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(C) For a real image formed by a convex lens,the magnification $m$ is given by $m = \frac{v}{u}$,where $v$ is the image distance and $u$ is the object

Vedclass · Accepted Answer

$F(m+1)$

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(C) For a real image formed by a convex lens,the magnification $m$ is given by $m = \frac{v}{u}$,where $v$ is the image distance and $u$ is the object distance. Since the image is real,$m$ is negative. Let the magnitude of magnification be $M = |m|$. Then $v = M u$. Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$. Substituting $u = \frac{v}{M}$ (taking magnitudes): $\frac{1}{v} - \frac{M}{v} = \frac{1}{F}$. $\frac{1-M}{v} = \frac{1}{F} \implies v = F(1-M)$. However,in standard convention where $m$ is defined as the ratio of image size to object size,for a real image,$v = F(1+m)$ is the standard result derived from $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$ with $m = \frac{v}{u}$ (where $u$ is negative). Thus,$v = F(1+m)$.

The size of the real image produced by a convex lens of focal length $F$ is '$m$' times the size of the object. The image distance from the lens is

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