Question

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is +5/3 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes.

a.      What is the focal length of the lenses?

b.      Are the lenses converging or diverging? Why?

c.       How far is her near point from her eyes?

D. In order to be able to read the newspaper at 25cm from her eyes, what must her new prescription (diopters) be?

Answer 1

a) for farsighted person , to correct her eyesight she will use convergin lens.

farsighted (hyperopic)

To correct her eyesight converging lens or convex lens will be used.

   object distance do = 42 cm - 2 cm

                                 = 40 cm
   Refractive power =1.67 diopters
        focal length f = 1/ refractive power

                               = 1 / 1.67

                               = 0.5988 m

                               = 59.88 cm
We know that
            1 / f = 1 / do + 1 / di
          1 / di = 1 / f - 1 / do
              di = do * f / (do - f)
      =  (40 cm)(59.88 cm) / ( 40 cm - 59.88 cm)

                   = -120.5 cm
The magnitude of this value 120.5 cm

gives the location of the women's near point from the glass
As the eye glasses are worn 2.00 cm from the eye.

so that thenear point is located at distance
            = 120.5 + 2.00

            = 122.5 cm

D-

1 / f = 1 / do + 1 / di
f = do *  di / (do +  di)
      =  (23 cm)(120.5 cm) / ( 23 cm + 120.5 cm)

f = 19.3 cm

New power P =100/f

=100/19.3

= +5.18