What focal length is needed for a 0.60x magnification with a diverging lens?

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To achieve a magnification of 0.60x with a diverging lens for an object 1.40 meters away, the formula M = -di/do is applied, resulting in di being -0.84 meters. The lens formula 1/do + 1/di = 1/f is then used, leading to the calculation of the focal length f. The final calculation indicates that the required focal length is approximately 1.42 meters. There is a clarification regarding the sign convention in the calculations. The discussion emphasizes the importance of correctly interpreting the signs in lens equations.
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Homework Statement


What focus is needed for in order for a diverging lens to produce a magnification of 0.60x on an object located 1.40 meters away?

Homework Equations


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

The Attempt at a Solution


-di / 1.4 = 0.6
1/1.4 + 1/.84 =1/f
-di=0.84
1.42
 
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Taylor Marks said:

Homework Statement


What focus is needed for in order for a diverging lens to produce a magnification of 0.60x on an object located 1.40 meters away?

Homework Equations


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

The Attempt at a Solution


-di / 1.4 = 0.6
1/1.4 + 1/.84 =1/f
-di=0.84
1.42
Is that + sign correct?
 
Just a small remark - rather say what focal length is needed...
 
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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