(a) Equivalent focal length $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$ $\ = \frac{1}{20} + \frac{1}{30}$ $\therefore F = \frac{20 \times 30}{20 + 30}$ $= \frac{600}{50} = 12cm$
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