$A$ plano-convex lens of refractive index $1.5$ and radius of curvature $30 \ cm$ is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens should an object be placed in order to have a real image of the same size as the object? $(... \ cm)$

$A$ plano-convex lens $(f=20 \,cm)$ is silvered at its plane surface. What will be its new focal length (in $\,cm$)?

The diagram shows a silvered equiconvex lens. An object of length $1\,cm$ is placed at a distance of $30\,cm$ in front of the lens. What will be the final image properties? The refractive index of the lens is $\mu$ and the refractive index of the medium in which the lens has been placed is $2\mu$. Both surfaces have a radius of curvature $R$.

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

An object is placed $0.1 \,m$ in front of a convex lens of focal length $20 \,cm$ made of a material of refractive index $1.5$. The surface of the lens away from the object is silvered. If the radius of curvature of the silvered surface is $22 \,cm$,then the distance of the final image from the silvered surface is (in $\,cm$)

An object is placed in front of a symmetrical convex lens with refractive index $1.5$ and radius of curvature $40 \, cm$. The surface of the lens further away from the object is silvered. Under auto-collimation condition,the object distance is.......$cm$.