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(C) Given: Focal length $f = -20 \, cm$,object distance $u = -30 \, cm$,and object length $L_o = 2 \, mm$.Using the mirror formula: $\frac{1}{v} + \fr

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$8$

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(C) Given: Focal length $f = -20 \, cm$,object distance $u = -30 \, cm$,and object length $L_o = 2 \, mm$.Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-30} = \frac{-3 + 2}{60} = -\frac{1}{60}$.So,$v = -60 \, cm$.The transverse magnification is $m = -\frac{v}{u} = -\frac{-60}{-30} = -2$.The longitudinal magnification $m_L$ for a small object is given by $m_L = -m^2$.$m_L = -(-2)^2 = -4$.The length of the image $L_I = |m_L| 	imes L_o = |-4| 	imes 2 \, mm = 8 \, mm$.

$A$ small rod of length $2 \, mm$ lies along the principal axis at a distance of $30 \, cm$ from a concave mirror of focal length $20 \, cm$. The axial length of the image of the rod is.....$mm$.

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