1. Pressure inside a soap bubble
   For a soap bubble of radius $r$, the excess pressure inside is $\Delta P = \frac{4T}{r}$ where $T$ is the surface tension.

So for the two bubbles:

• Smaller bubble of radius $a$: $P_1 = P_0 + \frac{4T}{a}$
• Larger bubble of radius $b$: $P_2 = P_0 + \frac{4T}{b}$

Since $b>a$, we have $\frac{4T}{a} > \frac{4T}{b}$ so pressure is greater in the smaller bubble.

2. Common surface between the two bubbles
   When two bubbles coalesce, a common interface is formed between them. The pressure difference across this common surface is $P_1 - P_2 = \frac{4T}{a} - \frac{4T}{b}$

If the radius of curvature of the common surface is $R$, then for a soap film, $P_1 - P_2 = \frac{4T}{R}$

Therefore, $\frac{4T}{R} = \frac{4T}{a} - \frac{4T}{b}$

Cancelling $4T$, $\frac{1}{R} = \frac{1}{a} - \frac{1}{b}$

3. Simplify $\frac{1}{R} = \frac{b-a}{ab}$ Hence, $R = \frac{ab}{b-a}$

4. Match with options This corresponds to: $\boxed{\frac{ab}{b-a}}$ which is Option D.