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

are there options? cause you ain’t. real specific