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(B) $1$. Focal length of convex lens $(f_1)$:Using the lens maker's formula: $\frac{1}{f_1} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$

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$0.8$

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(B) $1$. Focal length of convex lens $(f_1)$:Using the lens maker's formula: $\frac{1}{f_1} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} ight]$$\frac{1}{f_1} = (1.5 - 1) \left[ \frac{1}{20} - \left( \frac{1}{-20} ight) ight] = 0.5 	imes \left( \frac{2}{20} ight) = \frac{1}{20}$$\Rightarrow f_1 = +20 \ cm$$2$. Focal length of concave lens $(f_2)$:$\frac{1}{f_2} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} ight]$$\frac{1}{f_2} = (1.5 - 1) \left[ -\frac{1}{20} - \frac{1}{20} ight] = 0.5 	imes \left( -\frac{2}{20} ight) = -\frac{1}{20}$$\Rightarrow f_2 = -20 \ cm$$3$. For lens $L_1$:Object distance $u_1 = -10 \ cm$,focal length $f_1 = +20 \ cm$.Using $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \Rightarrow \frac{1}{v_1} - \frac{1}{-10} = \frac{1}{20}$$\frac{1}{v_1} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20} \Rightarrow v_1 = -20 \ cm$Magnification $m_1 = \frac{v_1}{u_1} = \frac{-20}{-10} = 2$$4$. For lens $L_2$:The image formed by $L_1$ acts as an object for $L_2$. The distance between lenses is $10 \ cm$.Object distance $u_2 = -(20 + 10) = -30 \ cm$,focal length $f_2 = -20 \ cm$.Using $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \Rightarrow \frac{1}{v_2} - \frac{1}{-30} = \frac{1}{-20}$$\frac{1}{v_2} = -\frac{1}{20} - \frac{1}{30} = \frac{-3 - 2}{60} = -\frac{5}{60} = -\frac{1}{12} \Rightarrow v_2 = -12 \ cm$Magnification $m_2 = \frac{v_2}{u_2} = \frac{-12}{-30} = \frac{2}{5} = 0.4$$5$. Total magnification:$m = m_1 	imes m_2 = 2 	imes 0.4 = 0.8$

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