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

[Xyius]

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?

 

[PeterO]

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??

 

[Xyius]

Hmm.. that seems to make sense. So is my answer of 15cm correct? Why is the magnification smaller though?

 

[Xyius]

Can anyone help me with this? What is incorrect about my logic?

 

[rwisz]

I would suggest approaching this problem by first understanding the principles of optics and refraction. Refraction is the change in direction of a wave as it passes from one medium to another. In this case, light is passing from the air into the glass bowl and then into the air again.

To determine the apparent position and magnification of the fish's eye, we need to consider the refraction of light at the air-glass interface. The angle of refraction (the angle at which light bends) is dependent on the index of refraction of the two materials and the angle of incidence (the angle at which light hits the interface). The index of refraction of air is approximately 1, and the index of refraction of glass is around 1.5.

For the first part of the problem, where the fish's eye is at the center of the bowl, the light rays coming from the fish's eye will be perpendicular to the surface of the bowl. Therefore, there will be no refraction and the fish's eye will appear to be in its actual position.

For the second part of the problem, where the fish's eye is halfway between the center and the glass, we can use the thin lens equation to determine the apparent position and magnification. The thin lens equation is 1/s + 1/s' = 1/f, where s is the object distance, s' is the image distance, and f is the focal length.

Since the glass is thin enough to be neglected, we can use the radius of the bowl (15 cm) as the focal length. The object distance (s) would be the distance from the fish's eye to the center of the bowl (15 cm) and the image distance (s') would be the distance from the fish's eye to the observer (15 cm). Plugging these values into the thin lens equation, we get a magnification of -1, meaning the image would be inverted and the same size as the object. The apparent position would also be 15 cm from the observer, which is the actual position of the fish's eye.

In conclusion, understanding the principles of optics and refraction can help us solve problems involving light passing through different mediums. By using the thin lens equation and considering the index of refraction of the materials involved, we can determine the apparent position and magnification of the fish's eye in a spherical glass bowl.