(II) A fish is swimming in water inside a thin spherical glass bowl of uniform thickness. Assuming the radius of curvature of the bowl is 32.0 cm, locate the image of the fish if the fish is located: (a) at the center of the bowl; (b) 20.0 cm from the side of the bowl between the observer and the center of the bowl. Assume the fish is small.

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1. Determine the focal length of the spherical glass bowl using the lensmaker's equation for a thin lens: \( f = \frac{R}{2} \), where \( R \) is the radius of curvature. Given \( R = 32.0 \, \text{cm} \), calculate \( f = 16.0 \, \text{cm} \).
2. Analyze case (a): When the fish is at the center of the bowl, it is located at the radius of curvature from any point on the bowl. Use the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) where \( d_o = R = 32.0 \, \text{cm} \) and solve for \( d_i \) (image distance from the bowl's surface).
3. For case (a), substitute the values into the mirror equation and solve for \( d_i \): \( \frac{1}{16.0} = \frac{1}{32.0} + \frac{1}{d_i} \). Solving this gives \( d_i = 32.0 \, \text{cm} \), indicating the image is formed at the center, coinciding with the object's position.
4. Analyze case (b): When the fish is 20.0 cm from the side of the bowl, calculate the object distance from the center of the bowl, \( d_o = R - 20.0 \, \text{cm} = 12.0 \, \text{cm} \). Use the mirror equation again to find \( d_i \).
5. For case (b), substitute \( d_o = 12.0 \, \text{cm} \) into the mirror equation: \( \frac{1}{16.0} = \frac{1}{12.0} + \frac{1}{d_i} \). Solving this equation gives \( d_i = -48.0 \, \text{cm} \), indicating the image is formed 48.0 cm from the center on the opposite side of the bowl.