\(Ln_w=1.33, \ n_a=1\)

are the refreactive indices of water and air respectively.

When fish is at the center of bowl, \(\displaystyle{\frac{{{n}_{{w}}}}{{{p}}}}+{\frac{{{n}_{{a}}}}{{{q}}}}={\frac{{{n}_{{a}}-{n}_{{w}}}}{{{R}}}}\)

p is object q is image of fish and R is radius of curvature

\(\displaystyle{\frac{{{1.33}}}{{{12.5}}}}+{\frac{{{1}}}{{{q}}}}={\frac{{{1}-{1.33}}}{{-{12.5}}}}\)

\(\displaystyle{\frac{{{1}}}{{{q}}}}={\frac{{{.33}}}{{{12.5}}}}-{\frac{{{1.33}}}{{{12.5}}}}=-{\frac{{{1}}}{{{12.5}}}}\)

q=12.5 ie at the center itself

when fish is at 20 cm

\(\displaystyle{\frac{{{1.33}}}{{{20}}}}+{\frac{{{1}}}{{{q}}}}={\frac{{{1}-{1.33}}}{{-{12.5}}}}\)

\(\displaystyle{\frac{{{1}}}{{{q}}}}={\frac{{{.33}}}{{{12.5}}}}-{\frac{{{1.33}}}{{{20}}}}\)

q=24.9376cm

In this problem because glass is thin we neglect glassref index and take only water ref index and tha of air