A thin, symmetric double convex lens of power P is cut into three parts A,B and C as shown.The power of

The concave lens is a diverging lens, its focal length is negative, the image formed is always virtual and diminished.

The convex lens is the converging lens, its focal length is positive, the image formed is real, virtual, enlarged, or diminished based upon the position of the object.

The focal length is the distance between the optical center and the image.

The power of lens is $P=\frac{1}{f}$ .

It already has the power of a lens is $P=\frac{1}{f}$ .

Having said that the lens is cut into as shown,

It can be seen that here the radius of the two surfaces will be the same. Hence the focal length of the part $\text{A}$ is the same as that of a complete lens.

Considering the fact that the power of the lens part $\text{A}$ remains same. the power of $\text{A}$ is $\text{P}$

It already has the power of a lens is $P=\frac{1}{f}$

Having said that the lens is cut into as shown as,

Considering the fact that the radius of one surface will same while that of another part will be $\infty$ .

Hence the focal length of this part will be double that of the complete lens. So the power of part $\text{B}$ is $\frac{\text{P}}{2}$ .