Define magnification for a lens and derive its expression.
(N/A) The ratio of the size of the image obtained by a lens to the size of the object is called linear magnification $(m)$.
In figures $(a)$ and $(b)$,convex and concave lenses are shown respectively.
Let the object height be $AB = h$ and the image height be $A'B' = h'$.
Let the object distance be $BP = u$ and the image distance be $B'P = v$.
Right-angled triangles $\triangle ABP$ and $\triangle A'B'P$ are similar.
Therefore,$\frac{A'B'}{AB} = \frac{B'P}{BP}$.
Applying the sign convention: $AB = h$,$A'B' = -h'$ (for real image),$BP = -u$,$B'P = v$.
Substituting these values: $\frac{-h'}{h} = \frac{v}{-u}$.
Therefore,$\frac{h'}{h} = \frac{v}{u}$.
Thus,magnification $m = \frac{h'}{h} = \frac{v}{u}$.
Magnification is negative for a real image and positive for a virtual image.
In figures $(a)$ and $(b)$,convex and concave lenses are shown respectively.
Let the object height be $AB = h$ and the image height be $A'B' = h'$.
Let the object distance be $BP = u$ and the image distance be $B'P = v$.
Right-angled triangles $\triangle ABP$ and $\triangle A'B'P$ are similar.
Therefore,$\frac{A'B'}{AB} = \frac{B'P}{BP}$.
Applying the sign convention: $AB = h$,$A'B' = -h'$ (for real image),$BP = -u$,$B'P = v$.
Substituting these values: $\frac{-h'}{h} = \frac{v}{-u}$.
Therefore,$\frac{h'}{h} = \frac{v}{u}$.
Thus,magnification $m = \frac{h'}{h} = \frac{v}{u}$.
Magnification is negative for a real image and positive for a virtual image.
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