An equiconvex lens has power $P$. It is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be

$A$ 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 $; cm$)

The radius of curvature of the convex surface of a plano-convex lens is $12 \ cm$ and its refractive index is $1.5$. Find the focal length of this lens in $cm$ when the plane surface is silvered.

An upright object is placed at a distance of $40\, cm$ in front of a convergent lens of focal length $20\, cm$. $A$ convergent mirror of focal length $10\, cm$ is placed at a distance of $60\, cm$ on the other side of the lens. The position and size of the final image will be

$A$ thin convex lens $L$ (refractive index $= 1.5$) is placed on a plane mirror $M$. When a pin is placed at $A$,such that $OA = 18\, cm$,its real inverted image is formed at $A$ itself,as shown in the figure. When a liquid of refractive index $\mu_l$ is put between the lens and the mirror,the pin has to be moved to $A'$,such that $OA' = 27\, cm$,to get its inverted real image at $A'$ itself. The value of $\mu_l$ will be

$A$ thin convex lens of focal length $f$ is placed on a plane mirror as shown in the figure. When an object is kept at a distance $a$ from the lens-mirror combination,its image is formed at a distance $a/3$ in front of the combination. The value of $a$ is