Question

where, $u_1$, $u_2$ and $v_1$, $v_2$ are the respective object and image distances of the ends of the object from the pole, such that $\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f}$ and $\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f}$

However, if the object has infinitesimal size $du$ and the corresponding image size is $dv$, then we have

$m_{axial}=-\frac{dv}{du}$

Further since we know that $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ i.e., $v^{-1}+u^{-1}=f^{-1}$. Taking the derivative of this equation with respect to $u$, we get

$\frac{d}{du}(v^{-1})+\frac{d}{du}(u^{-1})=\frac{d}{du}(f^{-1})$

$\Rightarrow -v^{-2}\frac{dv}{du}-u^{-2}=0$

$\Rightarrow m_{axial}=\frac{dv}{du}=-\frac{v^2}{u^2}=-\left(\frac{f}{f-u}\right)^2=-\left(\frac{f-v}{f}\right)^2$

ILLUSTRATION 18

Answer 1

The provided text is an illustration explaining the concept and derivation of axial magnification for an infinitesimal object using the mirror formula and differentiation. It derives the relationship $dv/du = -v^2/u^2$. Based on the definition $m_{axial} = -dv/du$, the axial magnification is $m_{axial} = v^2/u^2$. This is equal to the square of the transverse magnification, i.e., $m_{axial} = m_t^2$. The formulas provided at the end of the illustration, $m_{axial}=-\left(\frac{f}{f-u}\right)^2=-\left(\frac{f-v}{f}\right)^2$, likely contain a sign error based on the definition and derivation within the text; the correct formulas for $m_{axial} = v^2/u^2$ should have a positive sign outside the square.

Answer 2

The text derives the axial magnification for an infinitesimal object using differentiation of the mirror formula. Starting from $1/v + 1/u = 1/f$, differentiation with respect to $u$ yields $dv/du = -v^2/u^2$. If axial magnification is defined as $m_{axial} = -dv/du$, then $m_{axial} = v^2/u^2$. This is equal to the square of the transverse magnification $m_t = -v/u$. The formulas for $m_{axial}$ in terms of $f$ and $u$ or $f$ and $v$ are derived by substituting the expressions for $v/u$ from the mirror formula into $m_{axial} = (v/u)^2$. The final formulas presented in the text contain an extraneous negative sign. The magnitude of axial magnification is $(v/u)^2 = (f/(u-f))^2 = ((v-f)/f)^2$. The negative sign in $dv/du$ indicates axial inversion.