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Question

(C) Transverse magnification is given by $m = -\frac{v}{u}$.Longitudinal magnification $(m_L)$ is defined as the ratio of the change in image position

Vedclass · Accepted Answer

$m^2$

Vedclass · Answer

(C) Transverse magnification is given by $m = -\frac{v}{u}$.Longitudinal magnification $(m_L)$ is defined as the ratio of the change in image position to the change in object position: $m_L = \frac{dv}{du}$.From the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Differentiating both sides with respect to $u$:$0 = -\frac{1}{v^2} \frac{dv}{du} - \frac{1}{u^2}$.Rearranging the terms:$\frac{dv}{du} = -\frac{v^2}{u^2}$.Since $m = -\frac{v}{u}$,it follows that $m^2 = \frac{v^2}{u^2}$.Therefore,$m_L = -m^2$. In terms of magnitude,the longitudinal magnification is $m^2$.

Let the transverse magnification produced by a spherical mirror be $m$. Then,for the same position of the object,the longitudinal magnification will be:

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