A transparent thin film of uniform thickness and refractive index $n_1=1.4$ is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index $n_2=1.5$, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $f_1$ from the film, while rays of light traversing from glass to air get focused at distance  $f_2$ from the film. Then

Method 1:
   $$\begin{gathered} \frac{1.4}{v}-\frac{1}{\infty }=\frac{1.4-1}{R} \\ \frac{1.4}{v}=\frac{0.4}{R};v=\frac{7R}{2} \end{gathered}$$      
Film to glass:  $\frac{1.5}{v^{\text{'}}}-\frac{1.4}{\frac{7R}{2}}=\frac{1.5-1.4}{+R}$
   $$\begin{gathered} \frac{1.5}{v^{\text{'}}}-\frac{0.4}{R}=\frac{0.1}{R} \\ \frac{1.5}{v^{\text{'}}}=\frac{0.5}{R};v^{\text{'}}=3R\Rightarrow f_1=3R \end{gathered}$$                  
Glass to film:  $\frac{1.4}{v}-\frac{1.5}{\infty }=\frac{1.4-1.5}{-R}$
   $$\begin{gathered} \frac{1.4}{v}=\frac{-0.1}{-R} \\ v=-14R \end{gathered}$$                   
Film to air:  $\frac{1}{v^{\text{'}}}-\frac{1.4}{14R}=\frac{1-1.4}{-R}$  
 $$\begin{gathered} \frac{1}{v^{\text{'}}}-\frac{1}{10R}=\frac{+0.4}{+R} \\ \frac{1}{v^{\text{'}}}=\frac{0.4}{R}+\frac{0.1}{R}=\frac{0.5}{R} \\ {v}^{\text{'}}=\frac{R}{0.5}=2R\Rightarrow f_2=2R \end{gathered}$$                                           
Method 2: For air to glass
          $\frac{1.5}{f_1}=\frac{1.4-1}{R}+\frac{1.5-1.4}{R}\therefore f_1=3R$
For glass to air.
$\frac{1}{f_2}=\frac{1.4-1.5}{-R}+\frac{1-1.4}{-R}\therefore f_2=2R$