c
Using \(\frac{1}{f}=\left(\frac{\mu_{2}}{\mu_{1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

\(\frac{1}{f}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\dots ( 1)\)

and \(\frac{1}{f_{l}}=\left(\frac{1.5}{1.42}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\dots (2)\)

equation \((1) /(2)\)

we get \(\frac{f_{ળ}}{f}=\frac{0.5}{0.056}\)

\(=8.93 \approx 9\)