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If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together, what will be their resultant power ?
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$-0.75 D$
$f_1=+80 \mathrm{~cm}, f_2=-50 \mathrm{~cm}$.
Resultant focal length
$\begin{aligned} F & =\frac{f_1 f_2}{f_1+f_2} \\ & =\frac{80 \times-50}{80-50}=\frac{-4000}{30} \\ & =-\frac{4000}{3} \mathrm{~cm} .\end{aligned}$
Resultant power $\begin{aligned} P & =\frac{100}{F}=\frac{100}{-400 / 3} \\ & =-\frac{3}{4} \\ & =-0.75 \mathrm{D} .\end{aligned}$
Resultant focal length
$\begin{aligned} F & =\frac{f_1 f_2}{f_1+f_2} \\ & =\frac{80 \times-50}{80-50}=\frac{-4000}{30} \\ & =-\frac{4000}{3} \mathrm{~cm} .\end{aligned}$
Resultant power $\begin{aligned} P & =\frac{100}{F}=\frac{100}{-400 / 3} \\ & =-\frac{3}{4} \\ & =-0.75 \mathrm{D} .\end{aligned}$
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