Calculating Image Position and Height Using Two Lenses?

Click For Summary

Homework Help Overview

The discussion revolves around calculating the image position and height using two lenses, with a specific focus on a 2 cm tall object placed in relation to the lenses. The problem involves applying the thin lens formula and comparing results with ray-tracing techniques.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the application of the thin lens formula to determine image distances and heights. There are attempts to reconcile calculated values with ray diagram results, and questions arise regarding the accuracy of the ray diagram in relation to the theoretical calculations.

Discussion Status

Some participants have provided guidance on the use of ray diagrams, while others express confusion about the results obtained from the thin lens formula. Multiple interpretations of the image position are being explored, and there is an ongoing examination of the relationship between calculated values and graphical representations.

Contextual Notes

Participants note discrepancies between calculated image positions and those observed in ray diagrams, leading to discussions about the assumptions made in the calculations and the potential for errors in diagram construction.

CalebtheCoward
Messages
8
Reaction score
0

Homework Statement


A 2 cm tall object is 20 cm to the left of a lens with a focal length of 10 cm. A second lens with a focal length of 15 cm is 30 cm to the right of the first lens. Calculate the image position and height. Compare with ray-tracing.

Homework Equations


1/f = 1/do + 1/di, m = -di/do and hi/ho; f: focal point, do: object distance, di: image distance, hi: image height, ho: object height.

The Attempt at a Solution


1/dia = 1/10 - 1/20, dia=20 cm; 1/dib = 1/15 - 1/(-dia+30), dib=-30 cm. (a denotes first lens, b second).
And this is where I stumble. Looking in the back of the book, the correct answer seems to be: dib=30 cm, hib=6.0 cm. Doing the previous problems, the measurements are relative from the second lens, and after comparing the result with my ray diagram 30 cm, or -30 cm for that matter, is no where near my second image convergence point.
 
Physics news on Phys.org
You seem to be right. The second lens form the image at -30cm with a magnification of x3. So the image formed by the 2nd lens will be inverted and 6cm high.
 
Have you tried making a ray diagram?
 
No, these are just from the thin lens formula. Do you have a problem with constructing the diagram?
 
I must be, because when I construct the diagram, the location of the second image is neither 30 cm in front of or behind the second lens, which doesn't lend our answer much levity.
 
ray diagram

The image op the tip of the arrow, p, will seem to be located at p' if you extend the rays backwards.
That is if one looks to the left through the lens the rays seem to originate from the point p'.
 

Attachments

  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K