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An object is located $4 \mathrm{m}$ from the first of two thin converging lenses of focal lengths $2 \mathrm{m}$ and $1 \mathrm{m},$ respectively. The lenses are separated by $3 \mathrm{m}$. The final image formed by the second lens is located from the source at a distance of
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The correct answer is:
$5.5 \mathrm{m}$
From the lens formula (for first lens)
$$
\begin{aligned}
\frac{1}{f_{1}} &=\frac{1}{v_{1}}-\frac{1}{u_{1}} \Rightarrow \frac{1}{2}=\frac{1}{v_{1}}-\frac{1}{(-4)} \\
\Rightarrow \quad \frac{1}{2}+\frac{1}{4} &=\frac{1}{v_{1}}=\frac{3}{4} \\
\Rightarrow \quad v_{1} &=\frac{4}{3}, u_{2}=3-\frac{4}{3}=5 / 3
\end{aligned}
$$
This image will be treated as the source for second lens, then again from lens formula, we have
$\frac{1}{f_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}4$
$\Rightarrow \quad \frac{1}{1}=\frac{1}{v_{2}}+5 / 3$
$\Rightarrow \quad 1-5 / 3=\frac{1}{v_{2}} \Rightarrow v_{2}=-3 / 2$
[By Eq. (i)]
This is the final image distance from 2 nd lens. So, the overall distance of image from the primary source(or object)
Let $d=4+3-1.5=5.5$
$$
\begin{aligned}
\frac{1}{f_{1}} &=\frac{1}{v_{1}}-\frac{1}{u_{1}} \Rightarrow \frac{1}{2}=\frac{1}{v_{1}}-\frac{1}{(-4)} \\
\Rightarrow \quad \frac{1}{2}+\frac{1}{4} &=\frac{1}{v_{1}}=\frac{3}{4} \\
\Rightarrow \quad v_{1} &=\frac{4}{3}, u_{2}=3-\frac{4}{3}=5 / 3
\end{aligned}
$$
This image will be treated as the source for second lens, then again from lens formula, we have
$\frac{1}{f_{2}}=\frac{1}{v_{2}}-\frac{1}{u_{2}}4$
$\Rightarrow \quad \frac{1}{1}=\frac{1}{v_{2}}+5 / 3$
$\Rightarrow \quad 1-5 / 3=\frac{1}{v_{2}} \Rightarrow v_{2}=-3 / 2$
[By Eq. (i)]
This is the final image distance from 2 nd lens. So, the overall distance of image from the primary source(or object)
Let $d=4+3-1.5=5.5$
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