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

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(B) For a convex lens,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ and magnification is $m = \frac{v}{u}$.For a real image,$u = -u_1$

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

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(B) For a convex lens,the lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ and magnification is $m = \frac{v}{u}$.For a real image,$u = -u_1$ and $v = v_1$. The magnification is $m = \frac{v_1}{u_1}$,so $v_1 = m u_1$.Using the lens formula: $\frac{1}{f} = \frac{1}{m u_1} - \frac{1}{-u_1} = \frac{1}{u_1} (\frac{1}{m} + 1) = \frac{1}{u_1} (\frac{1+m}{m})$.Thus,$\frac{u_1}{f} = \frac{1+m}{m} \dots (1)$.For a virtual image,$u = -u_2$ and $v = -v_2$. The size of the virtual image is $2$ times the real image,so $v_2 = 2 v_1 = 2 m u_1$.Using the lens formula: $\frac{1}{f} = \frac{1}{-2 m u_1} - \frac{1}{-u_2} = \frac{1}{u_2} - \frac{1}{2 m u_1}$.Multiplying by $f$: $1 = \frac{f}{u_2} - \frac{f}{2 m u_1}$.From $(1)$,$f = \frac{m u_1}{1+m}$. Substituting this into the equation for $u_2$ and solving for $f$ leads to the relation: $\frac{u_1-u_2}{f} = \frac{3}{2m}$,which gives $f = \frac{2 m (u_1-u_2)}{3}$.

$A$ 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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