Calculating Magnification of Christmas Tree Ornament

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The discussion focuses on calculating the magnification of a spherical Christmas tree ornament with a diameter of 4.76 cm, where an object is placed 11.9 cm away. The focal length is determined to be 1.19 cm using the formula f = 0.5 * r. The equation (1/di) + (1/do) = 1/f is applied to find the image distance (di), resulting in a value of approximately 1.32222 cm. The magnification (m) is calculated as -0.1111, indicating a reduced and inverted image. Participants question the type of mirror and its implications for the focal length, suggesting a need for clarification on the ornament's reflective properties.
Kris1120
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Homework Statement



A spherical Christmas tree ornament is
4.76 cm in diameter.
What is the magnification of the image of
an object placed 11.9 cm away from the ornament?

Homework Equations



f = .5* r

(1/di) + (1/do) = 1/f

m=-(di/do)

The Attempt at a Solution



f = .5*2.38 = 1.19 cm

(1/di) = (1/1.19) - (1/11.9) = 1.32222 cm

m =-(1.32222/11.9) = -0.1111
 
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Hi Kris1120,

Kris1120 said:

Homework Statement



A spherical Christmas tree ornament is
4.76 cm in diameter.
What is the magnification of the image of
an object placed 11.9 cm away from the ornament?

Homework Equations



f = .5* r

(1/di) + (1/do) = 1/f

m=-(di/do)

The Attempt at a Solution



f = .5*2.38 = 1.19 cm

I don't believe this is right. What type of mirror is this? What does that tell you about the focal length?
 
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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