(A) (i) The given lens is a convex lens as power of convex lens is positive.
(ii)
$\mathit{P}=\frac{1}{\mathit{f}}$
$+2=\frac{1}{\mathit{f}}$
$\mathit{f}=+0.50\mathrm{m}$
Its focal length is $+0.50\mathrm{m}$.
Using the relation $\frac{1}{\mathit{f}}=\frac{1}{\mathit{v}}-\frac{1}{\mathit{u}}$
$\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}+\frac{1}{\mathit{u}}$
$=\frac{1}{0.5}+\frac{-1}{1}=\frac{2+\left(-1\right)}{1}$
$\mathit{v}=1\mathrm{m}$
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 $\mathit{f}$ is used to represent the focal length.