The near point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near point of the normal eye is 25 cm.)

According to the question

Object Distance (u) = -25 cm


Image Distance (v) = -50 cm


By Lens Formula





F = 50 cm.


Since focal length is positive hence the lens is convex lens.


Power =


Focal length = 50 cm = 0.5 m


1 m = 100 cm.


Power =


Hence the Power is 2 Dioptre


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