plz help

Asked by narinder | 28th Oct, 2009, 10:44: PM

Expert Answer:

Since, the near point of the hypermetropic eye is1m (100cm), the lens used should be such that the rays of light starting from normal near point(25cm) appear to come from the near point of hypermetropic eye.

         Thus,

                   u = -25cm

                   v =-100cm

                 1/f =1/v -1/u

                   f = uv/u-v

                     = (-25) x (-100)/ (-25) – (-100)

                     =100/3

                     =33.3cm

              P = 100/f

                 = 100/33.3

                 = 3 D

Since the power is positive , lens must be convex.

Answered by  | 30th Oct, 2009, 02:16: PM

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