Creating an Image 2x Size of Object w/ Convex Lens: Half Focal Length Away

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A converging lens creates an image twice the size of the object when the object is placed half a focal length away from the lens. The relationship between object distance (do), image distance (di), and focal length (f) is clarified through the equations -di/do = hi/ho and 1/f = 1/di + 1/do. The confusion arises from interpreting f=2do; it actually indicates that do equals half the focal length, not two focal lengths. Thus, the correct placement of the object is at f/2, not 2f. Understanding this distinction resolves the initial misunderstanding about the object's position relative to the lens.
Gear2d
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Homework Statement


A converging lens (convex lens) creates an image that is 2 times the size of the object. The object is placed:


Homework Equations



-di/do = hi/ho
1/f = 1/di + 1/do

The Attempt at a Solution



I thought the answer was that the object is place two focal lengths in front of the lens. But the answer says its half a focal length in front of the lens.

I mean: -di/do = hi/ho = 2
-di = 2do
1/f = 1/di + 1/do
= -1/2do + 1/do = 1/2do
f = 2do

Isn't f=2do saying two focal lengths?
 
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Gear2d said:

Homework Statement


A converging lens (convex lens) creates an image that is 2 times the size of the object. The object is placed:


Homework Equations



-di/do = hi/ho
1/f = 1/di + 1/do

The Attempt at a Solution



I thought the answer was that the object is place two focal lengths in front of the lens. But the answer says its half a focal length in front of the lens.

I mean: -di/do = hi/ho = 2
-di = 2do
1/f = 1/di + 1/do
= -1/2do + 1/do = 1/2do
f = 2do

Isn't f=2do saying two focal lengths?

I'm not sure what you mean. If f=10 for example, then d_o=5, so d_o is half the focal length.
 
So simple, thanks. I guess I just missed an easy answer by thinking too much (or thinking to less).

Thanks again
 
Gear2d said:
Isn't f=2do saying two focal lengths?

No.

f=2do is saying do=f/2, or half a focal length.

(do=2f would be saying two focal lengths.)
 
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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