Calculating Distance and Magnification for Two Lenses | Homework Example

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SUMMARY

The discussion revolves around calculating the distance of the image formed by two lenses with focal lengths of 20 cm and 25 cm, and determining the magnification. The image distance for the first lens is calculated to be -60 cm, indicating a virtual image. The object distance for the second lens is found to be 130 cm, leading to an image distance of approximately 30.95 cm for the second lens. The overall magnification is calculated as -12/13, confirming the relationship between the two lenses.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/di + 1/do
  • Knowledge of magnification calculations, m = -di/do
  • Familiarity with the concepts of virtual and real images in optics
  • Basic grasp of focal lengths and their significance in lens systems
NEXT STEPS
  • Study the principles of optical systems involving multiple lenses
  • Learn about the effects of lens arrangement on image formation
  • Explore advanced magnification calculations in compound lens systems
  • Investigate the implications of virtual images in optical physics
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the behavior of light through multiple lenses.

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Homework Statement


I don't have the exact problem, but I have a sample that I made demonstrating my question:

There are two lens, the first of focal length 20cm and the second of focal length 25cm. the image is located 15cm in front of the first lens, and the distance between the two lens is 70cm.
Find the distance of the image of the second lens, and the magnification.

Homework Equations


1/f=1/di+1/do
m=-di/do

The Attempt at a Solution



after calculating using 1/20=1/15+1/di, the di comes out to be -60cm.

70-(-60)=130=position of object for second lens

1/25=1/130+1/di, the di is ~30.95

magnification of first lens: -(-60/15)=4
magnification of second lens: -(30/130)=-3/13

magnification of both=-3/13*4=-12/13


But my question is: the di of the first lens came out to be negative, signifying that the image is on the same side as the object, and is virtual. Do i use the 1/f=1/di+1/do equation one more time through the first lens with positive 60 because it goes through the first lens again to get to the second lens, or do I use it directly through the second lens? Thanks for any help
 
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But my question is: the di of the first lens came out to be negative, signifying that the image is on the same side as the object, and is virtual. Do i use the 1/f=1/di+1/do equation one more time through the first lens with positive 60 because it goes through the first lens again to get to the second lens, or do I use it directly through the second lens? Thanks for any help
No - what matters is the path the light travels and light does not come from the virtual image - that's why it is virtual.
 

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