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A beam of light converges at a point $P$. Now a lens is placed in the path of the convergent beam $12 \mathrm{~cm}$ from P. At what point does the beam converge if the lens is (a) a convex lens of focal length $20 \mathrm{~cm}$, and (b) a concave lens of focal length $16 \mathrm{~cm}$ ?
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Verified Answer
As a lens is placed in the path of the convengent beam, thepoint $\mathrm{P}$ would tie on the right of the lens and acts as the virtual object.
(a) For convex lens $\mathrm{f}=20 \mathrm{~cm}, \mathrm{u}=12 \mathrm{~cm}, \mathrm{v}=$ ?
$$
\begin{aligned}
&\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
&\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{v}}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60} \\
&\therefore \mathrm{v}=\frac{60}{80}=7.5 \mathrm{~cm} .
\end{aligned}
$$
The image formed (beam converge) is real, at $7.5 \mathrm{~cm}$ from the lens on its right side.
(b) For concave lens $\mathrm{f}=-16 \mathrm{~cm}, \mathrm{u}=12 \mathrm{~cm}, \mathrm{v}=$ ?
$$
\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{12}+\frac{1}{-16}=\frac{1}{12}-\frac{1}{16}=\frac{4-3}{48}=\frac{1}{48}
$$
$$
\therefore \mathrm{v}=48 \mathrm{~cm} \text {. }
$$
Image is real, and at $48 \mathrm{~cm}$ from the lens on the right side.
(a) For convex lens $\mathrm{f}=20 \mathrm{~cm}, \mathrm{u}=12 \mathrm{~cm}, \mathrm{v}=$ ?
$$
\begin{aligned}
&\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
&\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{v}}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60} \\
&\therefore \mathrm{v}=\frac{60}{80}=7.5 \mathrm{~cm} .
\end{aligned}
$$
The image formed (beam converge) is real, at $7.5 \mathrm{~cm}$ from the lens on its right side.
(b) For concave lens $\mathrm{f}=-16 \mathrm{~cm}, \mathrm{u}=12 \mathrm{~cm}, \mathrm{v}=$ ?
$$
\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{12}+\frac{1}{-16}=\frac{1}{12}-\frac{1}{16}=\frac{4-3}{48}=\frac{1}{48}
$$
$$
\therefore \mathrm{v}=48 \mathrm{~cm} \text {. }
$$
Image is real, and at $48 \mathrm{~cm}$ from the lens on the right side.
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