In the figure,an air lens of radii of curvature $10 \ cm$ ($R_1 = 10 \ cm$,$R_2 = -10 \ cm$) is cut in a cylinder of glass $(\mu = 1.5)$. The focal length and the nature of the lens is:

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

An achromatic combination is made with a lens of focal length $f$ and dispersive power $\omega$ with a lens having a dispersive power of $2\omega$. The focal length of the second lens will be:

$A$ thin convex lens is placed just above an empty vessel of depth $80 \,cm$. The image of a coin kept at the bottom of the vessel is thus formed $20 \,cm$ above the lens. If now water is poured in the vessel up to a height of $64 \,cm$, what will be the approximate new position of the image? Assume that the refractive index of water is $4/3$.

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 $2 \times 10^8 \, m/s$,the focal length of the lens is.......$cm$.

An object is placed in front of a convex lens,and an image of height $8 \, cm$ is formed. When the lens is moved to a new position,the image height becomes $2 \, cm$. Find the height of the object in $cm$.