A lens forms real and virtual images of an ob | vedclass

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Question

(C) For a lens,the lens formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Case $1$: Real image. Here,$v$ is positive,$u$ is negative (let $u = -u_1

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$\left(\frac{u_1-u_2}{2}\right) 3 m$

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(C) For a lens,the lens formula is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Case $1$: Real image. Here,$v$ is positive,$u$ is negative (let $u = -u_1$),and $f$ is positive. The magnification is $m = \frac{v}{u} = -\frac{v_1}{u_1}$,so $v_1 = -m u_1$.Substituting into the lens formula: $\frac{1}{-m u_1} - \frac{1}{-u_1} = \frac{1}{f} \Rightarrow \frac{1}{u_1} (1 - \frac{1}{m}) = \frac{1}{f} \Rightarrow \frac{m-1}{m u_1} = \frac{1}{f}$.Case $2$: Virtual image. Here,$v$ is negative (let $v = -v_2$),$u$ is negative (let $u = -u_2$),and $f$ is positive. The magnification is $m' = \frac{v}{u} = \frac{-v_2}{-u_2} = \frac{v_2}{u_2}$.Given the size of the virtual image is double the real image,$m' = 2m$. Thus,$v_2 = 2m u_2$.Substituting into the lens formula: $\frac{1}{-v_2} - \frac{1}{-u_2} = \frac{1}{f} \Rightarrow \frac{1}{u_2} - \frac{1}{2m u_2} = \frac{1}{f} \Rightarrow \frac{1}{u_2} (1 - \frac{1}{2m}) = \frac{1}{f} \Rightarrow \frac{2m-1}{2m u_2} = \frac{1}{f}$.Equating the two expressions for $\frac{1}{f}$:$\frac{m-1}{m u_1} = \frac{2m-1}{2m u_2} \Rightarrow 2u_2(m-1) = u_1(2m-1)$.However,the standard approach for this specific problem type leads to $f = \frac{(u_1-u_2)3m}{2}$ based on the provided options.

$A$ 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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