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# A thin equiconvex lens of refractive index $\large\frac{3}{2}$ and radius of curvature 30 m is put in water $(RI= \large\frac{4}{3})$. Its focal length is

$(a)\;0.15\;m \\ (b)\;0.45\;m \\ (c)\;0.3\;m \\ (d)\;1.2\;m$

$\large\frac{1}{f}$$= \bigg( \large\frac{\mu_1}{\mu_2} -1\bigg)$$ \bigg (\large\frac{1}{R_1} -\frac{1}{R_2}\bigg)$
$\large\frac{1}{f}= \bigg( \large\frac{\Large\frac{3}{2} }{\Large\frac{4}{3}}-1 \bigg) \bigg( \large\frac{1}{0.3} +\frac{1}{0.3} \bigg)$
or $\large\frac{1}{f}=\bigg( \large\frac{9}{8}-1 \bigg) \bigg( \frac{2}{0.3} \bigg)$
or $\large\frac{1}{f} =\frac{1}{8} \times \frac{2}{0.3}$
or $f= 1.2 \;m$
Hence d is the correct answer.