Question

Q.7 Make a diagram to show how hypermetropia is corrected. The near point of a
hypermetropic eye is 1 m.
What is the power of the lens required to correct this
defect ? Assume that the near point of the normal eye is 25 cm.

Answer 1

(a) The hyper-metropic correction is shown below by using convex lens correction.

(b) first calculate the focal length of the convex lens required in this case. For hypermetropic eye can see the nearby object kept at 25 cm (at near point if normal eye) clearly if the image of this object is formed at its own near point which is 1 meter here.

Object distance, u = -25 cm (Normal near point)
Image distance, v = -1 m (Near point of this defective eye) = -100 cm
Focal length, f= ? (To be calculated)
Putting these values in the lens formula,
1/v - 1/u = 1/f

or, 1/f = 1/-100 - 1/-25

or, 1/f = -1/100 + 1/25 = 3/100

or, f = 100/3 cm

we know, power of lens can be calculated as , P = 1/ focal length in metre or 100/Focal length in cm

= 100/(100/3) = +0.33D

thus , power of convex lens required is +0.33D .

Answer 2

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