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

$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)

In the displacement method of a lens,we obtain two images for different positions of the lens. One image is magnified,and the other is diminished. If $m$ is the magnification of the first image,then the focal length of the lens is given by:

The focal length of a thin biconvex lens is $20 \ cm$. When an object is moved from a distance of $25 \ cm$ in front of it to $50 \ cm$,the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is

The minimum distance between an object and its real image formed by a convex lens of focal length $f$ is (in $f$)

The diameter of a plano-convex lens is $6 \ cm$ and its thickness at the centre is $3 \ mm$. If the speed of light in the material of the lens is $2 \times 10^8 \ m/s$,find the focal length of the lens in $cm$.