$A$ convex lens forms a real and inverted image of a needle at a distance of $50\, cm$ from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also,find the power of the lens.

(N/A) When an object is placed at the centre of curvature,$2F_1$,of a convex lens,its image is formed at the centre of curvature,$2F_2$,on the other side of the lens. The image formed is inverted and of the same size as the object.
It is given that the image of the needle is formed at a distance of $50\, cm$ from the convex lens. Hence,the needle is placed in front of the lens at a distance of $50\, cm$.
Object distance,$u = -50\, cm$
Image distance,$v = 50\, cm$
Focal length $= f$
According to the lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{50} - \frac{1}{-50} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25}$
$f = 25\, cm = 0.25\, m$
Power of the lens,$P = \frac{1}{f(\text{in meters})} = \frac{1}{0.25} = +4\, D$
Hence,the power of the given lens is $+4\, D$.