A lens having refractive index 1.6 has focal length of 12 cm, when it is in air. Find the focal length of the lens when …

As we know the lens maker formula in a medium is given by
\(\frac{1}{f}=[\frac{{\mu }_{L}}{{\mu }_{m}}-1][\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}]\)

For air, refractive index of medium is unity
\({\mu }_{m}=1\)

Substituting values for air
\(\frac{1}{12}=[1.6-1][\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}]\)

Simplifying
\(\frac{1}{12}=\frac{6}{10}[\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}]\)

Hence
\([\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}]=\frac{10}{72}\)

For water as surrounding medium
\(\frac{1}{f}=[\frac{1.6}{1.28}-1][\frac{10}{72}]\)

Simplifying refractive index term
\(\frac{1}{f}=\frac{32}{128}\times \frac{10}{72}\)

\(\frac{1}{f}=\frac{1}{4}\times \frac{10}{72}\)
\(f=28.8\) cm

\(f=288\) mm