An object is placed at a distance of $4\,cm$ in front of a concave lens of focal length $12\,cm$. The nature and position of the image from the lens is

The distance between an object and the screen is $100 \ cm$. $A$ lens produces an image on the screen when placed at either of the positions $40 \ cm$ apart. The power of the lens is

$A$ parallel beam of white light falls on a convex lens. Images of blue,red,and green light are formed on the other side of the lens at distances $x$,$y$,and $z$ respectively from the pole of the lens. Then:

The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index $= 1.7$) are same. If the curvature of the plano-concave lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of the radius of curvature of the front and back surface of the biconvex lens is . . . . . . .

The radii of curvature of a double convex lens are $4 \ cm$ and $8 \ cm$. If the refractive index of the material of the lens is $1.5$,the focal length of the lens is nearly (in $cm$)

$A$ planoconvex lens has a maximum thickness of $6 \,cm$. When placed on a horizontal table with the curved surface in contact with the table surface, the apparent depth of the bottommost point of the lens is found to be $4 \,cm$. If the lens is inverted such that the plane face of the lens is in contact with the surface of the table, the apparent depth of the centre of the plane face is found to be $\left(\frac{17}{4}\right) \,cm$. The radius of curvature of the lens is (in $\,cm$)