Question

An eye has a near point distance of 0.75 m. What sort of lens in spectacles would be needed to reduce the near point distance of 0.25 m ? Also calculate the power of lens required. Is this eye long-sightedness or short-sightedness ?

Answer 1

the person is suffering from HYPERMETROPIA 

a suitable power of convex lens would be needed for this defect

f=0.25m

p=1/f

=1/0.75m

=1.66D

Answer 2

The Lens formula is given as:

                         $\frac {1} {f} = \frac {1} {v} - \frac {1} {u}$

The distance of the image = 0.75m

The distance of the object = 0.25m

                          $\frac {1} {f} = \frac {1} {-0.75} -\frac {1} {-0.25}$

                          $\frac{1} {f} = \frac {(-1+3)} {0.75}$

                          f = 2.67

Then the power of the eye is the inverse of the focal length of the eye.

                          $P = \frac {1} {f}$

                          $= \frac {1} {2.67}$

                          P = 0.37D