Consider an arrangement of two thin equiconvex lenses made of glass  $({\mathrm{\mu }}_{\mathrm{g}}=3/2)$ and each of focal length in air $10 \mathrm{cm}$ are separated by a distance $10 \mathrm{cm}$. The space between the two lenses is filled with water of refractive index ${\mathrm{\mu }}_{\mathrm{w}}=4/3$ . A small linear object $\text{'}\mathrm{AB}\text{'}$  of size $0.20 \mathrm{cm}$ is placed perpendicular to the optic axis at a distance $10 \mathrm{cm}$ from the first lens in air as shown. Then choose the correct option(s).

For the image formed by the first lens
$$\begin{gathered} \frac{{\mathrm{\mu }}_3}{\mathrm{v}}-\frac{{\mathrm{\mu }}_1}{\mathrm{u}}=\left(\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{{\mathrm{R}}_1}\right)+\left(\frac{{\mathrm{\mu }}_3-{\mathrm{\mu }}_2}{{\mathrm{R}}_2}\right) \\ \Rightarrow \frac{4}{3\mathrm{v}}-\frac{1}{(-10)}=\frac{(\frac{3}{2}-1)}{10}+\frac{(\frac{4}{3}-\frac{3}{2})}{-10} \\ \frac{4}{3\mathrm{v}}+\frac{1}{10}=\frac{1}{20}+\frac{1}{60} \\ \Rightarrow \mathrm{v}=-40\mathrm{cm} \\ \mathrm{Lateral} \mathrm{magnification}, {\mathrm{m}}_1=\frac{{\mathrm{\mu }}_1\mathrm{v}}{{\mathrm{\mu }}_3}\frac{\mathrm{v}}{\mathrm{u}}=\frac{1\times 3}{4}\left(\frac{-40}{-10}\right)=+3 \\ \mathrm{For} \mathrm{the} \mathrm{image} \mathrm{formed} \mathrm{by} \mathrm{the} \mathrm{second} \mathrm{lens} \end{gathered}$$
$$\begin{gathered} \frac{{\mathrm{\mu }}_3}{\mathrm{v}}-\frac{{\mathrm{\mu }}_1}{\mathrm{u}}=\left(\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{{\mathrm{R}}_1}\right)+\left(\frac{{\mathrm{\mu }}_3-{\mathrm{\mu }}_2}{{\mathrm{R}}_2}\right) \\ \Rightarrow \frac{1}{\mathrm{v}}-\frac{4}{3(-50)}=\left(\frac{\frac{3}{2}-\frac{4}{3}}{10}\right)+\left(\frac{1-\frac{3}{2}}{-10}\right)\Rightarrow \frac{1}{\mathrm{v}}+\frac{2}{75}=\frac{1}{60}+\frac{1}{20} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{15}-\frac{2}{75}\Rightarrow \mathrm{v}=+25\mathrm{cm} \end{gathered}$$
The position of final image formed is 25 cm right to the second lens. Lateral magnification, 
${\mathrm{m}}_2=\frac{{\mathrm{\mu }}_1}{{\mathrm{\mu }}_3}\frac{\mathrm{v}}{\mathrm{u}}=\frac{4}{3\times 1}\left(\frac{25}{-50}\right)=-\frac{2}{3}$
Overall magnification, $\mathrm{m}={\mathrm{m}}_1\times {\mathrm{m}}_2=(+3)\times (-\frac{2}{3})=-2$
The size of the final image for $=2\times 0.2\mathrm{cm}=0.4 \mathrm{cm}$