The figure shows an equiconvex lens of focal length $f$. If the lens is cut along $AB$,the focal length of each half will be

$A$ point source is placed $20 \,cm$ to the left of a concave lens of focal length $10 \,cm$.
$(a)$ Where is the image formed?
$(b)$ Where to the right of the lens would you place a concave mirror of focal length $5 \,cm$,so that the final image is coincident with the source?
$(c)$ Where would the final image be formed,if the concave mirror is replaced by a plane mirror at the same position?

Two identical thin plano-convex glass lenses (refractive index $1.5$) each having a radius of curvature of $20\, cm$ are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index $1.7$. The focal length of the combination is.......$cm$.

An equiconvex lens is cut into two parts along its $(i) \ XOX'$ axis and $(ii) \ YOY'$ axis as shown in the figure. Let $f, f', f''$ be the focal lengths of the original lens,each half in the first case,and each half in the second case,respectively. Which of the following statements is correct?

$A$ plano-convex lens,when silvered on the plane side,behaves like a concave mirror of focal length $30 \ cm$. However,when silvered on the convex side,it behaves like a concave mirror of focal length $10 \ cm$. The refractive index of its material is:

$A$ thin plano-convex lens made of glass of refractive index $1.5$ is immersed in a liquid of refractive index $1.2$. When the plane side of the lens is silver-coated for complete reflection,the lens immersed in the liquid behaves like a concave mirror of focal length $0.2 \ m$. The radius of curvature of the curved surface of the lens is: (in $m$)