the lenses in a students eyes have arefractive power of 52. 0 diopters when she is able to focus on the board if the distance between the ey lens and the retina is 2.00 cm find how

     p = 49.95 cm

This is the distance from the student to the stepped, in order to be able to reach this distance they must sit in the first row

Explanation:

In medicine it is very common to express the potential visual corrections that is

         P = 1 / f

where P is the power and f the focal length in meters

In this exercises give the power, let's find the focal length

        f = 1 / p

        f = 1/52

        f = 0.01923 m = 1.923 cm

For geometrical optics calculations the most used equation is the constructor equation

        1 / f = 1 / p + 1 / q

Where p and q are the distance to the object and the image, respectively

To be able to see an object clearly, its image must be on the retina,

       q = 2.00 cm

find the distance to the object

        1 / p = 1 / f - 1 / q

        1 / p = 1 / 1,923 - 1/2.00

        1 / p = 0.02002

         p = 49.95 cm

This is the distance from the student to the stepped, in order to be able to reach this distance they must sit in the first row

a straight line sloped