The refractive index of the material of a double convex lens is $1.5$ and its focal length is $5 \,cm$. If the radii of curvature are equal, the value of the radius of curvature (in $cm$) is

The refractive index of a converging lens is $1.4$. What will be the focal length of this lens if it is placed in a medium of the same refractive index? (Assume the radii of curvature of the faces of the lens are $R_{1}$ and $R_{2}$ respectively)

How is it convenient to obtain an image by a lens practically?

The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index $= 1.7$) are same. If the curvature of the plano-concave lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of the radius of curvature of the front and back surface of the biconvex lens is . . . . . . .

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:

$A$ convex lens is immersed in water. How will its power $P$ change?