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(B) Since $STU$ is a plane mirror,we can take the mirror image of the whole system about it. The final image is formed at the object position if the r

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$(A), (B)$

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(B) Since $STU$ is a plane mirror,we can take the mirror image of the whole system about it. The final image is formed at the object position if the rays return along the same path.The system acts as a combination of a lens and a mirror. The equivalent focal length $F_{eq}$ of the system is given by $\frac{1}{F_{eq}} = \frac{2}{f_L} + \frac{1}{f_M}$.For a plane mirror,$f_M = \infty$,so $\frac{1}{F_{eq}} = \frac{2}{f_L}$.The lens is formed by the glass-liquid interface. The power of the lens is $P = (\mu_g - \mu_l) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.Here,$R_1 = 9 \ cm$ and $R_2 = \infty$ (planar surface).So,$\frac{1}{f_L} = (1.6 - n) \left(\frac{1}{9} - 0\right) = \frac{1.6 - n}{9}$.Thus,$\frac{1}{F_{eq}} = 2 \left(\frac{1.6 - n}{9}\right) = \frac{3.2 - 2n}{9}$.For the image to form on the object,the object must be at the center of curvature of the equivalent mirror,i.e.,$h = 2F_{eq}$.Therefore,$\frac{2}{h} = \frac{3.2 - 2n}{9} \implies h = \frac{18}{3.2 - 2n} = \frac{9}{1.6 - n}$.Checking the options:$(A)$ For $n = 1.42, h = \frac{9}{1.6 - 1.42} = \frac{9}{0.18} = 50 \ cm$. (Correct)$(B)$ For $n = 1.35, h = \frac{9}{1.6 - 1.35} = \frac{9}{0.25} = 36 \ cm$. (Correct)$(C)$ For $n = 1.45, h = \frac{9}{1.6 - 1.45} = \frac{9}{0.15} = 60 \ cm$. (Incorrect)$(D)$ For $n = 1.48, h = \frac{9}{1.6 - 1.48} = \frac{9}{0.12} = 75 \ cm$. (Incorrect)Thus,options $(A)$ and $(B)$ are correct.

List-$I$	List-$II$
$P$. $e \rightarrow f$	$1$. $\mu_1 > \sqrt{2} \mu_2$
$Q$. $e \rightarrow g$	$2$. $\mu_2 > \mu_1$ and $\mu_2 > \mu_3$
$R$. $e \rightarrow h$	$3$. $\mu_1 = \mu_2$
$S$. $e \rightarrow i$	$4$. $\mu_2 < \mu_1 < \sqrt{2} \mu_2$ and $\mu_2 > \mu_3$

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