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

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SUMMARY

The discussion focuses on determining the apparent position and magnification of a goldfish viewed through a spherical glass fish-bowl with a diameter of 30 cm. The problem involves applying the Mirror Equation, specifically \(\frac{1}{s}+\frac{1}{s'}=\frac{-2}{R}\), to analyze two scenarios: when the fish is at the center of the bowl and when it is positioned halfway to the glass. The user reports that while the apparent position aligns with expectations, the calculated magnification results in a smaller image, leading to confusion regarding the sign convention used in the calculations.

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  • Basic principles of refraction and magnification
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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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