Question

A near sighted person wears eye glass of power 5.5D for distant vision. His doctor prescribes a correction of +.1D in near vision part of his bi focals, which is measured relative to the main part of the lens. Then, the focal length of his near part of the lens is

Answer 1

Answer:

I m not sure about my answer ,then sorry for the wrong answer.

Explanation:

For the near - sightedness,

  Power of lens = 5.5D

     so,

   focal lenght = 1/p

                        =1/5.5

                          = 10/55

                         =+18 cm (approx.)

 Power of lens which has to be corrected = 0.1 D

              so,

                     focal lenght = 1/p

                                          = 1/0.1

                                           =10m =1000cm

As the near vision part is measured to the main part of the lens.

So,

    the focal lenght of his near part of lens = 1000 - 18

                                                                       =9.82 m

Answer 2

ANSWER:

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.

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, 

f2 = 1/p2 

f = 1/ 4.5  = - 22.22 cm