Question

A woman wears bifocal glasses with the lenses 2.0 cm in front of her eyes. The upper half of each lens has power-0.500 diopter and corrects her far vision so that she can focus clearly on distant when looking through that half. The lower half of each lens has power +2.00 diopters and corrects her near vision when she looks through that half What are the far point and near point of her eyes?

Answer 1

Answer:

q = -2 m  and  q = -0.5 m

Explanation:

For this exercise we must use the equation of the optical constructor

        1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and the image, respectively

Let's start with the far vision point, in this case the power of the lens is

        P = -0.5D

power is defined as the inverse of the focal length in meter

      f = 1 / D

      f = -1 / 0.5

      f = - 2m

the object for the far vision point is at infinity p = infinity

     1 / f = 1 / p + i / q

      1 / q = 1 / f - 1 / p

      1 / q = -1/2 - 1 / ∞

       q = -2 m

The sign indicates that the image is on the same side as the object

Now let's lock the near view point

D = +2.00 D

f = 1 / D

f = 0.5m

the near mink point is p = 25 cm = 0.25 m

        1 / f = 1 / p + 1 / q

        1 / q = 1 / f - 1 / p

        1 / q = 1 / 0.5 - 1 / 0.25

        1 / q = -2

        q = -0.5 m

the sign indicates that the image is on the same side as the object in front of the lens