A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of \({f}_{1}\) in air. Anothe…

Using the lens maker's formula

\(\frac{1}{{f}_{1}}=\left({n}_{\text{glass }}-{n}_{\text{air }}\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)\)

Since one surface is flat, \({R}_{2}\) is infinity, so

\(\frac{1}{{R}_{2}}=0\)

For the first lens in air (with refractive index 1.5 and radius of curvature 2 cm ):

\(\frac{1}{{f}_{1}}=(1.5-1)\left(\frac{1}{2}\right)\\ {f}_{1}=4cm\)

For the second lens:

\(\frac{1}{{f}_{2}}=\left(\frac{1.5}{1.2}-1\right)\left(\frac{1}{3}-0\right)\\ {f}_{2}=12cm\)

Ratio of focal lengths:

\({f}_{1}:{f}_{2}=4:12=1:3\)