$A$ thin biconvex lens is prepared from glass $(mu = 1.5)$ where both curved surfaces have equal radii of $20 \text{ cm}$ each. The left side surface of the lens is silvered from the outside to make it reflecting. To have the position of the image and the object at the same place, the object should be placed from the lens at a distance of . . . . . . $\text{cm}$.

The two surfaces of a biconvex lens have the same radii of curvature. This lens is made of glass of refractive index $1.5$ and has a focal length $10 \ cm$ in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water (refractive index $= 4/3$), its focal length (in $cm$) is:

The final image of the object forms on the object itself. If the focal length of the lens is $30 \ cm$,then the radius of curvature of the mirror is .... $cm$.

$A$ plano-convex lens $(f = 20 \ cm)$ is silvered at the plane surface. Now,the new focal length $F$ will be........$cm$.

In an experiment,a convex lens of focal length $15 \, cm$ is placed coaxially on an optical bench in front of a convex mirror at a distance of $5 \, cm$ from it. It is found that an object and its image coincide if the object is placed at a distance of $20 \, cm$ from the lens. The focal length of the convex mirror is.......$cm$.

$A$ symmetric double convex lens is cut into two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was $4 \ D$,what will be the power of each cut lens (in $D$)?