Placing object between 2 spherical mirrors

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A concave mirror with a focal length of 13.6 cm and a convex mirror with a focal length of -7.00 cm are positioned 35.8 cm apart, with an object placed 17.9 cm from each mirror. The image produced by the concave mirror is calculated to be at 56.6 cm, which then serves as the object for the convex mirror. The distance for the convex mirror's object is found to be 20.8 cm after accounting for the separation distance. The final calculation yields an image distance of -5.24 cm for the convex mirror, indicating that the image forms behind it. It is confirmed that this scenario involves a virtual object, necessitating the use of a negative value for the object distance.
ilovejava
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Homework Statement


A concave mirror (f1 = 13.6 cm) and a convex mirror (f2 = −7.00 cm) are facing each other and are separated by a distance of 35.8 cm. An object is placed between the mirrors and is 17.9 cm from each mirror. Consider the light from the object that reflects first from the concave mirror and then from the convex mirror. What is the distance of the image (di2) produced by the convex mirror?

Homework Equations


1/do + 1/i = 1/f

The Attempt at a Solution


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The image created by the concave mirror will act as the object for the convex mirror. Plugging in the values for do and f for the concave mirror I got di = 56.6cm . Therefore, the do for the convex mirror will be 56.6-35.8 (distance mirrors are separated by) = 20.8 . Lastly, I just plugged this value and the focal length of the convex mirror into 1/do + 1/i = 1/f and got -5.24cm as di2
 
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Does the image at 56.6 cm appear in front of or behind of the convex mirror? You should draw a ray diagram.
 
kuruman said:
Does the image at 56.6 cm appear in front of or behind of the convex mirror? You should draw a ray diagram.
I drew one out and it appears that the image forms behind the convex mirror. So this would be a virtual object does that mean I should have made 20.8 negative?
 
ilovejava said:
I drew one out and it appears that the image forms behind the convex mirror. So this would be a virtual object does that mean I should have made 20.8 negative?
Yes.
 
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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