Explain how the image is formed by a thin convex lens and derive the lens maker's formula: $\frac{1}{f} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.

(N/A) The image formation by a double convex lens can be analyzed in two steps:
$1$. The first refracting surface (with radius $R_1$) forms an image $I_1$ of the object $O$. For this surface,the refraction formula is: $\frac{n_1}{OB} + \frac{n_2}{BI_1} = \frac{n_2 - n_1}{R_1}$.
$2$. The image $I_1$ acts as a virtual object for the second refracting surface (with radius $R_2$). This surface forms the final image at $I$. For this surface,the refraction formula is: $\frac{n_2}{DI_1} + \frac{n_1}{DI} = \frac{n_1 - n_2}{R_2}$.
$3$. Assuming the lens is thin,$B$ and $D$ are very close to the optical center $P$. Thus,$BI_1 \approx DI_1$. Adding the two equations:
$\frac{n_1}{OB} + \frac{n_1}{DI} = (n_2 - n_1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
$4$. Using sign conventions,$OB = -u$ and $DI = v$. Substituting these:
$n_1 \left( \frac{1}{v} - \frac{1}{u} \right) = (n_2 - n_1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
$5$. Dividing by $n_1$ and setting $n_{21} = \frac{n_2}{n_1}$:
$\frac{1}{v} - \frac{1}{u} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
Since $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we get the lens maker's formula: $\frac{1}{f} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.