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Three identical convex lenses each of focal length \(f\) are placed in a straight line separated by a distance \(f\) from each other. An object is located at \(\mathrm{f} / 2\) in front of the leftmost lens. Then,
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The correct answer is:
Final image will be at \(f / 2\) behind the rightmost lens and its magnification will be -1 .
Hint : 
For first lens
\(\begin{aligned}
& u=-\frac{f}{2} \\
& \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{\frac{-f}{2}} \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f} \Rightarrow \frac{1}{v}=\frac{-1}{f} \\
& v=-f \\
& m_1=\frac{v}{u}=\frac{-f}{\frac{-f}{2}}=2
\end{aligned}\)
For second lens
\(u=-(f+f)=-2 f\)
\(\begin{aligned}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-2 f}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2 f} \Rightarrow \frac{1}{v}=\frac{1}{2 f} \\
& v=2 f \\
& m_1=\frac{v}{u}=\frac{2 f}{-2 f}=-1
\end{aligned}\)
For third lens
\(\begin{aligned}
& u=f \\
& f=f \\
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{2}{f} \Rightarrow v=\frac{f}{2}
\end{aligned}\)
\(m_3=\frac{v}{u}=\frac{\frac{f}{2}}{f}=\frac{1}{2}\)
Total magnification \(=m_1 m_2 m_3=2 \times(-1) \times \frac{1}{2}=-1\)
For first lens
\(\begin{aligned}
& u=-\frac{f}{2} \\
& \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{v}-\frac{1}{\frac{-f}{2}} \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f} \Rightarrow \frac{1}{v}=\frac{-1}{f} \\
& v=-f \\
& m_1=\frac{v}{u}=\frac{-f}{\frac{-f}{2}}=2
\end{aligned}\)
For second lens
\(u=-(f+f)=-2 f\)
\(\begin{aligned}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-2 f}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2 f} \Rightarrow \frac{1}{v}=\frac{1}{2 f} \\
& v=2 f \\
& m_1=\frac{v}{u}=\frac{2 f}{-2 f}=-1
\end{aligned}\)
For third lens
\(\begin{aligned}
& u=f \\
& f=f \\
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u} \Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{2}{f} \Rightarrow v=\frac{f}{2}
\end{aligned}\)
\(m_3=\frac{v}{u}=\frac{\frac{f}{2}}{f}=\frac{1}{2}\)
Total magnification \(=m_1 m_2 m_3=2 \times(-1) \times \frac{1}{2}=-1\)
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