Focal length of a convex lens is $20 \ cm$ and its refractive index is $1.5$. It produces an erect,enlarged image if the distance of the object from the lens is: (in $cm$)

$A$ concave lens of focal length $f$ produces an image $(1/x)$ of the size of the object. The distance of the object from the lens is:

$A$ lens is made of glass having an index of refraction $1.5$. One side of the lens is flat and the other side is convex with a radius $R$. If an object is placed $60 \ cm$ towards the convex side of the lens, the image is formed at $120 \ cm$ on the other side of the lens. The value of $R$ is (in $cm$)

The two surfaces of a concave lens, made of glass of refractive index $1.5$, have the same radii of curvature $R$. It is now immersed in a medium of refractive index $1.75$. Then the lens:

$A$ convex lens of focal length $f$ is placed somewhere in between an object and a screen. The distance between the object and the screen is $x$. If the numerical value of the magnification produced by the lens is $m$, then the focal length of the lens is

An object $2.4 \ m$ in front of a lens forms a sharp image on a film $12 \ cm$ behind the lens. $A$ glass plate $1 \ cm$ thick,of refractive index $1.50$ is interposed between the lens and the film with its plane faces parallel to the film. At what distance (from the lens) should the object be shifted to be in sharp focus on the film (in $m$)?