Analyzing 2-Lens System: Image Position, Magnification & Nature

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SUMMARY

The discussion focuses on analyzing a two-lens optical system consisting of a convex lens with a focal length of 300mm and a concave lens with a focal length of -50mm. An object is positioned 6m from the convex lens, and the lenses are separated by 200mm. The final image's position, magnification, and nature are determined by first calculating the image formed by the convex lens and then using that image as the object for the concave lens. The magnification is calculated by multiplying the magnifications of both lenses.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/x + 1/y = 1/f
  • Knowledge of convex and concave lens properties
  • Ability to calculate magnification for optical systems
  • Familiarity with virtual and real objects in optics
NEXT STEPS
  • Study the principles of lens combinations in optical systems
  • Learn about the calculation of magnification for multiple lenses
  • Explore the concept of virtual objects in optics
  • Investigate the effects of lens aberrations on image quality
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the behavior of multi-lens systems in optical applications.

anizet
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Homework Statement


A convex lens (f1=300mm) is placed 200 mm from a concave lens (f2=-50mm). An object is placed 6m away form the convex lens. (the order is as follows: object - 6m - convex lens - 200mm - concave lens).
Determine the position, magnification and nature of a final image.


Homework Equations


for a single lens:
1/x+1/y=1/f (where x is distance from the lens to the object, y is the distance from the lens to the image),


The Attempt at a Solution


Should I analyze 2 lenses separately?
At first first lens, which gives the image at y1:
1/6m+1/y1=1/300mm
y1=~31.5cm, so it is placed behind the second lens! Should I now consider the second lens, starting from the point, where is the image? I mean:
1/x2+1/y2=1/f2
where x2=20cm-31.5 cm=-11.5cm ?
I'm a little bit confused..
And finally, how to calculate the magnification?
 
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anizet said:
Should I analyze 2 lenses separately?
At first first lens, which gives the image at y1:
1/6m+1/y1=1/300mm
y1=~31.5cm, so it is placed behind the second lens! Should I now consider the second lens, starting from the point, where is the image? I mean:
1/x2+1/y2=1/f2
where x2=20cm-31.5 cm=-11.5cm ?
Exactly right. Solve it in steps. The image from the first lens becomes the object for the second lens (virtual object, in this case, since it's behind the second lens).
And finally, how to calculate the magnification?
Find the magnification for each, then multiply them.
 

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