How to Obtain a Magnification of -2 from a Convex Lens?

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SUMMARY

To achieve a magnification of -2 using a convex lens with a negative focal length, the correct approach is to place a real object at a distance of 3f/2 from the lens. The relevant equations include the lens formula (1/do) + (1/di) = (1/f) and the magnification formula m = -di/do. Options involving virtual objects or incorrect distances were eliminated based on the requirement for an inverted image. The solution confirms that option C is the only viable choice.

PREREQUISITES
  • Understanding of lens formulas, specifically (1/do) + (1/di) = (1/f)
  • Knowledge of magnification concepts, particularly m = -di/do
  • Familiarity with the properties of convex lenses, including focal length
  • Ability to differentiate between real and virtual objects in optics
NEXT STEPS
  • Study the derivation of the lens formula and its applications in optics
  • Learn about the characteristics of images formed by convex lenses
  • Explore the concept of magnification in more complex optical systems
  • Investigate the effects of varying object distances on image properties
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the behavior of convex lenses and magnification principles.

Sylvia
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Homework Statement


You wish to obtain a magnification of -2 from a convex lens of focal length f. The only possible solution is to:

A) place a virtual object at a distance 2f/3 from the lens.
B) place a virtual object at a distance 5f/3 from the lens.
C) place a real object at a distance 3f/2 from the lens.
D) place a real object at a distance 2f/3 from the lens.
E) place a virtual object at a distance 3f/2 from the lens.

Homework Equations


(1/do) + (1/di) = (1/f)
m = -di/do
f is negative because it is a convex lens

The Attempt at a Solution


I'm not sure how to factor in focal length, I am unsure how to get the relationship between object distance and focal length
 
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Sylvia, you can attempt this quetion by eliminating approach.
Here's what you should consider,
First try by using real objects.
Note that for magnification of -2 you should have an inverted image.
We kniw for object between f and o , the image is virtual and erect hence,
Option D is ruled out.
Now considering option C
The objet will be between f and 2f
Hence image will be real and enlarged.
And also by putting magnification ( -v/u = -2) you can get your answer.
Hence option C seems correct to me.

If you want further solution letme know.
 
Sylvia said:

Homework Equations


(1/do) + (1/di) = (1/f)
m = -di/do
f is negative because it is a convex lens

The Attempt at a Solution


I'm not sure how to factor in focal length, I am unsure how to get the relationship between object distance and focal length

The focal length of convex lenses is positive.
You know how magnification defined. So you can express do in terms of di and substitute into the equation (1/do) + (1/di) = (1/f). Solve for do in terms of f.
 
Thank you both very much, I understand it now!
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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