NTSE Stage 2 SAT 2019 Exam Question Paper

The focal length of combination of lenses is given as,
$\frac{1}{\mathit{f}}=\frac{1}{{\mathit{f}}_1}+\frac{1}{{\mathit{f}}_2}-\frac{\mathit{d}}{{\mathit{f}}_1{\mathit{f}}_2}$
where, focal length of convex lens,${\mathit{f}}_1=10\mathit{c}\mathit{m}$
focal length of concave lens${\mathit{f}}_2=-10\mathit{c}\mathit{m}$
distance between lenses,$\mathit{d}=2\mathit{c}\mathit{m}$
Then, focal length of the combination ‘/ ’ will be given by,
$\frac{1}{\mathit{f}}=\frac{1}{10}-\frac{1}{10}+\frac{2}{100}$
$\frac{1}{\mathit{f}}=\frac{2}{100}$
$\mathit{f}=+50$cm (converging lens)