For case (1) 
v11.5−∝1=R1(1.5−1)f11−v11.5=∝1−1.5 ⇒f11=R1(1.5−1) ...(1) ⇒f11=2R11 For case (2) $\begin{aligned}
& \frac{1.5}{v_1}-\frac{1.2}{\infty}=\frac{(1.5-1.2)}{R_2} \
& \frac{1.2}{f_2}-\frac{1.5}{v_1}=\frac{(1.2-1.5)}{\propto} \
& \Rightarrow \frac{1.2}{f_2}=\frac{0.3}{R_2} \
& \Rightarrow \frac{1}{f_2}=\frac{1}{4 R_2}
\end{aligned}Soclearly,\frac{f_1}{f_2}=\frac{2 \times 2}{4 \times 3}=\frac{1}{3}$