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Two thin equiconvex lenses each of focal length $0.2 \mathrm{~m}$ are placed coaxially with their optic centres $0.5 \mathrm{~m}$ apart. Then the focal length of the combination is
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The correct answer is:
$-0.4 \mathrm{~m}$
Equivalent focal length $(F)$ of two lenses separated by distance $\mathrm{d}$ is given by
$$
\begin{aligned}
\frac{1}{F} &=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}} \\
&=\frac{1}{0.2}+\frac{1}{0.2}-\frac{0.5}{(0.2)(0.2)} \\
&=5+5-0.5 \times 5 \times 5 \\
&=10-12.5=-2.5 \\
\therefore \quad F &=-\frac{1}{2.5}=-0.4 \mathrm{~m}
\end{aligned}
$$
$$
\begin{aligned}
\frac{1}{F} &=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1} f_{2}} \\
&=\frac{1}{0.2}+\frac{1}{0.2}-\frac{0.5}{(0.2)(0.2)} \\
&=5+5-0.5 \times 5 \times 5 \\
&=10-12.5=-2.5 \\
\therefore \quad F &=-\frac{1}{2.5}=-0.4 \mathrm{~m}
\end{aligned}
$$
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