How Does Refraction Affect the Apparent Position of a Fish in a Bowl?

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Homework Help Overview

The discussion revolves around the concept of refraction as it pertains to the apparent position of a fish viewed through a spherical glass fish-bowl. The problem involves determining the apparent position and magnification of the fish's eye based on its actual position within the bowl, which has a diameter of 30 cm. Participants are exploring the implications of the thin glass assumption on the refraction effects.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply the mirror equation to find the apparent position and magnification of the fish. There are questions regarding the sign conventions used in the calculations and the reasoning behind the observed magnification being smaller than expected. Some participants are also questioning the assumptions made about light traveling through the bowl.

Discussion Status

The discussion is ongoing, with participants seeking clarification on their calculations and the underlying principles of refraction. Some have expressed tentative agreement on certain points, while others are still questioning the logic and assumptions involved in the problem.

Contextual Notes

Participants note the assumption that the glass is thin enough to neglect its effects on refraction, which may influence their interpretations and calculations. There is also mention of specific sign conventions from a textbook that may be causing confusion.

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Homework Statement


I asked for help on the concept of this problem in another forum section on this site. I went through the problem and I am getting an answer that KIND OF makes sense, but not really..

The problem says..
A small goldfish is viewed through a spherical glass fish-bowl 30 cm in diameter. Determine the apparent position and magnification of the fish’s eye when its actual position is a) at the center of the bowl and b) nearer to the observer, halfway from center to glass, along the line of sight. Assume that the glass is thin enough so that its effect on the refraction may be neglected.


Homework Equations


The Mirror Equation
\frac{1}{s}+\frac{1}{s'}=\frac{-2}{R}


The Attempt at a Solution



http://img217.imageshack.us/img217/2995/imgcqu.jpg

Yes I know its REALLY sloppy! I am getting an image that is closer to the observer, which makes sense, yet I am getting a magnification of a smaller fish which doesn't make sense. I followed the sign convention in my book. "s" is negative because "O" is to the right of "V". (So says my book.) Can anyone help?
 
Last edited by a moderator:
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Xyius said:

Homework Statement


I asked for help on the concept of this problem in another forum section on this site. I went through the problem and I am getting an answer that KIND OF makes sense, but not really..

The problem says..
A small goldfish is viewed through a spherical glass fish-bowl 30 cm in diameter. Determine the apparent position and magnification of the fish’s eye when its actual position is a) at the center of the bowl and b) nearer to the observer, halfway from center to glass, along the line of sight. Assume that the glass is thin enough so that its effect on the refraction may be neglected.


Homework Equations


The Mirror Equation
\frac{1}{s}+\frac{1}{s'}=\frac{-2}{R}


The Attempt at a Solution



http://img217.imageshack.us/img217/2995/imgcqu.jpg

Yes I know its REALLY sloppy! I am getting an image that is closer to the observer, which makes sense, yet I am getting a magnification of a smaller fish which doesn't make sense. I followed the sign convention in my book. "s" is negative because "O" is to the right of "V". (So says my book.) Can anyone help?

Case 1.
If the fish's eye is at the centre of the bowl, all light leaving the eye will be traveling along a radius of the spherical bowl of water, and thus hit the surface at right angles and pass straight through will it not??
 
Last edited by a moderator:
Hmm.. that seems to make sense. So is my answer of 15cm correct? Why is the magnification smaller though?
 
Can anyone help me with this? What is incorrect about my logic?
 

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