c
(c)

\(\mu_g=1.5 \quad P=5 D\)

\(5=\left(\mu_g-1\right)\left(\frac{1}{R}+\frac{1}{R}\right)\)

\(5 \times 2=\frac{2}{R}\)

\(R=\frac{1}{5}=0.2 \,m =20 \,cm \quad f_l < 0\)

\(\frac{1}{f_l}=\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac{2}{R}\right)\)

\(-\frac{1}{100}=\left(\frac{\mu_g}{\mu_l}-1\right) \frac{2}{20}\)

\(\frac{\mu_g}{\mu_l}-1=-\frac{1}{10}\)

\(\frac{1.5}{\mu_l}=\frac{9}{10}, \mu_l=\frac{5}{3}\)