Question

Let the transverse magnification produced by an spherical mirror is m. Then for the same position of objective mirror the longitudinal magnification will be :

• A. m
• B. $\sqrt{m}$
• C. $m^2$
• D. 1/m

Answer 1

Answer: (C)

$m^2$

$\begin{array}{c}\text{Transverse magnification}\\ m=-\frac{v}{u}-\left(i\right)\\ \text{longitudnal magnification}\\ m_L=\frac{\Delta v}{\Delta u}-\left(ii\right)\\ \text{now,}\\ \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\end{array}$

$\text{differentiating}$

$0=-\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}$

$\begin{array}{c}\Rightarrow \frac{\Delta v}{\Delta u}=\frac{v^2}{u^2}\\ \text{from}\left(i\right)\left(ii\right)\text{and}\left(iii\right)\\ m_L=m^2\end{array}$