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(B) Given: Radius of curvature $R = 10 \, cm$. Focal length $f = R/2 = 5 \, cm$. Since it is a concave mirror,$f = -5 \, cm$. The object distance for

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$3:1$

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(B) Given: Radius of curvature $R = 10 \, cm$. Focal length $f = R/2 = 5 \, cm$. Since it is a concave mirror,$f = -5 \, cm$. The object distance for the bend $Q$ is $u = -20 \, cm$.
$1$. For the upright portion $PQ$ (transverse magnification):
The transverse magnification $m_t = \frac{f}{f - u} = \frac{-5}{-5 - (-20)} = \frac{-5}{15} = -\frac{1}{3}$.
The length of the image of the upright portion is $L_1 = |m_t| 	imes L_0 = \frac{1}{3} L_0$.
$2$. For the horizontal portion $QR$ (longitudinal magnification):
The longitudinal magnification $m_l = -\left(\frac{f}{f - u}ight)^2 = -\left(\frac{-5}{-5 - (-20)}ight)^2 = -\left(\frac{-5}{15}ight)^2 = -\frac{1}{9}$.
The length of the image of the horizontal portion is $L_2 = |m_l| 	imes L_0 = \frac{1}{9} L_0$.
$3$. Ratio of the lengths of the images:
Ratio = $\frac{L_1}{L_2} = \frac{\frac{1}{3} L_0}{\frac{1}{9} L_0} = \frac{9}{3} = \frac{3}{1}$.
Thus,the ratio is $3:1$.

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