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A lens forms real and virtual images of an object, when the object is at $u_1$ and $u_2$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
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Verified Answer
The correct answer is:
$\left(\frac{u_1-u_2}{2}\right) 3 m$
Lens maker formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ Case 1 Real image $v$ and $f$ are positive, $u$ is negative, so
$$
\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f} \Rightarrow \frac{u_1}{v_1}+1=\frac{u_1}{f}
$$
Since, magnification for real image,
$$
m=\frac{-v_1}{u_1}
$$
Case 2 Virtual image, as image is formed infront of lens both $u$ and $v$ are negative, so
$$
-\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f} \quad \text { or } \quad \frac{-u_2}{v_2}-1=\frac{u_2}{f}
$$
Given, size of virtual image $=2 \times$ size of real image
Adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
-\frac{1}{m}-\frac{1}{2 m}+1-1 & =\frac{u_1}{f}-\frac{u_2}{f} \\
\frac{-3}{2 m} & =\frac{u_1-u_2}{f} \\
f & =\frac{\left(u_1-u_2\right) 3 m}{2}
\end{aligned}
$$
$$
\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f} \Rightarrow \frac{u_1}{v_1}+1=\frac{u_1}{f}
$$
Since, magnification for real image,
$$
m=\frac{-v_1}{u_1}
$$
Case 2 Virtual image, as image is formed infront of lens both $u$ and $v$ are negative, so
$$
-\frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f} \quad \text { or } \quad \frac{-u_2}{v_2}-1=\frac{u_2}{f}
$$
Given, size of virtual image $=2 \times$ size of real image
Adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
-\frac{1}{m}-\frac{1}{2 m}+1-1 & =\frac{u_1}{f}-\frac{u_2}{f} \\
\frac{-3}{2 m} & =\frac{u_1-u_2}{f} \\
f & =\frac{\left(u_1-u_2\right) 3 m}{2}
\end{aligned}
$$
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