Question

A near sighted person wears eye glass of power 5.5D for distant vision. His doctor prescribes a correction of + ID in near vision part of his bifocals, which is measured relative to the main part of the lens. Then, the focal length of his near vision part of the lens is

A. -18.18 cm B. -20 cm C. -22.22 cm D. + 20.22 cm

Answer 1

It is given that Power of Lens, P = 5.5 D 

Now, since this lens is used for Myopic Eye, so it will be concave lens and hence the sign will be negative. So P = - 5.5 D 

Now, for near vision power is to be increased by + 1 D. 

Therefore, Power = -5.5 D + 1 D 

= -4.5 D 

Now, we know, 

P = 1/f f = 1/p 

f = 1/ 4.5 

= - 22.22 cm 

Hence, Option C is correct