Understanding the Lens Equation: Fish & Pattern Location

[meadow]

i just need a little help with a relatively easy lens question

A small fish is cast into the center of a glass sphere of radius R=5 cm and n=1.5. Where will an observer see the fish? Where will the observer see a decorative pattern painted on the back side of a sphere?

So, I thought I could use the formula to find the image distance i=n-n1/R; and i subtracted 1.0 (n of air) from 1.5, then divided by the radius to find the distance of the image of the fish. For the background pattern, I used 10 cm as the value of "R", because it is the length of the diameter away from the observer.
Am I right in my thinking? I feel like I am just plugging and chugging, rather than really understanding the concept...(so i am not sure it i am even evaluating correctly)
Thank you.

 

[lightgrav]

Doesn't the image distance (measured from what?)
depend on the location of the object?

You COULD do this with Snell's Law, if you have too much time ...

 

[meadow]

I thought because the fish was placed in the center of the sphere, the distance of the object would be the distance of the radius. I don't, however, know the distance of the observer. Is there a way to explain where the observer would see the fish and the pattern without that distance? How should I approach this problem?

 

[lightgrav]

( i = n-n/R is obviously incorrect : a distance [meter] is not equal 1/R [1/m] )

If you don't like Snell ...
\frac{n_g}{d_{ob}} - \frac{n_a}{d_{im}} = \frac{n_g - n_a}{R}
should've been derived in your textbook ("single-surface refraction"):
distances are measured from the front of the glass, closest to person.

draw 2 rays from the fish ... are they deflected at the edge of the glass?

draw 2 rays from the back ... are they deflected at the front of the glass?