$A$ glass convex lens $(\mu_{g}=1.5)$ has a focal length of $8 \ cm$ when placed in air. What would be the focal length of the lens in $cm$ when it is immersed in water $(\mu_{w}=\frac{4}{3})$?

The power of a biconvex lens is $10 \; D$ and the radius of curvature of each surface is $10 \; cm$. The refractive index of the material of the lens is:

The refractive index of a material of a plano-concave lens is $5/3$, and the radius of curvature is $0.3 \,m$. The focal length of the lens in air is (in $\,m$)

$A$ 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 $= 4/3$)

Where should an object be placed on the axis of a convex lens of focal length $8 \ cm$ so as to achieve magnification of $-4$ (in $cm$)? (Distances are measured from the optic centre of the lens)

$A$ double convex lens made of glass has both radii of curvature of magnitude $20 \,cm$. Incident light rays parallel to the axis of the lens will converge to a point at a distance $L$ from the common pole $P$. The value of $L$ is [Refractive index of glass $= 1.5$].