(A) (i) The given lens is a convex lens as power of convex lens is positive.(ii)$P\;=\;\frac{1}{f}$$+2\;=\;\frac{1}{f}$$f\;=\;+0.50 \text{m}$Its focal length is $+0.50 \text{m}$.Using the relation $\frac{1}{f}\;=\;\frac{1}{v}\;-\;\frac{1}{u}$$\frac{1}{v}\;=\;\frac{1}{f}\;+\;\frac{1}{u}$$\;=\;\frac{1}{0.5}\;+\;\frac{-1}{1}\;=\;\frac{2+(-1)}{1}$$v\;=\;1 \text{m}$OR OR(B) (i) Principal focus: Several rays of light parallel to the principal axis are falling on a concave lens. These rays, after refraction from the lens, are appearing to diverge from a point on the principal axis. This point on the principal axis is called the principal focus of the concave lens.(ii) Focal length: The distance of the principal focus from the optical centre of a lens is called its focal length. The letter $f$ is used to represent the focal length.