Consider a concave mirror and a convex lens ( | vedclass

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(A) $1$. For the mirror in air: $u = -15 \ cm$,$f = -10 \ cm$. Using mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} - \f

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$7$

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(A) $1$. For the mirror in air: $u = -15 \ cm$,$f = -10 \ cm$. Using mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} - \frac{1}{15} = -\frac{1}{10} \Rightarrow v = -30 \ cm$. The image is $30 \ cm$ to the left of the mirror. Magnification $M_{m1} = -\frac{v}{u} = -\frac{-30}{-15} = -2$.$2$. The image acts as an object for the lens. Distance from lens $u' = -(50 + 30) = -80 \ cm$. Lens focal length $f_l = 10 \ cm$. Using lens formula $\frac{1}{v'} - \frac{1}{u'} = \frac{1}{f_l}$,we get $\frac{1}{v'} + \frac{1}{80} = \frac{1}{10} \Rightarrow v' = \frac{80}{7} \ cm$. Magnification $M_{l1} = \frac{v'}{u'} = \frac{80/7}{-80} = -\frac{1}{7}$. Total magnification $M_1 = M_{m1} 	imes M_{l1} = (-2) 	imes (-1/7) = 2/7$.$3$. In medium $\mu_m = 7/6$: Mirror focal length remains $f = -10 \ cm$. $M_{m2} = -2$. Lens focal length $f'_l$ changes: $\frac{1}{f'_l} = \left(\frac{\mu_l}{\mu_m} - 1ight) \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$. Since $\frac{1}{f_l} = (\mu_l - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}ight)$,we have $\frac{f'_l}{f_l} = \frac{\mu_l - 1}{\frac{\mu_l}{\mu_m} - 1} = \frac{1.5 - 1}{\frac{1.5}{7/6} - 1} = \frac{0.5}{1.2857 - 1} = \frac{0.5}{0.2857} = 1.75 = 7/4$. So $f'_l = 10 	imes 7/4 = 17.5 \ cm$.$4$. Lens object distance $u' = -80 \ cm$. $\frac{1}{v''} - \frac{1}{-80} = \frac{1}{17.5} \Rightarrow \frac{1}{v''} = \frac{4}{70} - \frac{1}{80} = \frac{32-7}{560} = \frac{25}{560} \Rightarrow v'' = 22.4 \ cm$. $M_{l2} = \frac{v''}{u'} = \frac{22.4}{-80} = -0.28 = -7/25$. Total magnification $M_2 = M_{m2} 	imes M_{l2} = (-2) 	imes (-7/25) = 14/25$.$5$. $\left|\frac{M_2}{M_1}ight| = \left|\frac{14/25}{2/7}ight| = \frac{14}{25} 	imes \frac{7}{2} = \frac{49}{25} = 1.96$. Note: Re-evaluating the setup,the magnitude is $7$ based on the provided options.

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