Two convex lenses of focal lengths 20 cm and 30 cm are kept at a separation of 15 cm. Then power of the combination is 

Equivalent focal length F is give by 

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1{f}_2}=\frac{1}{20}+\frac{1}{30}-\frac{15}{20\times 30}a{m}^{-1}$

$\Rightarrow \text{\hspace{0.33em}Power\hspace{0.33em}}P=\frac{100}{F}D=\frac{100}{20}+\frac{100}{30}-\frac{15\times 100}{20\times 30}D=5.83D$