A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?

a)   p = 95.66 cm, b) p = 93.13 cm

Explanation:

For this problem we use the  constructor equation

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

the power of the lens is

         P = 1 / f

         f = 1 / P

         f = 1 / 2.25

         f = 0.4444 m

the distance to the object is

the distance to the image is

          q = 85 -2

           q = 83 cm

we must have all the magnitudes in the same units

           f = 0.4444 m = 44.44 cm

we calculate

           1 / p = 0.010454

            p = 95.66 cm

b) if they were contact lenses

            q = 85 cm

             1 / p = 0.107375

             p = 93.13 cm

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

that would be because of both gravity and friction. so     c