# Lens problem! Please check my work.

## Homework Statement

An object is placed 12.8 cm to the left of a diverging lens of focal length -6.35 cm. A converging lens of focal length 12.8 cm is placed a distance of d to the right of the diverging lens. Find the distance d that places the final image at infinity

1/f=1/di+1/do

## The Attempt at a Solution

I understand that I have a few unknowns and need to solve this a couple of times to discover the di:

So [/B]1/f=1/di +1/do : MAIN EQUATION
Di: 1/(-6.635)=1/di+1/(12.8cm). gives me di of -4.244 cm
Do2: 4.244+25.6=29.844 cm
1/do2+1/di2=1/f2 (1/2984)+1/di2=1/12.8cm

Eventually I end up with -22.41 cm.

I don't know if I am supposed to solve this for 2nd time through lens. Therefore, I thought I ended up -22.41 cm. However, that is not right. Please help.

I have a small request: Please read the problem very carefully. I do not want to answer what kind of mirror this is if the problem already answers it. I have had to answer questions regarding variables that I clearly explained in the problem. Please refrain from it. I want to learn so if you ask me a question to help me better understand the problem than that is fine. BUT DO NOT ASK ME ABOUT WHAT THE VARIABLES ARE BECAUSE I FEEL LIKE I AM REPEATING MYSELF RATHER THAN LEARNING ANYTHING NEW.

Quantum Defect
Homework Helper
Gold Member

## Homework Statement

An object is placed 12.8 cm to the left of a diverging lens of focal length -6.35 cm. A converging lens of focal length 12.8 cm is placed a distance of d to the right of the diverging lens. Find the distance d that places the final image at infinity

1/f=1/di+1/do

## The Attempt at a Solution

I understand that I have a few unknowns and need to solve this a couple of times to discover the di:

So [/B]1/f=1/di +1/do : MAIN EQUATION
Di: 1/(-6.635)=1/di+1/(12.8cm). gives me di of -4.244 cm
Do2: 4.244+25.6=29.844 cm
1/do2+1/di2=1/f2 (1/2984)+1/di2=1/12.8cm

Eventually I end up with -22.41 cm.

I don't know if I am supposed to solve this for 2nd time through lens. Therefore, I thought I ended up -22.41 cm. However, that is not right. Please help.

I have a small request: Please read the problem very carefully. I do not want to answer what kind of mirror this is if the problem already answers it. I have had to answer questions regarding variables that I clearly explained in the problem. Please refrain from it. I want to learn so if you ask me a question to help me better understand the problem than that is fine. BUT DO NOT ASK ME ABOUT WHAT THE VARIABLES ARE BECAUSE I FEEL LIKE I AM REPEATING MYSELF RATHER THAN LEARNING ANYTHING NEW.

I think that it might help you if you drew a ray diagram. Remember that for a converging lens, an image at infinity is achieved by placing the object at its focus.

By ray tracing, once you find out where the virtual image is from the diverging lens, you need to place the left-hand-side focus of the converging lens on top of the image for the diverging lens.

Also, in the problem statement you say that the focal length of th ediverging lens is -6.35 cm, while in the work you show, you have a value of -6.635 cm. I suspect that this is a typo.

When you check your answer, you might think about what this lens combination is acting like. You have effectively increased the focal length of a short-focal length lens. Does this match your picture?