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An object is placed $30 \mathrm{cm}$ away from a convex lens of focal length $10 \mathrm{cm}$ and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved by $45 \mathrm{cm}$ to get $\underline{a}$ sharp image again. The magnitude of focal length of the concave lens is (in $\mathrm{cm}$ )
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Verified Answer
The correct answer is:
20
For the first condition
$$
\frac{1}{f_{1}}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{10}=\frac{1}{v}+\frac{1}{30}
$$
$\frac{1}{v}=\frac{1}{10}-\frac{1}{30}$
$v=\frac{30}{2}=15 \mathrm{cm}$
$$
\begin{aligned}
\text { where } f_{1}=10 \mathrm{cm} & \begin{array}{l}
u=-30 \mathrm{cm} \\
v & =?
\end{array}
\end{aligned}
$$
For the first condition
For the second condition when concave lens is placed
$$
\begin{array}{l}
v^{\prime}=(15+45) \mathrm{cm}=60 \mathrm{cm} \\
\frac{1}{F}=\frac{1}{v^{\prime}}-\frac{1}{u}
\end{array}
$$
(where $F=$ focal length of combination)
$\therefore$
$$
\begin{array}{l}
\frac{1}{F}=\frac{1}{60}+\frac{1}{30} \\
F=\frac{60}{3} \mathrm{cm}=20 \mathrm{cm}
\end{array}
$$
The magnitude of focal length of concave lens $\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \Rightarrow \frac{1}{20}=\frac{1}{10}+\frac{1}{f_{2}} \Rightarrow f_{2}=-20 \mathrm{cm}$
(Negative sign for concave lens)
$$
\frac{1}{f_{1}}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{10}=\frac{1}{v}+\frac{1}{30}
$$
$\frac{1}{v}=\frac{1}{10}-\frac{1}{30}$
$v=\frac{30}{2}=15 \mathrm{cm}$
$$
\begin{aligned}
\text { where } f_{1}=10 \mathrm{cm} & \begin{array}{l}
u=-30 \mathrm{cm} \\
v & =?
\end{array}
\end{aligned}
$$
For the first condition
For the second condition when concave lens is placed
$$
\begin{array}{l}
v^{\prime}=(15+45) \mathrm{cm}=60 \mathrm{cm} \\
\frac{1}{F}=\frac{1}{v^{\prime}}-\frac{1}{u}
\end{array}
$$
(where $F=$ focal length of combination)
$\therefore$
$$
\begin{array}{l}
\frac{1}{F}=\frac{1}{60}+\frac{1}{30} \\
F=\frac{60}{3} \mathrm{cm}=20 \mathrm{cm}
\end{array}
$$
The magnitude of focal length of concave lens $\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \Rightarrow \frac{1}{20}=\frac{1}{10}+\frac{1}{f_{2}} \Rightarrow f_{2}=-20 \mathrm{cm}$
(Negative sign for concave lens)
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