For a thin convex lens, the radii of curvature are $15 \,cm$ and $30 \,cm$ respectively. The focal length of the lens is $20 \,cm$. The refractive index of the material is:

$A$ 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 radius of curvature of the curved face of a plano-convex lens is $12 \,cm$ and its refractive index is $1.5$. Then,the focal length of the lens is (in $\,cm$)

$A$ convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens material. Then its focal length will

$A$ convex lens of refractive index $1.5$ and focal length $18 \, cm$ in air is immersed in water. The change in focal length of the lens will be $........... \, cm$. (Given refractive index of water $= 4/3$)

The focal length of a thin converging lens in air is $20 \ cm$. When the lens is immersed in a liquid,it behaves like a concave lens of power $1 \ D$. If the refractive index of the material of the lens is $1.5$,the refractive index of the liquid is