An object is placed at a distance of $1.50 \, m$ from a screen. $A$ convex lens is placed between them to form a real image four times larger than the object on the screen. The position of the lens is at a distance of ..... (in $, m$)

$A$ double convex thin lens made of glass (refractive index $\mu = 1.5$) has both radii of curvature of magnitude $20 \ cm$. Incident light rays parallel to the axis of the lens will converge at a distance $L$ such that $L = ...... \ cm$.

The radius of curvature of the convex surface of a thin plano-convex lens is $15 \, cm$ and the refractive index of its material is $1.6$. The power of the lens will be:

$A$ beam of light converges at a point $P$. Now a lens is placed in the path of the convergent beam $12 \, cm$ from $P$. At what point does the beam converge if the lens is
$(a)$ a convex lens of focal length $20 \, cm$,and
$(b)$ a concave lens of focal length $16 \, cm$?

$A$ convex lens is made from glass material having a refractive index of $1.4$ with the same radius of curvature on both sides. The ratio of its focal length to its radius of curvature is . . . . . . .

$A$ glass convex lens has a refractive index of $1.55$ with both faces having the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \,cm$ (in $\,cm$)?