To a fish in a fish bowl, do the objects outside look smaller? Based on refraction, does the world outside the fishbowl appear smaller to the fish than it actually is?
 A: Model the fish bowl as a thin glass shell filled with water.
An observer sits in the water and looks outside. We then compare observed size of far away objects to their size in air.
Assumptions:


*

*The glass shell is thin wrt. its dimensions. It therefore acts as a thin lens.

*$\mu_{glass}>\mu_{water}>\mu_{air}$.
For a bowl with flat surfaces, objects appear further away but of same size.  
Now consider a spherical bowl.
First let the bowl be empty. It acts,ever so slightly, as a diverging lens to the observer. Objects appear smaller than they actually are.
Filling the bowl with water, the refractive power of the inner surface is reduced. However compared to the outer surface, its more curved so intrinsically, its better at refracting. These two competing factors make the nature of optics depend on a threshold value $\mu_{th}$(typical case discussed further below):.


*

*For ${\mu_{water}}<\mu_{th}$, objects still appear smaller than they are though less so than the case of the empty bowl.

*For ${\mu_{water}}>\mu_{th}$, the glass shell is now,ever so slightly, converging. So depending on where the objects are wrt the bowl, images may appear magnified or diminished. Most objects would appear smaller (though inverted). Some close by objects would appear magnified and erect instead. Exactly at what proximity this happens depends on the focal length of the system and whether it lies within the bowl.
In either case objects would appear further away.
For the typical values of 
$ \mu_{glass}=1.5,\mu_{water}=1.3,\mu_{air}=1,\frac{R_{2}}{R_{1}}>0.9$.
where
$R_1$ and $R_2$ are the glass bowl's outer and inner radii resp.
the case in point would be 2. with focus lying entirely outside the bowl and so most objects would appear smaller to a fish.

The analysis is based on the following slightly modified lens equation along with some basic ray optics:
$\frac{\mu_2}{f} =\frac{\mu-1}{R_1}-\frac{\mu-\mu_2}{R_2}$
where
$\mu_2$ is the refractive index of the medium in contact with R$_2$(here water) and $\mu$, that of the lens(glass). (the indices are wrt the medium in contact with $R_1$(here air))
The condition $f<0\implies \mu_2<\mu_{th}=\mu(1-\frac{R_2}{R_1})+\frac{R_2}{R_1}$

A ray diagrams for the typical values. 


