Question

A person is unable to see objects nearer than 50 cm. He wants to read a book placed at a distance of 25 cm . Name the defect of vision he is suffering from. How can it be corrected ? Draw ray diagrams for (i) the defective eye ,(ii)its correction using a suitable corrective lens.

Answer 1

"The person has hypermetropia defect which is the farsightedness that means that the image formed at beyond the retina so it require to be a convergent lens to get focused at retina.

The corrective measure is:

u = - 25cm

v = -50 cm

1/f = 1/v – 1/v

1/f = 1/-50 – 1/-25

1/f = -1 +2/50

1/f = 1/50

P = 1/f = 1/50 = 0.02 cm = 0.0002 D

"