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

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SUMMARY

The discussion focuses on a physics problem involving refraction in a spherical fish bowl with a radius of 20 cm. The fish is initially positioned 10 cm from the front surface, leading to confusion about the sign convention for object distance (s). The consensus is that the object distance (s) should be considered positive, while the image distance (s') will be negative due to the viewer's position. This clarification resolves the apparent inconsistency in the solution manual regarding the sign of s.

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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.
 

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