May 24, 2008

### lenaosu

1. An object is placed in front of a converging lens in such a position that the lens (f = 11.0 cm) creates a real image located 27.0 cm from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens (f = 19.0 cm). A new, real image is formed. What is the image distance of this new image?

2. I tried using 1/f= 1/d0-1/di to solve for this. But it doesn't seem to be working.
Any suggestions.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 24, 2008

### alphysicist

Hi lenaosu,

The thin lens formula is

$$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$

Depending on the problem, any of these variables might turn out to be negative numbers. What numbers are you using, and what do you get?

3. May 24, 2008

### lenaosu

i used the thin lens equations twice.
1.) 1/f-1/di=1/do
1/11-1/27=1/di
di=18.6cm

2.) 1/f-1/do=1/di
1/19-1/18.6=1/di
di= -883.5cm

My calculations for my second image doesn't seem correct. My image should be to the left of my lens and not to the righ tof my lens.

4. May 24, 2008

### alphysicist

Since the object distance is less than the focal length for the second lens, it's image should be virtual, and so I think the fact that the final image length is negative is right.

However, you should not round your result from step 1 to three digits. Keep all the digits you can, and then round the final answer. Here it makes a relatively large difference.