Finding Distance d for Infinity Image: Diverging and Converging Lens Problem

Click For Summary
SUMMARY

The discussion focuses on calculating the distance d between a diverging lens with a focal length of -6.35 cm and a converging lens with a focal length of 12.8 cm, such that the final image is at infinity. The main equation used is the lens formula 1/f = 1/di + 1/do. The user initially calculated a distance di of -4.244 cm and subsequently attempted to find the distance d, resulting in an incorrect value of -22.41 cm. The discussion emphasizes the importance of correctly applying the lens formula and drawing ray diagrams for clarity.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/di + 1/do
  • Knowledge of diverging and converging lenses
  • Ability to perform ray tracing for optical systems
  • Familiarity with focal lengths and their significance in optics
NEXT STEPS
  • Study the principles of ray tracing for lens systems
  • Learn about the effects of combining diverging and converging lenses
  • Explore the concept of effective focal length in lens combinations
  • Practice solving lens problems using the lens formula with various configurations
USEFUL FOR

Students studying optics, physics educators, and anyone looking to deepen their understanding of lens systems and image formation.

premed_love
Messages
19
Reaction score
0

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

Homework Equations


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.
 
Physics news on Phys.org
premed_love said:

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

Homework Equations


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?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
12
Views
9K
Replies
8
Views
6K
Replies
1
Views
2K
  • · Replies 28 ·
Replies
28
Views
7K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K