Finding Focal Length of Diverging Lens with Parallel Rays

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SUMMARY

The focal length of a diverging lens can be determined using the relationship between the diverging lens and a converging lens positioned behind it. In this case, the converging lens has a focal length of 31 cm and is located 24 cm behind the diverging lens. When parallel rays enter the diverging lens, they create a virtual image that serves as the object for the converging lens. The correct application of the lens formula, 1/f = 1/do + 1/di, is essential, where 'do' represents the distance from the converging lens to the virtual image formed by the diverging lens.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/do + 1/di
  • Knowledge of the properties of diverging and converging lenses
  • Familiarity with virtual images and their characteristics
  • Basic principles of optics, including ray diagrams
NEXT STEPS
  • Study the derivation and application of the lens formula for both converging and diverging lenses
  • Learn how to construct ray diagrams for diverging and converging lenses
  • Explore the concept of virtual images and their significance in optical systems
  • Investigate the effects of varying distances between lenses on focal lengths and image formation
USEFUL FOR

Students studying optics, physics educators, and anyone seeking to understand the principles of lens systems and image formation in optical devices.

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Homework Statement



I'm trying to find the focal length of the diverging lens. The converging lens is 24cm behind a diverging lens. The focal length of the converging lens is 31cm. I know that parallel light strikes the diverging lens and then affter passing through the converging lens the light is again parallel.

The Attempt at a Solution



I know that the image for the diverging lens becomes the object for the converging lens. Does this mean that the distance between the diverging and converging lens is also the distance between the object and the converging lens?


If the rays are parallel when they enter the diverging lens then the distance between the object and the diverging lens is infinity. The distance between the image and the converging lens should also be infinity.
So for the converging lens:
1/f = 1/do + 1/infinity

I don't exactly know what to do with 'do'. I know its equal to the image of the diverging lens but I don't get the right answer when I substitute 1/fd - 1/infinity = 1/do in for 1/do in the converging equation.

I'd appreciate any suggestions!
Thank you so much..this is driving me crazy!
 
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one over a very large number is a very small number so you can ignore the term 1/infinity

another point is that for the diverging lens the object is at infinity so the term should be 1/di

since the image is virtual and forms to the left of the diverging lens the term should actually be -1/di
 
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