An object $O$ between two parallel (vertical) plane mirrors is approaching plane mirror $M_1$ as shown. If $V_1, V_2$ denote the magnitude of relative velocity of image with respect to object in mirror $M_1, M_2$ respectively and $V_{1, 2}$ denotes relative velocity of approach of the images formed due to direct reflection by each mirror, then

A plane mirror in $y$-$z$ plane is moving with $-2 \hat{\mathrm{i}} {\mathrm{ms}}^{-1}$ as shown. The object shown now starts moving with $(3\hat{\mathrm{i}}+\hat{\mathrm{j}}-4\hat{\mathrm{k}}) \mathrm{m} {\mathrm{s}}^{-1}$. What is the velocity of the image?

An equi-biconvex lens $(\mu =1.5)$ of focal length $20 \mathrm{cm}$ is cut into two parts and placed in contact as shown in the figure. A liquid of refractive index $1.3$ is filled between the two parts of lens. The equivalent focal length of the combination is (in $\mathrm{cm}$)

A convex lens of focal length $20 \mathrm{cm}$ and another convex lens of focal length $40 \mathrm{cm}$ are placed co-axially. What should be the distance $d$ (in $\mathrm{cm}$) between lenses so that final image of object $O$ is formed on $O$ itself?