A short linear object O is placed on the prin | vedclass

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(A) For a convex mirror,the focal length is $f = +f$ and the object distance is $u = -f/3$.Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac

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$4/3$

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(A) For a convex mirror,the focal length is $f = +f$ and the object distance is $u = -f/3$.Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{f} - \frac{1}{-f/3} = \frac{1}{f} + \frac{3}{f} = \frac{4}{f}$.Thus,$v = f/4$.The transverse magnification $m_T$ is given by $m_T = -v/u = -(f/4) / (-f/3) = 3/4$.The longitudinal magnification $m_L$ for a small object is given by $m_L = -m_T^2$ (or $|m_L| = m_T^2$).The ratio of transverse magnification to longitudinal magnification is $\frac{m_T}{|m_L|} = \frac{m_T}{m_T^2} = \frac{1}{m_T}$.Substituting the value of $m_T$,we get $\frac{1}{3/4} = 4/3$.

$A$ short linear object $O$ is placed on the principal axis of a convex mirror of focal length $f$ as shown in the figure. The ratio of transverse magnification to longitudinal magnification is

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