Refraction at a spherical surface

[sghaussi]

hello again:

A Spherical Fish Bowl. A small tropical fish is at the center of a water-filled spherical fish bowl 28.0 cm in diameter.

Find the apparent position of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored.

i'm dealing with the formula : (n_a/s) + (n_b/s_prime) = 0

n_a = index of refraction of air (1.00) ?
n_b = index of refraction of water (1.33)?
s = position of fish = 14 cm ?
s_prime = position of image?

I tried plugging into the above formula, however I keep getting the wrong answer. Also, I tried switching n_a with n_b but that also didn't work. I'm thinking that the position of my fish isn't 14 cm? What am I doing wrong?

 

[SpaceTiger]

i'm dealing with the formula : (n_a/s) + (n_b/s_prime) = 0

That equation is only for a flat surface. More generally, it's:

$$\frac{n_1}{s}+\frac{n_2}{s'}=\frac{n_2-n_1}{R}$$

where R is the radius of curvature of the bowl.

 

[thepatientmental]

Hello there,

The formula you are using is correct, but the position of the fish is not 14 cm. In this problem, we are dealing with a spherical surface, so the position of the fish needs to be measured from the center of the sphere, not from the surface. Since the fish is at the center of the bowl, its position would be 0 cm.

So, the correct formula to use would be: (n_a/s) + (n_b/s_prime) = 0, where s = 0 cm.

Plugging in the values, we get: (1.00/0) + (1.33/s_prime) = 0
Solving for s_prime, we get: s_prime = 0 cm.

This means that the image of the fish would also be at the center of the bowl, at a distance of 0 cm from the surface. This is because light rays passing through a spherical surface are not bent, but instead continue in a straight line.

I hope this helps clarify the problem for you. Let me know if you have any further questions.