$A$ biconvex lens of focal length $15 \,cm$ is in front of a plane mirror. The distance between the lens and the mirror is $10 \,cm$. $A$ small object is kept at a distance of $30 \,cm$ from the lens. The final image is

$A$ plano-convex lens $(\mu = 1.5)$ has a radius of curvature of $10 \; cm$. It is silvered on its plane surface. Find the focal length after silvering. (in $; cm$)

Two thin similar convex lenses are joined together front to front with the rear portion silvered such that a sharp image of an object at infinity is formed $20 \, cm$ from the mirror. When the air between the lenses is replaced by water $(\mu_w = 4/3)$,then the image formed by the mirror is at a distance of ........ $cm$.

Consider a concave mirror and a convex lens (refractive index $=1.5$) of focal length $10 \ cm$ each,separated by a distance of $50 \ cm$ in air (refractive index $=1$) as shown in the figure. An object is placed at a distance of $15 \ cm$ from the mirror. Its erect image formed by this combination has magnification $M_1$. When the setup is kept in a medium of refractive index $7/6$,the magnification becomes $M_2$. The magnitude $\left|\frac{M_2}{M_1}\right|$ is

One face of a rectangular glass plate $6 \, cm$ thick is silvered. An object held $8 \, cm$ in front of the first face forms an image $12 \, cm$ behind the silvered face. The refractive index of the glass is

$A$ double convex lens of focal length $F$ is cut into two equal parts along the vertical axis. The focal length of each part will be