b
\(\frac{1}{f_{\mathrm{a}}}-\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{R_{2}}\right)\)

\(\frac{1}{\mathrm{f}_{\mathrm{m}}}=\left(\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)

\(\frac{1}{f_{m}}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

Dividing \((i)\) by \( (ii), \frac{f_{\mathrm{m}}}{f_{\mathrm{a}}}=\left(\frac{1.5-1}{\frac{1.5}{1.6}-1}\right)=-8\)

\( \mathrm{P}_{\mathrm{a}}=-5=\frac{\mu}{\mathrm{f}_{\mathrm{a}}}=\frac{1}{\mathrm{f}_{\mathrm{a}}}  \Rightarrow \mathrm{f}_{\mathrm{a}}=-\frac{1}{5} \)

\( \Rightarrow  \mathrm{f}_{\mathrm{m}}=-8 \times \mathrm{f}_{\mathrm{a}}=-8 \times-\frac{1}{5}=\frac{8}{5} \)

\(\mathrm{P}_{\mathrm{m}}= \frac{\mu}{\mathrm{f}_{\mathrm{m}}}=\frac{1.6}{8} \times 5=1 \mathrm{\,D} \)