Placement of an object and magnification

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To produce a sharp image on the screen with a 0.05 m focal length lens placed 1 m away, the lens must be positioned 0.05 m from the screen. This results in an image distance of 0.95 m from the lens. The corresponding magnification calculated is -0.05, indicating a reduced inverted image. The sum of the object and image distances must equal 1 m, confirming the placement is correct. The calculations align with the lens maker equation and magnification formula.
Violagirl
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Homework Statement



An object is 1 m from a screen. At what points may a 0.05-m focal length lens be placed so as to produce a sharp image on the screen? What are the corresponding magnifications?

Homework Equations



Lens maker equation:

1/s + 1/s' = 1/f

Magnification equation:

m = -s'/s = -f/s


The Attempt at a Solution



For the first question:

1/s' = 1/f -1/s = 1/0.05 m-1 = -.95 m.

1 m - .95 m = 0.05 m away from the screen.

Magnification:

m = -0.05 m / 1 m = -0.05

Is this right?
 
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Use the fact that the sum of the object and image distances is 100 cm.
 
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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