Question

The near point of hypermetropic eye is 1m.What is the power of lens required to correct this defect.Assume that the near point of the normal eye is 25cm.

Answer 1

the suitable lens for correcting this defect is convex lens and the power of lens required to correct this defect is calculated to be as: v=-100 ,u=-25 power is cm, 1/f(cm) f=1/v-1/u , 100/(100/3)=3diopter

Answer 2

Answer:Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye. The normal near point of the eye is 25 cm , i.e. for normal individuals to see clearly, an object must be at a distance of atleast 25 cm. For a hypermetropic eye with a near point of 1m (100 cm), to see an object placed at 25 cm, the virtual image needs to be for med at 100 cm. Hence applying the formula,

1/focal length (f)= 1/ object distance (u) + 1/ image distance (v)

Since the image formed is virtual a - (minus) sign is assigned

Hence it becomes,

1/f = 1/u -1/v

which in this situation is,

1/f = 1/25 -1/100

1/f = 4/100 - 1/100 = 3/100

Hence f = 100/3 = 33.3 cm

Diopteric power of the eye P = 100 / f = 100/ 33.3= 3 D

Hence the answer is 3D