Describe the strength of a lens needed to correct a myopic (near-sighted) eye which has a far pint of P = 1.0 m (far points means the greatest distance for which an object can be imaged on the retina). As above, take the image (lens to retina) distance to be Q = 0.02 m. Solution 1/f=1/o +1/i o=object distance i=image distance f=focal length POWER of LENS=1/f D 1/f=1/1+1/.02=51 D.