A certain child’s near point is 10.0 cm; her far point (with eyes relaxed) is 125 cm. Each eye lens is 2.00 cm from the retina. (a) Between what limits, measured in diopters, does the power of this lens–cornea combination vary? (b) Calculate the power of the eyeglass lens the child should use for relaxed distance vision. Is the lens converging or diverging?

a)    50.8D  and 60D , b)   P = 0.8D ,  divergent

Explanation:

a) We can solve this exercise using the constructor equation

      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.

the image is formed on the retina whereby q = 2.00 cm

let's find the focal length for the near point of view, where p = 10 cm

     1 / f = 1/10 + 1/2

     1 / f = 0.6

     f = 1.67 cm

power of len is defined as the inverse of the focal length in meters

     P = 1 / f

     P = 1 / 1.67 10⁻²

     P = 0.6 10² D

we carry out the same calculation for the far vision point p = 125 cm

     1 / f = 1/125 + 1/2

     1 / f = 0.508

     f = 1.9685 cm

     P = 1/1.9685  10⁻² D

     P = 0.508 10² D

therefore the lens formed by the lens has a power between

    50.8D  and 60D

b) For most children, the nearest point of view is 10 cm and the far point of view is infinite, so we see that in this case the nearest point is normal, but the far point is very small. Therefore, to relax the eye in this point of vision, we must use a lens that places the objects in the far point of vision, so in this case the image is = 125 cm.

       1 / f = 1 / p + 1 / q

       1 / f = 1 / inf + 1/125

        1 / f = 0.008

         P = 0.8D

As the image is formed on the same side where the object in front is divergent

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