A fish is 10 cm from the front surface of a spherical fish bowl

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The discussion revolves around a physics problem involving a fish in a spherical fish bowl and the calculation of its apparent position when viewed from outside. The main confusion arises over the sign convention for the object distance (s), where one participant questions why it is taken as -10 cm instead of +10 cm. Clarification is sought on whether the viewing perspective affects the sign, with some asserting that the book's approach is inconsistent. The consensus suggests that the object distance should be positive, while the image distance (s') will be negative due to the nature of refraction. This highlights the importance of understanding sign conventions in optics problems.
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Homework Statement



A fish is 10 cm from the front surface of a spherical fish bowl
of radius 20 cm. (a) How far behind the surface of the bowl does the fish appear
to someone viewing the fish from in front of the bowl? (b) By what distance does
the fish’s apparent location change (relative to the front surface of the bowl) when
it swims away to 30 cm from the front surface?

Homework Equations


(n1/s)+(n2/s')=(n2-n1)/r


The Attempt at a Solution


This problem was in the solution manual. It is one part that I don't understand. for s (object distance) they had it as -10cm, not +10cm. Why is that? In the book, the definition is that "s is positive for objects on the incident-light side of the surface" for refraction. Is this one different because the person is viewing from the other side the fish is on thus flipping the signs?
 
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can someone please answer my question.
 


charlies1902 said:
This problem was in the solution manual. It is one part that I don't understand. for s (object distance) they had it as -10cm, not +10cm. Why is that? In the book, the definition is that "s is positive for objects on the incident-light side of the surface" for refraction. Is this one different because the person is viewing from the other side the fish is on thus flipping the signs?
I'd say that the book is inconsistent. The position of the person viewing is irrelevant. s should be +; s' will turn out to be negative.
 
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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